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let 's take a look at a new type of trigonometry problem . interestingly , these problems ca n't be solved with sine , cosine , or tangent . a problem : in the triangle below , what is the measure of angle $ l $ ? what we know : relative to $ \angle l $ , we know the lengths of the opposite and adjacent sides , so we can write : $ \tan ( l ) = \dfrac { \text { opposite } } { \text { adjacent } } = \dfrac { 35 } { 65 } $ but this does n't help us find the measure of $ \angle l $ . we 're stuck ! what we need : we need new mathematical tools to solve problems like these . our old friends sine , cosine , and tangent aren ’ t up to the task . they take angles and give side ratios , but we need functions that take side ratios and give angles . we need inverse trig functions ! the inverse trigonometric functions we already know about inverse operations . for example , addition and subtraction are inverse operations , and multiplication and division are inverse operations . each operation does the opposite of its inverse . the idea is the same in trigonometry . inverse trig functions do the opposite of the “ regular ” trig functions . for example : inverse sine $ ( \sin^ { -1 } ) $ does the opposite of the sine . inverse cosine $ ( \cos^ { -1 } ) $ does the opposite of the cosine . inverse tangent $ ( \tan^ { -1 } ) $ does the opposite of the tangent . in general , if you know the trig ratio but not the angle , you can use the corresponding inverse trig function to find the angle . this is expressed mathematically in the statements below . trigonometric functions input angles and output side ratios | |inverse trigonometric functions input side ratios and output angles : - : | : -| : - : $ \sin ( \theta ) =\dfrac { \text { opposite } } { \text { hypotenuse } } $ | $ \rightarrow $ | $ \sin^ { -1 } \left ( \dfrac { \text { opposite } } { \text { hypotenuse } } \right ) =\theta $ $ \cos ( \theta ) =\dfrac { \text { adjacent } } { \text { hypotenuse } } $ | $ \rightarrow $ | $ \cos^ { -1 } \left ( \dfrac { \text { adjacent } } { \text { hypotenuse } } \right ) =\theta $ $ \tan ( \theta ) =\dfrac { \text { opposite } } { \text { adjacent } } $ | $ \rightarrow $ | $ \tan^ { -1 } \left ( \dfrac { \text { opposite } } { \text { adjacent } } \right ) =\theta $ misconception alert ! the expression $ \sin^ { -1 } ( x ) $ is not the same as $ \dfrac { 1 } { \sin ( x ) } $ . in other words , the $ \small { -1 } $ is not an exponent . instead , it simply means inverse function . however , there is an alternate notation that avoids this pitfall ! the inverse sine can also be expressed as $ \arcsin $ , the inverse cosine as $ \arccos $ , and the inverse tangent as $ \arctan $ . this notation is common in computer programming languages , but not in mathematics . solving the introductory problem in the introductory problem , we were given the opposite and adjacent side lengths , so we can use inverse tangent to find the angle . $ \begin { align } { m\angle l } & amp ; =\tan^ { -1 } \left ( \dfrac { \text { } \blued { \text { opposite } } } { \text { } \maroonc { \text { adjacent } } \text { } } \right ) \quad\small { \gray { \text { define . } } } \\ m\angle l & amp ; =\tan^ { -1 } \left ( \dfrac { \blued { 35 } } { \maroonc { 65 } } \right ) \quad\small { \gray { \text { substitute values . } } } \\ m\angle l & amp ; \approx 28.30^\circ \quad\small { \gray { \text { evaluate with a calculator . } } } \end { align } $ now let 's try some practice problems .
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a problem : in the triangle below , what is the measure of angle $ l $ ? what we know : relative to $ \angle l $ , we know the lengths of the opposite and adjacent sides , so we can write : $ \tan ( l ) = \dfrac { \text { opposite } } { \text { adjacent } } = \dfrac { 35 } { 65 } $ but this does n't help us find the measure of $ \angle l $ . we 're stuck !
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how am i supposed to know what is the adjacent and the opposite if they keep switching positions of the problems enlisted on this website ?
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let 's take a look at a new type of trigonometry problem . interestingly , these problems ca n't be solved with sine , cosine , or tangent . a problem : in the triangle below , what is the measure of angle $ l $ ? what we know : relative to $ \angle l $ , we know the lengths of the opposite and adjacent sides , so we can write : $ \tan ( l ) = \dfrac { \text { opposite } } { \text { adjacent } } = \dfrac { 35 } { 65 } $ but this does n't help us find the measure of $ \angle l $ . we 're stuck ! what we need : we need new mathematical tools to solve problems like these . our old friends sine , cosine , and tangent aren ’ t up to the task . they take angles and give side ratios , but we need functions that take side ratios and give angles . we need inverse trig functions ! the inverse trigonometric functions we already know about inverse operations . for example , addition and subtraction are inverse operations , and multiplication and division are inverse operations . each operation does the opposite of its inverse . the idea is the same in trigonometry . inverse trig functions do the opposite of the “ regular ” trig functions . for example : inverse sine $ ( \sin^ { -1 } ) $ does the opposite of the sine . inverse cosine $ ( \cos^ { -1 } ) $ does the opposite of the cosine . inverse tangent $ ( \tan^ { -1 } ) $ does the opposite of the tangent . in general , if you know the trig ratio but not the angle , you can use the corresponding inverse trig function to find the angle . this is expressed mathematically in the statements below . trigonometric functions input angles and output side ratios | |inverse trigonometric functions input side ratios and output angles : - : | : -| : - : $ \sin ( \theta ) =\dfrac { \text { opposite } } { \text { hypotenuse } } $ | $ \rightarrow $ | $ \sin^ { -1 } \left ( \dfrac { \text { opposite } } { \text { hypotenuse } } \right ) =\theta $ $ \cos ( \theta ) =\dfrac { \text { adjacent } } { \text { hypotenuse } } $ | $ \rightarrow $ | $ \cos^ { -1 } \left ( \dfrac { \text { adjacent } } { \text { hypotenuse } } \right ) =\theta $ $ \tan ( \theta ) =\dfrac { \text { opposite } } { \text { adjacent } } $ | $ \rightarrow $ | $ \tan^ { -1 } \left ( \dfrac { \text { opposite } } { \text { adjacent } } \right ) =\theta $ misconception alert ! the expression $ \sin^ { -1 } ( x ) $ is not the same as $ \dfrac { 1 } { \sin ( x ) } $ . in other words , the $ \small { -1 } $ is not an exponent . instead , it simply means inverse function . however , there is an alternate notation that avoids this pitfall ! the inverse sine can also be expressed as $ \arcsin $ , the inverse cosine as $ \arccos $ , and the inverse tangent as $ \arctan $ . this notation is common in computer programming languages , but not in mathematics . solving the introductory problem in the introductory problem , we were given the opposite and adjacent side lengths , so we can use inverse tangent to find the angle . $ \begin { align } { m\angle l } & amp ; =\tan^ { -1 } \left ( \dfrac { \text { } \blued { \text { opposite } } } { \text { } \maroonc { \text { adjacent } } \text { } } \right ) \quad\small { \gray { \text { define . } } } \\ m\angle l & amp ; =\tan^ { -1 } \left ( \dfrac { \blued { 35 } } { \maroonc { 65 } } \right ) \quad\small { \gray { \text { substitute values . } } } \\ m\angle l & amp ; \approx 28.30^\circ \quad\small { \gray { \text { evaluate with a calculator . } } } \end { align } $ now let 's try some practice problems .
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we need inverse trig functions ! the inverse trigonometric functions we already know about inverse operations . for example , addition and subtraction are inverse operations , and multiplication and division are inverse operations .
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how to find an angle in the triangle using inverse trigonometric ratios ?
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let 's take a look at a new type of trigonometry problem . interestingly , these problems ca n't be solved with sine , cosine , or tangent . a problem : in the triangle below , what is the measure of angle $ l $ ? what we know : relative to $ \angle l $ , we know the lengths of the opposite and adjacent sides , so we can write : $ \tan ( l ) = \dfrac { \text { opposite } } { \text { adjacent } } = \dfrac { 35 } { 65 } $ but this does n't help us find the measure of $ \angle l $ . we 're stuck ! what we need : we need new mathematical tools to solve problems like these . our old friends sine , cosine , and tangent aren ’ t up to the task . they take angles and give side ratios , but we need functions that take side ratios and give angles . we need inverse trig functions ! the inverse trigonometric functions we already know about inverse operations . for example , addition and subtraction are inverse operations , and multiplication and division are inverse operations . each operation does the opposite of its inverse . the idea is the same in trigonometry . inverse trig functions do the opposite of the “ regular ” trig functions . for example : inverse sine $ ( \sin^ { -1 } ) $ does the opposite of the sine . inverse cosine $ ( \cos^ { -1 } ) $ does the opposite of the cosine . inverse tangent $ ( \tan^ { -1 } ) $ does the opposite of the tangent . in general , if you know the trig ratio but not the angle , you can use the corresponding inverse trig function to find the angle . this is expressed mathematically in the statements below . trigonometric functions input angles and output side ratios | |inverse trigonometric functions input side ratios and output angles : - : | : -| : - : $ \sin ( \theta ) =\dfrac { \text { opposite } } { \text { hypotenuse } } $ | $ \rightarrow $ | $ \sin^ { -1 } \left ( \dfrac { \text { opposite } } { \text { hypotenuse } } \right ) =\theta $ $ \cos ( \theta ) =\dfrac { \text { adjacent } } { \text { hypotenuse } } $ | $ \rightarrow $ | $ \cos^ { -1 } \left ( \dfrac { \text { adjacent } } { \text { hypotenuse } } \right ) =\theta $ $ \tan ( \theta ) =\dfrac { \text { opposite } } { \text { adjacent } } $ | $ \rightarrow $ | $ \tan^ { -1 } \left ( \dfrac { \text { opposite } } { \text { adjacent } } \right ) =\theta $ misconception alert ! the expression $ \sin^ { -1 } ( x ) $ is not the same as $ \dfrac { 1 } { \sin ( x ) } $ . in other words , the $ \small { -1 } $ is not an exponent . instead , it simply means inverse function . however , there is an alternate notation that avoids this pitfall ! the inverse sine can also be expressed as $ \arcsin $ , the inverse cosine as $ \arccos $ , and the inverse tangent as $ \arctan $ . this notation is common in computer programming languages , but not in mathematics . solving the introductory problem in the introductory problem , we were given the opposite and adjacent side lengths , so we can use inverse tangent to find the angle . $ \begin { align } { m\angle l } & amp ; =\tan^ { -1 } \left ( \dfrac { \text { } \blued { \text { opposite } } } { \text { } \maroonc { \text { adjacent } } \text { } } \right ) \quad\small { \gray { \text { define . } } } \\ m\angle l & amp ; =\tan^ { -1 } \left ( \dfrac { \blued { 35 } } { \maroonc { 65 } } \right ) \quad\small { \gray { \text { substitute values . } } } \\ m\angle l & amp ; \approx 28.30^\circ \quad\small { \gray { \text { evaluate with a calculator . } } } \end { align } $ now let 's try some practice problems .
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interestingly , these problems ca n't be solved with sine , cosine , or tangent . a problem : in the triangle below , what is the measure of angle $ l $ ? what we know : relative to $ \angle l $ , we know the lengths of the opposite and adjacent sides , so we can write : $ \tan ( l ) = \dfrac { \text { opposite } } { \text { adjacent } } = \dfrac { 35 } { 65 } $ but this does n't help us find the measure of $ \angle l $ .
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how do i find an angle in the triangle ?
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let 's take a look at a new type of trigonometry problem . interestingly , these problems ca n't be solved with sine , cosine , or tangent . a problem : in the triangle below , what is the measure of angle $ l $ ? what we know : relative to $ \angle l $ , we know the lengths of the opposite and adjacent sides , so we can write : $ \tan ( l ) = \dfrac { \text { opposite } } { \text { adjacent } } = \dfrac { 35 } { 65 } $ but this does n't help us find the measure of $ \angle l $ . we 're stuck ! what we need : we need new mathematical tools to solve problems like these . our old friends sine , cosine , and tangent aren ’ t up to the task . they take angles and give side ratios , but we need functions that take side ratios and give angles . we need inverse trig functions ! the inverse trigonometric functions we already know about inverse operations . for example , addition and subtraction are inverse operations , and multiplication and division are inverse operations . each operation does the opposite of its inverse . the idea is the same in trigonometry . inverse trig functions do the opposite of the “ regular ” trig functions . for example : inverse sine $ ( \sin^ { -1 } ) $ does the opposite of the sine . inverse cosine $ ( \cos^ { -1 } ) $ does the opposite of the cosine . inverse tangent $ ( \tan^ { -1 } ) $ does the opposite of the tangent . in general , if you know the trig ratio but not the angle , you can use the corresponding inverse trig function to find the angle . this is expressed mathematically in the statements below . trigonometric functions input angles and output side ratios | |inverse trigonometric functions input side ratios and output angles : - : | : -| : - : $ \sin ( \theta ) =\dfrac { \text { opposite } } { \text { hypotenuse } } $ | $ \rightarrow $ | $ \sin^ { -1 } \left ( \dfrac { \text { opposite } } { \text { hypotenuse } } \right ) =\theta $ $ \cos ( \theta ) =\dfrac { \text { adjacent } } { \text { hypotenuse } } $ | $ \rightarrow $ | $ \cos^ { -1 } \left ( \dfrac { \text { adjacent } } { \text { hypotenuse } } \right ) =\theta $ $ \tan ( \theta ) =\dfrac { \text { opposite } } { \text { adjacent } } $ | $ \rightarrow $ | $ \tan^ { -1 } \left ( \dfrac { \text { opposite } } { \text { adjacent } } \right ) =\theta $ misconception alert ! the expression $ \sin^ { -1 } ( x ) $ is not the same as $ \dfrac { 1 } { \sin ( x ) } $ . in other words , the $ \small { -1 } $ is not an exponent . instead , it simply means inverse function . however , there is an alternate notation that avoids this pitfall ! the inverse sine can also be expressed as $ \arcsin $ , the inverse cosine as $ \arccos $ , and the inverse tangent as $ \arctan $ . this notation is common in computer programming languages , but not in mathematics . solving the introductory problem in the introductory problem , we were given the opposite and adjacent side lengths , so we can use inverse tangent to find the angle . $ \begin { align } { m\angle l } & amp ; =\tan^ { -1 } \left ( \dfrac { \text { } \blued { \text { opposite } } } { \text { } \maroonc { \text { adjacent } } \text { } } \right ) \quad\small { \gray { \text { define . } } } \\ m\angle l & amp ; =\tan^ { -1 } \left ( \dfrac { \blued { 35 } } { \maroonc { 65 } } \right ) \quad\small { \gray { \text { substitute values . } } } \\ m\angle l & amp ; \approx 28.30^\circ \quad\small { \gray { \text { evaluate with a calculator . } } } \end { align } $ now let 's try some practice problems .
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we 're stuck ! what we need : we need new mathematical tools to solve problems like these . our old friends sine , cosine , and tangent aren ’ t up to the task .
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can you prove even the most basic mathematical truths without using math ?
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let 's take a look at a new type of trigonometry problem . interestingly , these problems ca n't be solved with sine , cosine , or tangent . a problem : in the triangle below , what is the measure of angle $ l $ ? what we know : relative to $ \angle l $ , we know the lengths of the opposite and adjacent sides , so we can write : $ \tan ( l ) = \dfrac { \text { opposite } } { \text { adjacent } } = \dfrac { 35 } { 65 } $ but this does n't help us find the measure of $ \angle l $ . we 're stuck ! what we need : we need new mathematical tools to solve problems like these . our old friends sine , cosine , and tangent aren ’ t up to the task . they take angles and give side ratios , but we need functions that take side ratios and give angles . we need inverse trig functions ! the inverse trigonometric functions we already know about inverse operations . for example , addition and subtraction are inverse operations , and multiplication and division are inverse operations . each operation does the opposite of its inverse . the idea is the same in trigonometry . inverse trig functions do the opposite of the “ regular ” trig functions . for example : inverse sine $ ( \sin^ { -1 } ) $ does the opposite of the sine . inverse cosine $ ( \cos^ { -1 } ) $ does the opposite of the cosine . inverse tangent $ ( \tan^ { -1 } ) $ does the opposite of the tangent . in general , if you know the trig ratio but not the angle , you can use the corresponding inverse trig function to find the angle . this is expressed mathematically in the statements below . trigonometric functions input angles and output side ratios | |inverse trigonometric functions input side ratios and output angles : - : | : -| : - : $ \sin ( \theta ) =\dfrac { \text { opposite } } { \text { hypotenuse } } $ | $ \rightarrow $ | $ \sin^ { -1 } \left ( \dfrac { \text { opposite } } { \text { hypotenuse } } \right ) =\theta $ $ \cos ( \theta ) =\dfrac { \text { adjacent } } { \text { hypotenuse } } $ | $ \rightarrow $ | $ \cos^ { -1 } \left ( \dfrac { \text { adjacent } } { \text { hypotenuse } } \right ) =\theta $ $ \tan ( \theta ) =\dfrac { \text { opposite } } { \text { adjacent } } $ | $ \rightarrow $ | $ \tan^ { -1 } \left ( \dfrac { \text { opposite } } { \text { adjacent } } \right ) =\theta $ misconception alert ! the expression $ \sin^ { -1 } ( x ) $ is not the same as $ \dfrac { 1 } { \sin ( x ) } $ . in other words , the $ \small { -1 } $ is not an exponent . instead , it simply means inverse function . however , there is an alternate notation that avoids this pitfall ! the inverse sine can also be expressed as $ \arcsin $ , the inverse cosine as $ \arccos $ , and the inverse tangent as $ \arctan $ . this notation is common in computer programming languages , but not in mathematics . solving the introductory problem in the introductory problem , we were given the opposite and adjacent side lengths , so we can use inverse tangent to find the angle . $ \begin { align } { m\angle l } & amp ; =\tan^ { -1 } \left ( \dfrac { \text { } \blued { \text { opposite } } } { \text { } \maroonc { \text { adjacent } } \text { } } \right ) \quad\small { \gray { \text { define . } } } \\ m\angle l & amp ; =\tan^ { -1 } \left ( \dfrac { \blued { 35 } } { \maroonc { 65 } } \right ) \quad\small { \gray { \text { substitute values . } } } \\ m\angle l & amp ; \approx 28.30^\circ \quad\small { \gray { \text { evaluate with a calculator . } } } \end { align } $ now let 's try some practice problems .
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interestingly , these problems ca n't be solved with sine , cosine , or tangent . a problem : in the triangle below , what is the measure of angle $ l $ ? what we know : relative to $ \angle l $ , we know the lengths of the opposite and adjacent sides , so we can write : $ \tan ( l ) = \dfrac { \text { opposite } } { \text { adjacent } } = \dfrac { 35 } { 65 } $ but this does n't help us find the measure of $ \angle l $ . we 're stuck !
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without an angle inside the triangle , how can you tell exactly which side is adjacent and which is opposite ?
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let 's take a look at a new type of trigonometry problem . interestingly , these problems ca n't be solved with sine , cosine , or tangent . a problem : in the triangle below , what is the measure of angle $ l $ ? what we know : relative to $ \angle l $ , we know the lengths of the opposite and adjacent sides , so we can write : $ \tan ( l ) = \dfrac { \text { opposite } } { \text { adjacent } } = \dfrac { 35 } { 65 } $ but this does n't help us find the measure of $ \angle l $ . we 're stuck ! what we need : we need new mathematical tools to solve problems like these . our old friends sine , cosine , and tangent aren ’ t up to the task . they take angles and give side ratios , but we need functions that take side ratios and give angles . we need inverse trig functions ! the inverse trigonometric functions we already know about inverse operations . for example , addition and subtraction are inverse operations , and multiplication and division are inverse operations . each operation does the opposite of its inverse . the idea is the same in trigonometry . inverse trig functions do the opposite of the “ regular ” trig functions . for example : inverse sine $ ( \sin^ { -1 } ) $ does the opposite of the sine . inverse cosine $ ( \cos^ { -1 } ) $ does the opposite of the cosine . inverse tangent $ ( \tan^ { -1 } ) $ does the opposite of the tangent . in general , if you know the trig ratio but not the angle , you can use the corresponding inverse trig function to find the angle . this is expressed mathematically in the statements below . trigonometric functions input angles and output side ratios | |inverse trigonometric functions input side ratios and output angles : - : | : -| : - : $ \sin ( \theta ) =\dfrac { \text { opposite } } { \text { hypotenuse } } $ | $ \rightarrow $ | $ \sin^ { -1 } \left ( \dfrac { \text { opposite } } { \text { hypotenuse } } \right ) =\theta $ $ \cos ( \theta ) =\dfrac { \text { adjacent } } { \text { hypotenuse } } $ | $ \rightarrow $ | $ \cos^ { -1 } \left ( \dfrac { \text { adjacent } } { \text { hypotenuse } } \right ) =\theta $ $ \tan ( \theta ) =\dfrac { \text { opposite } } { \text { adjacent } } $ | $ \rightarrow $ | $ \tan^ { -1 } \left ( \dfrac { \text { opposite } } { \text { adjacent } } \right ) =\theta $ misconception alert ! the expression $ \sin^ { -1 } ( x ) $ is not the same as $ \dfrac { 1 } { \sin ( x ) } $ . in other words , the $ \small { -1 } $ is not an exponent . instead , it simply means inverse function . however , there is an alternate notation that avoids this pitfall ! the inverse sine can also be expressed as $ \arcsin $ , the inverse cosine as $ \arccos $ , and the inverse tangent as $ \arctan $ . this notation is common in computer programming languages , but not in mathematics . solving the introductory problem in the introductory problem , we were given the opposite and adjacent side lengths , so we can use inverse tangent to find the angle . $ \begin { align } { m\angle l } & amp ; =\tan^ { -1 } \left ( \dfrac { \text { } \blued { \text { opposite } } } { \text { } \maroonc { \text { adjacent } } \text { } } \right ) \quad\small { \gray { \text { define . } } } \\ m\angle l & amp ; =\tan^ { -1 } \left ( \dfrac { \blued { 35 } } { \maroonc { 65 } } \right ) \quad\small { \gray { \text { substitute values . } } } \\ m\angle l & amp ; \approx 28.30^\circ \quad\small { \gray { \text { evaluate with a calculator . } } } \end { align } $ now let 's try some practice problems .
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trigonometric functions input angles and output side ratios | |inverse trigonometric functions input side ratios and output angles : - : | : -| : - : $ \sin ( \theta ) =\dfrac { \text { opposite } } { \text { hypotenuse } } $ | $ \rightarrow $ | $ \sin^ { -1 } \left ( \dfrac { \text { opposite } } { \text { hypotenuse } } \right ) =\theta $ $ \cos ( \theta ) =\dfrac { \text { adjacent } } { \text { hypotenuse } } $ | $ \rightarrow $ | $ \cos^ { -1 } \left ( \dfrac { \text { adjacent } } { \text { hypotenuse } } \right ) =\theta $ $ \tan ( \theta ) =\dfrac { \text { opposite } } { \text { adjacent } } $ | $ \rightarrow $ | $ \tan^ { -1 } \left ( \dfrac { \text { opposite } } { \text { adjacent } } \right ) =\theta $ misconception alert ! the expression $ \sin^ { -1 } ( x ) $ is not the same as $ \dfrac { 1 } { \sin ( x ) } $ . in other words , the $ \small { -1 } $ is not an exponent .
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how do you know when to use each sin , cos , or tan when calculating ?
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let 's take a look at a new type of trigonometry problem . interestingly , these problems ca n't be solved with sine , cosine , or tangent . a problem : in the triangle below , what is the measure of angle $ l $ ? what we know : relative to $ \angle l $ , we know the lengths of the opposite and adjacent sides , so we can write : $ \tan ( l ) = \dfrac { \text { opposite } } { \text { adjacent } } = \dfrac { 35 } { 65 } $ but this does n't help us find the measure of $ \angle l $ . we 're stuck ! what we need : we need new mathematical tools to solve problems like these . our old friends sine , cosine , and tangent aren ’ t up to the task . they take angles and give side ratios , but we need functions that take side ratios and give angles . we need inverse trig functions ! the inverse trigonometric functions we already know about inverse operations . for example , addition and subtraction are inverse operations , and multiplication and division are inverse operations . each operation does the opposite of its inverse . the idea is the same in trigonometry . inverse trig functions do the opposite of the “ regular ” trig functions . for example : inverse sine $ ( \sin^ { -1 } ) $ does the opposite of the sine . inverse cosine $ ( \cos^ { -1 } ) $ does the opposite of the cosine . inverse tangent $ ( \tan^ { -1 } ) $ does the opposite of the tangent . in general , if you know the trig ratio but not the angle , you can use the corresponding inverse trig function to find the angle . this is expressed mathematically in the statements below . trigonometric functions input angles and output side ratios | |inverse trigonometric functions input side ratios and output angles : - : | : -| : - : $ \sin ( \theta ) =\dfrac { \text { opposite } } { \text { hypotenuse } } $ | $ \rightarrow $ | $ \sin^ { -1 } \left ( \dfrac { \text { opposite } } { \text { hypotenuse } } \right ) =\theta $ $ \cos ( \theta ) =\dfrac { \text { adjacent } } { \text { hypotenuse } } $ | $ \rightarrow $ | $ \cos^ { -1 } \left ( \dfrac { \text { adjacent } } { \text { hypotenuse } } \right ) =\theta $ $ \tan ( \theta ) =\dfrac { \text { opposite } } { \text { adjacent } } $ | $ \rightarrow $ | $ \tan^ { -1 } \left ( \dfrac { \text { opposite } } { \text { adjacent } } \right ) =\theta $ misconception alert ! the expression $ \sin^ { -1 } ( x ) $ is not the same as $ \dfrac { 1 } { \sin ( x ) } $ . in other words , the $ \small { -1 } $ is not an exponent . instead , it simply means inverse function . however , there is an alternate notation that avoids this pitfall ! the inverse sine can also be expressed as $ \arcsin $ , the inverse cosine as $ \arccos $ , and the inverse tangent as $ \arctan $ . this notation is common in computer programming languages , but not in mathematics . solving the introductory problem in the introductory problem , we were given the opposite and adjacent side lengths , so we can use inverse tangent to find the angle . $ \begin { align } { m\angle l } & amp ; =\tan^ { -1 } \left ( \dfrac { \text { } \blued { \text { opposite } } } { \text { } \maroonc { \text { adjacent } } \text { } } \right ) \quad\small { \gray { \text { define . } } } \\ m\angle l & amp ; =\tan^ { -1 } \left ( \dfrac { \blued { 35 } } { \maroonc { 65 } } \right ) \quad\small { \gray { \text { substitute values . } } } \\ m\angle l & amp ; \approx 28.30^\circ \quad\small { \gray { \text { evaluate with a calculator . } } } \end { align } $ now let 's try some practice problems .
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interestingly , these problems ca n't be solved with sine , cosine , or tangent . a problem : in the triangle below , what is the measure of angle $ l $ ? what we know : relative to $ \angle l $ , we know the lengths of the opposite and adjacent sides , so we can write : $ \tan ( l ) = \dfrac { \text { opposite } } { \text { adjacent } } = \dfrac { 35 } { 65 } $ but this does n't help us find the measure of $ \angle l $ .
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how can we solve for right triangle which has angle of 30 degree , 60 degree and 90 degree ?
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let 's take a look at a new type of trigonometry problem . interestingly , these problems ca n't be solved with sine , cosine , or tangent . a problem : in the triangle below , what is the measure of angle $ l $ ? what we know : relative to $ \angle l $ , we know the lengths of the opposite and adjacent sides , so we can write : $ \tan ( l ) = \dfrac { \text { opposite } } { \text { adjacent } } = \dfrac { 35 } { 65 } $ but this does n't help us find the measure of $ \angle l $ . we 're stuck ! what we need : we need new mathematical tools to solve problems like these . our old friends sine , cosine , and tangent aren ’ t up to the task . they take angles and give side ratios , but we need functions that take side ratios and give angles . we need inverse trig functions ! the inverse trigonometric functions we already know about inverse operations . for example , addition and subtraction are inverse operations , and multiplication and division are inverse operations . each operation does the opposite of its inverse . the idea is the same in trigonometry . inverse trig functions do the opposite of the “ regular ” trig functions . for example : inverse sine $ ( \sin^ { -1 } ) $ does the opposite of the sine . inverse cosine $ ( \cos^ { -1 } ) $ does the opposite of the cosine . inverse tangent $ ( \tan^ { -1 } ) $ does the opposite of the tangent . in general , if you know the trig ratio but not the angle , you can use the corresponding inverse trig function to find the angle . this is expressed mathematically in the statements below . trigonometric functions input angles and output side ratios | |inverse trigonometric functions input side ratios and output angles : - : | : -| : - : $ \sin ( \theta ) =\dfrac { \text { opposite } } { \text { hypotenuse } } $ | $ \rightarrow $ | $ \sin^ { -1 } \left ( \dfrac { \text { opposite } } { \text { hypotenuse } } \right ) =\theta $ $ \cos ( \theta ) =\dfrac { \text { adjacent } } { \text { hypotenuse } } $ | $ \rightarrow $ | $ \cos^ { -1 } \left ( \dfrac { \text { adjacent } } { \text { hypotenuse } } \right ) =\theta $ $ \tan ( \theta ) =\dfrac { \text { opposite } } { \text { adjacent } } $ | $ \rightarrow $ | $ \tan^ { -1 } \left ( \dfrac { \text { opposite } } { \text { adjacent } } \right ) =\theta $ misconception alert ! the expression $ \sin^ { -1 } ( x ) $ is not the same as $ \dfrac { 1 } { \sin ( x ) } $ . in other words , the $ \small { -1 } $ is not an exponent . instead , it simply means inverse function . however , there is an alternate notation that avoids this pitfall ! the inverse sine can also be expressed as $ \arcsin $ , the inverse cosine as $ \arccos $ , and the inverse tangent as $ \arctan $ . this notation is common in computer programming languages , but not in mathematics . solving the introductory problem in the introductory problem , we were given the opposite and adjacent side lengths , so we can use inverse tangent to find the angle . $ \begin { align } { m\angle l } & amp ; =\tan^ { -1 } \left ( \dfrac { \text { } \blued { \text { opposite } } } { \text { } \maroonc { \text { adjacent } } \text { } } \right ) \quad\small { \gray { \text { define . } } } \\ m\angle l & amp ; =\tan^ { -1 } \left ( \dfrac { \blued { 35 } } { \maroonc { 65 } } \right ) \quad\small { \gray { \text { substitute values . } } } \\ m\angle l & amp ; \approx 28.30^\circ \quad\small { \gray { \text { evaluate with a calculator . } } } \end { align } $ now let 's try some practice problems .
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they take angles and give side ratios , but we need functions that take side ratios and give angles . we need inverse trig functions ! the inverse trigonometric functions we already know about inverse operations . for example , addition and subtraction are inverse operations , and multiplication and division are inverse operations .
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how can you evaluate the inverse with calculator ?
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let 's take a look at a new type of trigonometry problem . interestingly , these problems ca n't be solved with sine , cosine , or tangent . a problem : in the triangle below , what is the measure of angle $ l $ ? what we know : relative to $ \angle l $ , we know the lengths of the opposite and adjacent sides , so we can write : $ \tan ( l ) = \dfrac { \text { opposite } } { \text { adjacent } } = \dfrac { 35 } { 65 } $ but this does n't help us find the measure of $ \angle l $ . we 're stuck ! what we need : we need new mathematical tools to solve problems like these . our old friends sine , cosine , and tangent aren ’ t up to the task . they take angles and give side ratios , but we need functions that take side ratios and give angles . we need inverse trig functions ! the inverse trigonometric functions we already know about inverse operations . for example , addition and subtraction are inverse operations , and multiplication and division are inverse operations . each operation does the opposite of its inverse . the idea is the same in trigonometry . inverse trig functions do the opposite of the “ regular ” trig functions . for example : inverse sine $ ( \sin^ { -1 } ) $ does the opposite of the sine . inverse cosine $ ( \cos^ { -1 } ) $ does the opposite of the cosine . inverse tangent $ ( \tan^ { -1 } ) $ does the opposite of the tangent . in general , if you know the trig ratio but not the angle , you can use the corresponding inverse trig function to find the angle . this is expressed mathematically in the statements below . trigonometric functions input angles and output side ratios | |inverse trigonometric functions input side ratios and output angles : - : | : -| : - : $ \sin ( \theta ) =\dfrac { \text { opposite } } { \text { hypotenuse } } $ | $ \rightarrow $ | $ \sin^ { -1 } \left ( \dfrac { \text { opposite } } { \text { hypotenuse } } \right ) =\theta $ $ \cos ( \theta ) =\dfrac { \text { adjacent } } { \text { hypotenuse } } $ | $ \rightarrow $ | $ \cos^ { -1 } \left ( \dfrac { \text { adjacent } } { \text { hypotenuse } } \right ) =\theta $ $ \tan ( \theta ) =\dfrac { \text { opposite } } { \text { adjacent } } $ | $ \rightarrow $ | $ \tan^ { -1 } \left ( \dfrac { \text { opposite } } { \text { adjacent } } \right ) =\theta $ misconception alert ! the expression $ \sin^ { -1 } ( x ) $ is not the same as $ \dfrac { 1 } { \sin ( x ) } $ . in other words , the $ \small { -1 } $ is not an exponent . instead , it simply means inverse function . however , there is an alternate notation that avoids this pitfall ! the inverse sine can also be expressed as $ \arcsin $ , the inverse cosine as $ \arccos $ , and the inverse tangent as $ \arctan $ . this notation is common in computer programming languages , but not in mathematics . solving the introductory problem in the introductory problem , we were given the opposite and adjacent side lengths , so we can use inverse tangent to find the angle . $ \begin { align } { m\angle l } & amp ; =\tan^ { -1 } \left ( \dfrac { \text { } \blued { \text { opposite } } } { \text { } \maroonc { \text { adjacent } } \text { } } \right ) \quad\small { \gray { \text { define . } } } \\ m\angle l & amp ; =\tan^ { -1 } \left ( \dfrac { \blued { 35 } } { \maroonc { 65 } } \right ) \quad\small { \gray { \text { substitute values . } } } \\ m\angle l & amp ; \approx 28.30^\circ \quad\small { \gray { \text { evaluate with a calculator . } } } \end { align } $ now let 's try some practice problems .
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they take angles and give side ratios , but we need functions that take side ratios and give angles . we need inverse trig functions ! the inverse trigonometric functions we already know about inverse operations .
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i dont understand how to expess the inverse of a function on a trig calculator ?
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let 's take a look at a new type of trigonometry problem . interestingly , these problems ca n't be solved with sine , cosine , or tangent . a problem : in the triangle below , what is the measure of angle $ l $ ? what we know : relative to $ \angle l $ , we know the lengths of the opposite and adjacent sides , so we can write : $ \tan ( l ) = \dfrac { \text { opposite } } { \text { adjacent } } = \dfrac { 35 } { 65 } $ but this does n't help us find the measure of $ \angle l $ . we 're stuck ! what we need : we need new mathematical tools to solve problems like these . our old friends sine , cosine , and tangent aren ’ t up to the task . they take angles and give side ratios , but we need functions that take side ratios and give angles . we need inverse trig functions ! the inverse trigonometric functions we already know about inverse operations . for example , addition and subtraction are inverse operations , and multiplication and division are inverse operations . each operation does the opposite of its inverse . the idea is the same in trigonometry . inverse trig functions do the opposite of the “ regular ” trig functions . for example : inverse sine $ ( \sin^ { -1 } ) $ does the opposite of the sine . inverse cosine $ ( \cos^ { -1 } ) $ does the opposite of the cosine . inverse tangent $ ( \tan^ { -1 } ) $ does the opposite of the tangent . in general , if you know the trig ratio but not the angle , you can use the corresponding inverse trig function to find the angle . this is expressed mathematically in the statements below . trigonometric functions input angles and output side ratios | |inverse trigonometric functions input side ratios and output angles : - : | : -| : - : $ \sin ( \theta ) =\dfrac { \text { opposite } } { \text { hypotenuse } } $ | $ \rightarrow $ | $ \sin^ { -1 } \left ( \dfrac { \text { opposite } } { \text { hypotenuse } } \right ) =\theta $ $ \cos ( \theta ) =\dfrac { \text { adjacent } } { \text { hypotenuse } } $ | $ \rightarrow $ | $ \cos^ { -1 } \left ( \dfrac { \text { adjacent } } { \text { hypotenuse } } \right ) =\theta $ $ \tan ( \theta ) =\dfrac { \text { opposite } } { \text { adjacent } } $ | $ \rightarrow $ | $ \tan^ { -1 } \left ( \dfrac { \text { opposite } } { \text { adjacent } } \right ) =\theta $ misconception alert ! the expression $ \sin^ { -1 } ( x ) $ is not the same as $ \dfrac { 1 } { \sin ( x ) } $ . in other words , the $ \small { -1 } $ is not an exponent . instead , it simply means inverse function . however , there is an alternate notation that avoids this pitfall ! the inverse sine can also be expressed as $ \arcsin $ , the inverse cosine as $ \arccos $ , and the inverse tangent as $ \arctan $ . this notation is common in computer programming languages , but not in mathematics . solving the introductory problem in the introductory problem , we were given the opposite and adjacent side lengths , so we can use inverse tangent to find the angle . $ \begin { align } { m\angle l } & amp ; =\tan^ { -1 } \left ( \dfrac { \text { } \blued { \text { opposite } } } { \text { } \maroonc { \text { adjacent } } \text { } } \right ) \quad\small { \gray { \text { define . } } } \\ m\angle l & amp ; =\tan^ { -1 } \left ( \dfrac { \blued { 35 } } { \maroonc { 65 } } \right ) \quad\small { \gray { \text { substitute values . } } } \\ m\angle l & amp ; \approx 28.30^\circ \quad\small { \gray { \text { evaluate with a calculator . } } } \end { align } $ now let 's try some practice problems .
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let 's take a look at a new type of trigonometry problem . interestingly , these problems ca n't be solved with sine , cosine , or tangent . a problem : in the triangle below , what is the measure of angle $ l $ ?
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i do n't understand where 's the ambiguity if let 's say the ratio of the adjacent side to the hipothenuse is 1/2 , why ca n't we conclude that the angle is 60 degrees given by the cos ( 60 ) value ?
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let 's take a look at a new type of trigonometry problem . interestingly , these problems ca n't be solved with sine , cosine , or tangent . a problem : in the triangle below , what is the measure of angle $ l $ ? what we know : relative to $ \angle l $ , we know the lengths of the opposite and adjacent sides , so we can write : $ \tan ( l ) = \dfrac { \text { opposite } } { \text { adjacent } } = \dfrac { 35 } { 65 } $ but this does n't help us find the measure of $ \angle l $ . we 're stuck ! what we need : we need new mathematical tools to solve problems like these . our old friends sine , cosine , and tangent aren ’ t up to the task . they take angles and give side ratios , but we need functions that take side ratios and give angles . we need inverse trig functions ! the inverse trigonometric functions we already know about inverse operations . for example , addition and subtraction are inverse operations , and multiplication and division are inverse operations . each operation does the opposite of its inverse . the idea is the same in trigonometry . inverse trig functions do the opposite of the “ regular ” trig functions . for example : inverse sine $ ( \sin^ { -1 } ) $ does the opposite of the sine . inverse cosine $ ( \cos^ { -1 } ) $ does the opposite of the cosine . inverse tangent $ ( \tan^ { -1 } ) $ does the opposite of the tangent . in general , if you know the trig ratio but not the angle , you can use the corresponding inverse trig function to find the angle . this is expressed mathematically in the statements below . trigonometric functions input angles and output side ratios | |inverse trigonometric functions input side ratios and output angles : - : | : -| : - : $ \sin ( \theta ) =\dfrac { \text { opposite } } { \text { hypotenuse } } $ | $ \rightarrow $ | $ \sin^ { -1 } \left ( \dfrac { \text { opposite } } { \text { hypotenuse } } \right ) =\theta $ $ \cos ( \theta ) =\dfrac { \text { adjacent } } { \text { hypotenuse } } $ | $ \rightarrow $ | $ \cos^ { -1 } \left ( \dfrac { \text { adjacent } } { \text { hypotenuse } } \right ) =\theta $ $ \tan ( \theta ) =\dfrac { \text { opposite } } { \text { adjacent } } $ | $ \rightarrow $ | $ \tan^ { -1 } \left ( \dfrac { \text { opposite } } { \text { adjacent } } \right ) =\theta $ misconception alert ! the expression $ \sin^ { -1 } ( x ) $ is not the same as $ \dfrac { 1 } { \sin ( x ) } $ . in other words , the $ \small { -1 } $ is not an exponent . instead , it simply means inverse function . however , there is an alternate notation that avoids this pitfall ! the inverse sine can also be expressed as $ \arcsin $ , the inverse cosine as $ \arccos $ , and the inverse tangent as $ \arctan $ . this notation is common in computer programming languages , but not in mathematics . solving the introductory problem in the introductory problem , we were given the opposite and adjacent side lengths , so we can use inverse tangent to find the angle . $ \begin { align } { m\angle l } & amp ; =\tan^ { -1 } \left ( \dfrac { \text { } \blued { \text { opposite } } } { \text { } \maroonc { \text { adjacent } } \text { } } \right ) \quad\small { \gray { \text { define . } } } \\ m\angle l & amp ; =\tan^ { -1 } \left ( \dfrac { \blued { 35 } } { \maroonc { 65 } } \right ) \quad\small { \gray { \text { substitute values . } } } \\ m\angle l & amp ; \approx 28.30^\circ \quad\small { \gray { \text { evaluate with a calculator . } } } \end { align } $ now let 's try some practice problems .
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however , there is an alternate notation that avoids this pitfall ! the inverse sine can also be expressed as $ \arcsin $ , the inverse cosine as $ \arccos $ , and the inverse tangent as $ \arctan $ . this notation is common in computer programming languages , but not in mathematics .
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do we have to remember the value of arctan or arccos or arcsine ?
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let 's take a look at a new type of trigonometry problem . interestingly , these problems ca n't be solved with sine , cosine , or tangent . a problem : in the triangle below , what is the measure of angle $ l $ ? what we know : relative to $ \angle l $ , we know the lengths of the opposite and adjacent sides , so we can write : $ \tan ( l ) = \dfrac { \text { opposite } } { \text { adjacent } } = \dfrac { 35 } { 65 } $ but this does n't help us find the measure of $ \angle l $ . we 're stuck ! what we need : we need new mathematical tools to solve problems like these . our old friends sine , cosine , and tangent aren ’ t up to the task . they take angles and give side ratios , but we need functions that take side ratios and give angles . we need inverse trig functions ! the inverse trigonometric functions we already know about inverse operations . for example , addition and subtraction are inverse operations , and multiplication and division are inverse operations . each operation does the opposite of its inverse . the idea is the same in trigonometry . inverse trig functions do the opposite of the “ regular ” trig functions . for example : inverse sine $ ( \sin^ { -1 } ) $ does the opposite of the sine . inverse cosine $ ( \cos^ { -1 } ) $ does the opposite of the cosine . inverse tangent $ ( \tan^ { -1 } ) $ does the opposite of the tangent . in general , if you know the trig ratio but not the angle , you can use the corresponding inverse trig function to find the angle . this is expressed mathematically in the statements below . trigonometric functions input angles and output side ratios | |inverse trigonometric functions input side ratios and output angles : - : | : -| : - : $ \sin ( \theta ) =\dfrac { \text { opposite } } { \text { hypotenuse } } $ | $ \rightarrow $ | $ \sin^ { -1 } \left ( \dfrac { \text { opposite } } { \text { hypotenuse } } \right ) =\theta $ $ \cos ( \theta ) =\dfrac { \text { adjacent } } { \text { hypotenuse } } $ | $ \rightarrow $ | $ \cos^ { -1 } \left ( \dfrac { \text { adjacent } } { \text { hypotenuse } } \right ) =\theta $ $ \tan ( \theta ) =\dfrac { \text { opposite } } { \text { adjacent } } $ | $ \rightarrow $ | $ \tan^ { -1 } \left ( \dfrac { \text { opposite } } { \text { adjacent } } \right ) =\theta $ misconception alert ! the expression $ \sin^ { -1 } ( x ) $ is not the same as $ \dfrac { 1 } { \sin ( x ) } $ . in other words , the $ \small { -1 } $ is not an exponent . instead , it simply means inverse function . however , there is an alternate notation that avoids this pitfall ! the inverse sine can also be expressed as $ \arcsin $ , the inverse cosine as $ \arccos $ , and the inverse tangent as $ \arctan $ . this notation is common in computer programming languages , but not in mathematics . solving the introductory problem in the introductory problem , we were given the opposite and adjacent side lengths , so we can use inverse tangent to find the angle . $ \begin { align } { m\angle l } & amp ; =\tan^ { -1 } \left ( \dfrac { \text { } \blued { \text { opposite } } } { \text { } \maroonc { \text { adjacent } } \text { } } \right ) \quad\small { \gray { \text { define . } } } \\ m\angle l & amp ; =\tan^ { -1 } \left ( \dfrac { \blued { 35 } } { \maroonc { 65 } } \right ) \quad\small { \gray { \text { substitute values . } } } \\ m\angle l & amp ; \approx 28.30^\circ \quad\small { \gray { \text { evaluate with a calculator . } } } \end { align } $ now let 's try some practice problems .
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interestingly , these problems ca n't be solved with sine , cosine , or tangent . a problem : in the triangle below , what is the measure of angle $ l $ ? what we know : relative to $ \angle l $ , we know the lengths of the opposite and adjacent sides , so we can write : $ \tan ( l ) = \dfrac { \text { opposite } } { \text { adjacent } } = \dfrac { 35 } { 65 } $ but this does n't help us find the measure of $ \angle l $ .
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is there an article out there that shows you an angle measure and one side ?
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let 's take a look at a new type of trigonometry problem . interestingly , these problems ca n't be solved with sine , cosine , or tangent . a problem : in the triangle below , what is the measure of angle $ l $ ? what we know : relative to $ \angle l $ , we know the lengths of the opposite and adjacent sides , so we can write : $ \tan ( l ) = \dfrac { \text { opposite } } { \text { adjacent } } = \dfrac { 35 } { 65 } $ but this does n't help us find the measure of $ \angle l $ . we 're stuck ! what we need : we need new mathematical tools to solve problems like these . our old friends sine , cosine , and tangent aren ’ t up to the task . they take angles and give side ratios , but we need functions that take side ratios and give angles . we need inverse trig functions ! the inverse trigonometric functions we already know about inverse operations . for example , addition and subtraction are inverse operations , and multiplication and division are inverse operations . each operation does the opposite of its inverse . the idea is the same in trigonometry . inverse trig functions do the opposite of the “ regular ” trig functions . for example : inverse sine $ ( \sin^ { -1 } ) $ does the opposite of the sine . inverse cosine $ ( \cos^ { -1 } ) $ does the opposite of the cosine . inverse tangent $ ( \tan^ { -1 } ) $ does the opposite of the tangent . in general , if you know the trig ratio but not the angle , you can use the corresponding inverse trig function to find the angle . this is expressed mathematically in the statements below . trigonometric functions input angles and output side ratios | |inverse trigonometric functions input side ratios and output angles : - : | : -| : - : $ \sin ( \theta ) =\dfrac { \text { opposite } } { \text { hypotenuse } } $ | $ \rightarrow $ | $ \sin^ { -1 } \left ( \dfrac { \text { opposite } } { \text { hypotenuse } } \right ) =\theta $ $ \cos ( \theta ) =\dfrac { \text { adjacent } } { \text { hypotenuse } } $ | $ \rightarrow $ | $ \cos^ { -1 } \left ( \dfrac { \text { adjacent } } { \text { hypotenuse } } \right ) =\theta $ $ \tan ( \theta ) =\dfrac { \text { opposite } } { \text { adjacent } } $ | $ \rightarrow $ | $ \tan^ { -1 } \left ( \dfrac { \text { opposite } } { \text { adjacent } } \right ) =\theta $ misconception alert ! the expression $ \sin^ { -1 } ( x ) $ is not the same as $ \dfrac { 1 } { \sin ( x ) } $ . in other words , the $ \small { -1 } $ is not an exponent . instead , it simply means inverse function . however , there is an alternate notation that avoids this pitfall ! the inverse sine can also be expressed as $ \arcsin $ , the inverse cosine as $ \arccos $ , and the inverse tangent as $ \arctan $ . this notation is common in computer programming languages , but not in mathematics . solving the introductory problem in the introductory problem , we were given the opposite and adjacent side lengths , so we can use inverse tangent to find the angle . $ \begin { align } { m\angle l } & amp ; =\tan^ { -1 } \left ( \dfrac { \text { } \blued { \text { opposite } } } { \text { } \maroonc { \text { adjacent } } \text { } } \right ) \quad\small { \gray { \text { define . } } } \\ m\angle l & amp ; =\tan^ { -1 } \left ( \dfrac { \blued { 35 } } { \maroonc { 65 } } \right ) \quad\small { \gray { \text { substitute values . } } } \\ m\angle l & amp ; \approx 28.30^\circ \quad\small { \gray { \text { evaluate with a calculator . } } } \end { align } $ now let 's try some practice problems .
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a problem : in the triangle below , what is the measure of angle $ l $ ? what we know : relative to $ \angle l $ , we know the lengths of the opposite and adjacent sides , so we can write : $ \tan ( l ) = \dfrac { \text { opposite } } { \text { adjacent } } = \dfrac { 35 } { 65 } $ but this does n't help us find the measure of $ \angle l $ . we 're stuck !
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in trigonometry , whether in pre-cal or ap physics , is the x always adjacent and the y always opposite in every case ?
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let 's take a look at a new type of trigonometry problem . interestingly , these problems ca n't be solved with sine , cosine , or tangent . a problem : in the triangle below , what is the measure of angle $ l $ ? what we know : relative to $ \angle l $ , we know the lengths of the opposite and adjacent sides , so we can write : $ \tan ( l ) = \dfrac { \text { opposite } } { \text { adjacent } } = \dfrac { 35 } { 65 } $ but this does n't help us find the measure of $ \angle l $ . we 're stuck ! what we need : we need new mathematical tools to solve problems like these . our old friends sine , cosine , and tangent aren ’ t up to the task . they take angles and give side ratios , but we need functions that take side ratios and give angles . we need inverse trig functions ! the inverse trigonometric functions we already know about inverse operations . for example , addition and subtraction are inverse operations , and multiplication and division are inverse operations . each operation does the opposite of its inverse . the idea is the same in trigonometry . inverse trig functions do the opposite of the “ regular ” trig functions . for example : inverse sine $ ( \sin^ { -1 } ) $ does the opposite of the sine . inverse cosine $ ( \cos^ { -1 } ) $ does the opposite of the cosine . inverse tangent $ ( \tan^ { -1 } ) $ does the opposite of the tangent . in general , if you know the trig ratio but not the angle , you can use the corresponding inverse trig function to find the angle . this is expressed mathematically in the statements below . trigonometric functions input angles and output side ratios | |inverse trigonometric functions input side ratios and output angles : - : | : -| : - : $ \sin ( \theta ) =\dfrac { \text { opposite } } { \text { hypotenuse } } $ | $ \rightarrow $ | $ \sin^ { -1 } \left ( \dfrac { \text { opposite } } { \text { hypotenuse } } \right ) =\theta $ $ \cos ( \theta ) =\dfrac { \text { adjacent } } { \text { hypotenuse } } $ | $ \rightarrow $ | $ \cos^ { -1 } \left ( \dfrac { \text { adjacent } } { \text { hypotenuse } } \right ) =\theta $ $ \tan ( \theta ) =\dfrac { \text { opposite } } { \text { adjacent } } $ | $ \rightarrow $ | $ \tan^ { -1 } \left ( \dfrac { \text { opposite } } { \text { adjacent } } \right ) =\theta $ misconception alert ! the expression $ \sin^ { -1 } ( x ) $ is not the same as $ \dfrac { 1 } { \sin ( x ) } $ . in other words , the $ \small { -1 } $ is not an exponent . instead , it simply means inverse function . however , there is an alternate notation that avoids this pitfall ! the inverse sine can also be expressed as $ \arcsin $ , the inverse cosine as $ \arccos $ , and the inverse tangent as $ \arctan $ . this notation is common in computer programming languages , but not in mathematics . solving the introductory problem in the introductory problem , we were given the opposite and adjacent side lengths , so we can use inverse tangent to find the angle . $ \begin { align } { m\angle l } & amp ; =\tan^ { -1 } \left ( \dfrac { \text { } \blued { \text { opposite } } } { \text { } \maroonc { \text { adjacent } } \text { } } \right ) \quad\small { \gray { \text { define . } } } \\ m\angle l & amp ; =\tan^ { -1 } \left ( \dfrac { \blued { 35 } } { \maroonc { 65 } } \right ) \quad\small { \gray { \text { substitute values . } } } \\ m\angle l & amp ; \approx 28.30^\circ \quad\small { \gray { \text { evaluate with a calculator . } } } \end { align } $ now let 's try some practice problems .
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$ \begin { align } { m\angle l } & amp ; =\tan^ { -1 } \left ( \dfrac { \text { } \blued { \text { opposite } } } { \text { } \maroonc { \text { adjacent } } \text { } } \right ) \quad\small { \gray { \text { define . } } } \\ m\angle l & amp ; =\tan^ { -1 } \left ( \dfrac { \blued { 35 } } { \maroonc { 65 } } \right ) \quad\small { \gray { \text { substitute values . } } } \\ m\angle l & amp ; \approx 28.30^\circ \quad\small { \gray { \text { evaluate with a calculator .
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hi , when we are asked to solve eg ) tan angle abc do we give the answer as fractions or simplify into decimals ?
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let 's take a look at a new type of trigonometry problem . interestingly , these problems ca n't be solved with sine , cosine , or tangent . a problem : in the triangle below , what is the measure of angle $ l $ ? what we know : relative to $ \angle l $ , we know the lengths of the opposite and adjacent sides , so we can write : $ \tan ( l ) = \dfrac { \text { opposite } } { \text { adjacent } } = \dfrac { 35 } { 65 } $ but this does n't help us find the measure of $ \angle l $ . we 're stuck ! what we need : we need new mathematical tools to solve problems like these . our old friends sine , cosine , and tangent aren ’ t up to the task . they take angles and give side ratios , but we need functions that take side ratios and give angles . we need inverse trig functions ! the inverse trigonometric functions we already know about inverse operations . for example , addition and subtraction are inverse operations , and multiplication and division are inverse operations . each operation does the opposite of its inverse . the idea is the same in trigonometry . inverse trig functions do the opposite of the “ regular ” trig functions . for example : inverse sine $ ( \sin^ { -1 } ) $ does the opposite of the sine . inverse cosine $ ( \cos^ { -1 } ) $ does the opposite of the cosine . inverse tangent $ ( \tan^ { -1 } ) $ does the opposite of the tangent . in general , if you know the trig ratio but not the angle , you can use the corresponding inverse trig function to find the angle . this is expressed mathematically in the statements below . trigonometric functions input angles and output side ratios | |inverse trigonometric functions input side ratios and output angles : - : | : -| : - : $ \sin ( \theta ) =\dfrac { \text { opposite } } { \text { hypotenuse } } $ | $ \rightarrow $ | $ \sin^ { -1 } \left ( \dfrac { \text { opposite } } { \text { hypotenuse } } \right ) =\theta $ $ \cos ( \theta ) =\dfrac { \text { adjacent } } { \text { hypotenuse } } $ | $ \rightarrow $ | $ \cos^ { -1 } \left ( \dfrac { \text { adjacent } } { \text { hypotenuse } } \right ) =\theta $ $ \tan ( \theta ) =\dfrac { \text { opposite } } { \text { adjacent } } $ | $ \rightarrow $ | $ \tan^ { -1 } \left ( \dfrac { \text { opposite } } { \text { adjacent } } \right ) =\theta $ misconception alert ! the expression $ \sin^ { -1 } ( x ) $ is not the same as $ \dfrac { 1 } { \sin ( x ) } $ . in other words , the $ \small { -1 } $ is not an exponent . instead , it simply means inverse function . however , there is an alternate notation that avoids this pitfall ! the inverse sine can also be expressed as $ \arcsin $ , the inverse cosine as $ \arccos $ , and the inverse tangent as $ \arctan $ . this notation is common in computer programming languages , but not in mathematics . solving the introductory problem in the introductory problem , we were given the opposite and adjacent side lengths , so we can use inverse tangent to find the angle . $ \begin { align } { m\angle l } & amp ; =\tan^ { -1 } \left ( \dfrac { \text { } \blued { \text { opposite } } } { \text { } \maroonc { \text { adjacent } } \text { } } \right ) \quad\small { \gray { \text { define . } } } \\ m\angle l & amp ; =\tan^ { -1 } \left ( \dfrac { \blued { 35 } } { \maroonc { 65 } } \right ) \quad\small { \gray { \text { substitute values . } } } \\ m\angle l & amp ; \approx 28.30^\circ \quad\small { \gray { \text { evaluate with a calculator . } } } \end { align } $ now let 's try some practice problems .
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they take angles and give side ratios , but we need functions that take side ratios and give angles . we need inverse trig functions ! the inverse trigonometric functions we already know about inverse operations . for example , addition and subtraction are inverse operations , and multiplication and division are inverse operations .
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why we say inverse of sin , cos , tan ?
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let 's take a look at a new type of trigonometry problem . interestingly , these problems ca n't be solved with sine , cosine , or tangent . a problem : in the triangle below , what is the measure of angle $ l $ ? what we know : relative to $ \angle l $ , we know the lengths of the opposite and adjacent sides , so we can write : $ \tan ( l ) = \dfrac { \text { opposite } } { \text { adjacent } } = \dfrac { 35 } { 65 } $ but this does n't help us find the measure of $ \angle l $ . we 're stuck ! what we need : we need new mathematical tools to solve problems like these . our old friends sine , cosine , and tangent aren ’ t up to the task . they take angles and give side ratios , but we need functions that take side ratios and give angles . we need inverse trig functions ! the inverse trigonometric functions we already know about inverse operations . for example , addition and subtraction are inverse operations , and multiplication and division are inverse operations . each operation does the opposite of its inverse . the idea is the same in trigonometry . inverse trig functions do the opposite of the “ regular ” trig functions . for example : inverse sine $ ( \sin^ { -1 } ) $ does the opposite of the sine . inverse cosine $ ( \cos^ { -1 } ) $ does the opposite of the cosine . inverse tangent $ ( \tan^ { -1 } ) $ does the opposite of the tangent . in general , if you know the trig ratio but not the angle , you can use the corresponding inverse trig function to find the angle . this is expressed mathematically in the statements below . trigonometric functions input angles and output side ratios | |inverse trigonometric functions input side ratios and output angles : - : | : -| : - : $ \sin ( \theta ) =\dfrac { \text { opposite } } { \text { hypotenuse } } $ | $ \rightarrow $ | $ \sin^ { -1 } \left ( \dfrac { \text { opposite } } { \text { hypotenuse } } \right ) =\theta $ $ \cos ( \theta ) =\dfrac { \text { adjacent } } { \text { hypotenuse } } $ | $ \rightarrow $ | $ \cos^ { -1 } \left ( \dfrac { \text { adjacent } } { \text { hypotenuse } } \right ) =\theta $ $ \tan ( \theta ) =\dfrac { \text { opposite } } { \text { adjacent } } $ | $ \rightarrow $ | $ \tan^ { -1 } \left ( \dfrac { \text { opposite } } { \text { adjacent } } \right ) =\theta $ misconception alert ! the expression $ \sin^ { -1 } ( x ) $ is not the same as $ \dfrac { 1 } { \sin ( x ) } $ . in other words , the $ \small { -1 } $ is not an exponent . instead , it simply means inverse function . however , there is an alternate notation that avoids this pitfall ! the inverse sine can also be expressed as $ \arcsin $ , the inverse cosine as $ \arccos $ , and the inverse tangent as $ \arctan $ . this notation is common in computer programming languages , but not in mathematics . solving the introductory problem in the introductory problem , we were given the opposite and adjacent side lengths , so we can use inverse tangent to find the angle . $ \begin { align } { m\angle l } & amp ; =\tan^ { -1 } \left ( \dfrac { \text { } \blued { \text { opposite } } } { \text { } \maroonc { \text { adjacent } } \text { } } \right ) \quad\small { \gray { \text { define . } } } \\ m\angle l & amp ; =\tan^ { -1 } \left ( \dfrac { \blued { 35 } } { \maroonc { 65 } } \right ) \quad\small { \gray { \text { substitute values . } } } \\ m\angle l & amp ; \approx 28.30^\circ \quad\small { \gray { \text { evaluate with a calculator . } } } \end { align } $ now let 's try some practice problems .
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they take angles and give side ratios , but we need functions that take side ratios and give angles . we need inverse trig functions ! the inverse trigonometric functions we already know about inverse operations . for example , addition and subtraction are inverse operations , and multiplication and division are inverse operations .
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why just inverse words are used ?
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let 's take a look at a new type of trigonometry problem . interestingly , these problems ca n't be solved with sine , cosine , or tangent . a problem : in the triangle below , what is the measure of angle $ l $ ? what we know : relative to $ \angle l $ , we know the lengths of the opposite and adjacent sides , so we can write : $ \tan ( l ) = \dfrac { \text { opposite } } { \text { adjacent } } = \dfrac { 35 } { 65 } $ but this does n't help us find the measure of $ \angle l $ . we 're stuck ! what we need : we need new mathematical tools to solve problems like these . our old friends sine , cosine , and tangent aren ’ t up to the task . they take angles and give side ratios , but we need functions that take side ratios and give angles . we need inverse trig functions ! the inverse trigonometric functions we already know about inverse operations . for example , addition and subtraction are inverse operations , and multiplication and division are inverse operations . each operation does the opposite of its inverse . the idea is the same in trigonometry . inverse trig functions do the opposite of the “ regular ” trig functions . for example : inverse sine $ ( \sin^ { -1 } ) $ does the opposite of the sine . inverse cosine $ ( \cos^ { -1 } ) $ does the opposite of the cosine . inverse tangent $ ( \tan^ { -1 } ) $ does the opposite of the tangent . in general , if you know the trig ratio but not the angle , you can use the corresponding inverse trig function to find the angle . this is expressed mathematically in the statements below . trigonometric functions input angles and output side ratios | |inverse trigonometric functions input side ratios and output angles : - : | : -| : - : $ \sin ( \theta ) =\dfrac { \text { opposite } } { \text { hypotenuse } } $ | $ \rightarrow $ | $ \sin^ { -1 } \left ( \dfrac { \text { opposite } } { \text { hypotenuse } } \right ) =\theta $ $ \cos ( \theta ) =\dfrac { \text { adjacent } } { \text { hypotenuse } } $ | $ \rightarrow $ | $ \cos^ { -1 } \left ( \dfrac { \text { adjacent } } { \text { hypotenuse } } \right ) =\theta $ $ \tan ( \theta ) =\dfrac { \text { opposite } } { \text { adjacent } } $ | $ \rightarrow $ | $ \tan^ { -1 } \left ( \dfrac { \text { opposite } } { \text { adjacent } } \right ) =\theta $ misconception alert ! the expression $ \sin^ { -1 } ( x ) $ is not the same as $ \dfrac { 1 } { \sin ( x ) } $ . in other words , the $ \small { -1 } $ is not an exponent . instead , it simply means inverse function . however , there is an alternate notation that avoids this pitfall ! the inverse sine can also be expressed as $ \arcsin $ , the inverse cosine as $ \arccos $ , and the inverse tangent as $ \arctan $ . this notation is common in computer programming languages , but not in mathematics . solving the introductory problem in the introductory problem , we were given the opposite and adjacent side lengths , so we can use inverse tangent to find the angle . $ \begin { align } { m\angle l } & amp ; =\tan^ { -1 } \left ( \dfrac { \text { } \blued { \text { opposite } } } { \text { } \maroonc { \text { adjacent } } \text { } } \right ) \quad\small { \gray { \text { define . } } } \\ m\angle l & amp ; =\tan^ { -1 } \left ( \dfrac { \blued { 35 } } { \maroonc { 65 } } \right ) \quad\small { \gray { \text { substitute values . } } } \\ m\angle l & amp ; \approx 28.30^\circ \quad\small { \gray { \text { evaluate with a calculator . } } } \end { align } $ now let 's try some practice problems .
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our old friends sine , cosine , and tangent aren ’ t up to the task . they take angles and give side ratios , but we need functions that take side ratios and give angles . we need inverse trig functions !
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because you are inputting side ratios for inverses , do you have to switch from degrees to radians on a calculator ?
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let 's take a look at a new type of trigonometry problem . interestingly , these problems ca n't be solved with sine , cosine , or tangent . a problem : in the triangle below , what is the measure of angle $ l $ ? what we know : relative to $ \angle l $ , we know the lengths of the opposite and adjacent sides , so we can write : $ \tan ( l ) = \dfrac { \text { opposite } } { \text { adjacent } } = \dfrac { 35 } { 65 } $ but this does n't help us find the measure of $ \angle l $ . we 're stuck ! what we need : we need new mathematical tools to solve problems like these . our old friends sine , cosine , and tangent aren ’ t up to the task . they take angles and give side ratios , but we need functions that take side ratios and give angles . we need inverse trig functions ! the inverse trigonometric functions we already know about inverse operations . for example , addition and subtraction are inverse operations , and multiplication and division are inverse operations . each operation does the opposite of its inverse . the idea is the same in trigonometry . inverse trig functions do the opposite of the “ regular ” trig functions . for example : inverse sine $ ( \sin^ { -1 } ) $ does the opposite of the sine . inverse cosine $ ( \cos^ { -1 } ) $ does the opposite of the cosine . inverse tangent $ ( \tan^ { -1 } ) $ does the opposite of the tangent . in general , if you know the trig ratio but not the angle , you can use the corresponding inverse trig function to find the angle . this is expressed mathematically in the statements below . trigonometric functions input angles and output side ratios | |inverse trigonometric functions input side ratios and output angles : - : | : -| : - : $ \sin ( \theta ) =\dfrac { \text { opposite } } { \text { hypotenuse } } $ | $ \rightarrow $ | $ \sin^ { -1 } \left ( \dfrac { \text { opposite } } { \text { hypotenuse } } \right ) =\theta $ $ \cos ( \theta ) =\dfrac { \text { adjacent } } { \text { hypotenuse } } $ | $ \rightarrow $ | $ \cos^ { -1 } \left ( \dfrac { \text { adjacent } } { \text { hypotenuse } } \right ) =\theta $ $ \tan ( \theta ) =\dfrac { \text { opposite } } { \text { adjacent } } $ | $ \rightarrow $ | $ \tan^ { -1 } \left ( \dfrac { \text { opposite } } { \text { adjacent } } \right ) =\theta $ misconception alert ! the expression $ \sin^ { -1 } ( x ) $ is not the same as $ \dfrac { 1 } { \sin ( x ) } $ . in other words , the $ \small { -1 } $ is not an exponent . instead , it simply means inverse function . however , there is an alternate notation that avoids this pitfall ! the inverse sine can also be expressed as $ \arcsin $ , the inverse cosine as $ \arccos $ , and the inverse tangent as $ \arctan $ . this notation is common in computer programming languages , but not in mathematics . solving the introductory problem in the introductory problem , we were given the opposite and adjacent side lengths , so we can use inverse tangent to find the angle . $ \begin { align } { m\angle l } & amp ; =\tan^ { -1 } \left ( \dfrac { \text { } \blued { \text { opposite } } } { \text { } \maroonc { \text { adjacent } } \text { } } \right ) \quad\small { \gray { \text { define . } } } \\ m\angle l & amp ; =\tan^ { -1 } \left ( \dfrac { \blued { 35 } } { \maroonc { 65 } } \right ) \quad\small { \gray { \text { substitute values . } } } \\ m\angle l & amp ; \approx 28.30^\circ \quad\small { \gray { \text { evaluate with a calculator . } } } \end { align } $ now let 's try some practice problems .
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we need inverse trig functions ! the inverse trigonometric functions we already know about inverse operations . for example , addition and subtraction are inverse operations , and multiplication and division are inverse operations .
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how do you know whether to have your calculator in degree or radian mode ?
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let 's take a look at a new type of trigonometry problem . interestingly , these problems ca n't be solved with sine , cosine , or tangent . a problem : in the triangle below , what is the measure of angle $ l $ ? what we know : relative to $ \angle l $ , we know the lengths of the opposite and adjacent sides , so we can write : $ \tan ( l ) = \dfrac { \text { opposite } } { \text { adjacent } } = \dfrac { 35 } { 65 } $ but this does n't help us find the measure of $ \angle l $ . we 're stuck ! what we need : we need new mathematical tools to solve problems like these . our old friends sine , cosine , and tangent aren ’ t up to the task . they take angles and give side ratios , but we need functions that take side ratios and give angles . we need inverse trig functions ! the inverse trigonometric functions we already know about inverse operations . for example , addition and subtraction are inverse operations , and multiplication and division are inverse operations . each operation does the opposite of its inverse . the idea is the same in trigonometry . inverse trig functions do the opposite of the “ regular ” trig functions . for example : inverse sine $ ( \sin^ { -1 } ) $ does the opposite of the sine . inverse cosine $ ( \cos^ { -1 } ) $ does the opposite of the cosine . inverse tangent $ ( \tan^ { -1 } ) $ does the opposite of the tangent . in general , if you know the trig ratio but not the angle , you can use the corresponding inverse trig function to find the angle . this is expressed mathematically in the statements below . trigonometric functions input angles and output side ratios | |inverse trigonometric functions input side ratios and output angles : - : | : -| : - : $ \sin ( \theta ) =\dfrac { \text { opposite } } { \text { hypotenuse } } $ | $ \rightarrow $ | $ \sin^ { -1 } \left ( \dfrac { \text { opposite } } { \text { hypotenuse } } \right ) =\theta $ $ \cos ( \theta ) =\dfrac { \text { adjacent } } { \text { hypotenuse } } $ | $ \rightarrow $ | $ \cos^ { -1 } \left ( \dfrac { \text { adjacent } } { \text { hypotenuse } } \right ) =\theta $ $ \tan ( \theta ) =\dfrac { \text { opposite } } { \text { adjacent } } $ | $ \rightarrow $ | $ \tan^ { -1 } \left ( \dfrac { \text { opposite } } { \text { adjacent } } \right ) =\theta $ misconception alert ! the expression $ \sin^ { -1 } ( x ) $ is not the same as $ \dfrac { 1 } { \sin ( x ) } $ . in other words , the $ \small { -1 } $ is not an exponent . instead , it simply means inverse function . however , there is an alternate notation that avoids this pitfall ! the inverse sine can also be expressed as $ \arcsin $ , the inverse cosine as $ \arccos $ , and the inverse tangent as $ \arctan $ . this notation is common in computer programming languages , but not in mathematics . solving the introductory problem in the introductory problem , we were given the opposite and adjacent side lengths , so we can use inverse tangent to find the angle . $ \begin { align } { m\angle l } & amp ; =\tan^ { -1 } \left ( \dfrac { \text { } \blued { \text { opposite } } } { \text { } \maroonc { \text { adjacent } } \text { } } \right ) \quad\small { \gray { \text { define . } } } \\ m\angle l & amp ; =\tan^ { -1 } \left ( \dfrac { \blued { 35 } } { \maroonc { 65 } } \right ) \quad\small { \gray { \text { substitute values . } } } \\ m\angle l & amp ; \approx 28.30^\circ \quad\small { \gray { \text { evaluate with a calculator . } } } \end { align } $ now let 's try some practice problems .
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let 's take a look at a new type of trigonometry problem . interestingly , these problems ca n't be solved with sine , cosine , or tangent .
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when is secant , cosecants , and cotangents discussed ?
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let 's take a look at a new type of trigonometry problem . interestingly , these problems ca n't be solved with sine , cosine , or tangent . a problem : in the triangle below , what is the measure of angle $ l $ ? what we know : relative to $ \angle l $ , we know the lengths of the opposite and adjacent sides , so we can write : $ \tan ( l ) = \dfrac { \text { opposite } } { \text { adjacent } } = \dfrac { 35 } { 65 } $ but this does n't help us find the measure of $ \angle l $ . we 're stuck ! what we need : we need new mathematical tools to solve problems like these . our old friends sine , cosine , and tangent aren ’ t up to the task . they take angles and give side ratios , but we need functions that take side ratios and give angles . we need inverse trig functions ! the inverse trigonometric functions we already know about inverse operations . for example , addition and subtraction are inverse operations , and multiplication and division are inverse operations . each operation does the opposite of its inverse . the idea is the same in trigonometry . inverse trig functions do the opposite of the “ regular ” trig functions . for example : inverse sine $ ( \sin^ { -1 } ) $ does the opposite of the sine . inverse cosine $ ( \cos^ { -1 } ) $ does the opposite of the cosine . inverse tangent $ ( \tan^ { -1 } ) $ does the opposite of the tangent . in general , if you know the trig ratio but not the angle , you can use the corresponding inverse trig function to find the angle . this is expressed mathematically in the statements below . trigonometric functions input angles and output side ratios | |inverse trigonometric functions input side ratios and output angles : - : | : -| : - : $ \sin ( \theta ) =\dfrac { \text { opposite } } { \text { hypotenuse } } $ | $ \rightarrow $ | $ \sin^ { -1 } \left ( \dfrac { \text { opposite } } { \text { hypotenuse } } \right ) =\theta $ $ \cos ( \theta ) =\dfrac { \text { adjacent } } { \text { hypotenuse } } $ | $ \rightarrow $ | $ \cos^ { -1 } \left ( \dfrac { \text { adjacent } } { \text { hypotenuse } } \right ) =\theta $ $ \tan ( \theta ) =\dfrac { \text { opposite } } { \text { adjacent } } $ | $ \rightarrow $ | $ \tan^ { -1 } \left ( \dfrac { \text { opposite } } { \text { adjacent } } \right ) =\theta $ misconception alert ! the expression $ \sin^ { -1 } ( x ) $ is not the same as $ \dfrac { 1 } { \sin ( x ) } $ . in other words , the $ \small { -1 } $ is not an exponent . instead , it simply means inverse function . however , there is an alternate notation that avoids this pitfall ! the inverse sine can also be expressed as $ \arcsin $ , the inverse cosine as $ \arccos $ , and the inverse tangent as $ \arctan $ . this notation is common in computer programming languages , but not in mathematics . solving the introductory problem in the introductory problem , we were given the opposite and adjacent side lengths , so we can use inverse tangent to find the angle . $ \begin { align } { m\angle l } & amp ; =\tan^ { -1 } \left ( \dfrac { \text { } \blued { \text { opposite } } } { \text { } \maroonc { \text { adjacent } } \text { } } \right ) \quad\small { \gray { \text { define . } } } \\ m\angle l & amp ; =\tan^ { -1 } \left ( \dfrac { \blued { 35 } } { \maroonc { 65 } } \right ) \quad\small { \gray { \text { substitute values . } } } \\ m\angle l & amp ; \approx 28.30^\circ \quad\small { \gray { \text { evaluate with a calculator . } } } \end { align } $ now let 's try some practice problems .
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let 's take a look at a new type of trigonometry problem . interestingly , these problems ca n't be solved with sine , cosine , or tangent . a problem : in the triangle below , what is the measure of angle $ l $ ?
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when do you know to divide when solving using sine , cosine , and tangent vs when to multiply when solving sine , cosine , and tangent ?
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let 's take a look at a new type of trigonometry problem . interestingly , these problems ca n't be solved with sine , cosine , or tangent . a problem : in the triangle below , what is the measure of angle $ l $ ? what we know : relative to $ \angle l $ , we know the lengths of the opposite and adjacent sides , so we can write : $ \tan ( l ) = \dfrac { \text { opposite } } { \text { adjacent } } = \dfrac { 35 } { 65 } $ but this does n't help us find the measure of $ \angle l $ . we 're stuck ! what we need : we need new mathematical tools to solve problems like these . our old friends sine , cosine , and tangent aren ’ t up to the task . they take angles and give side ratios , but we need functions that take side ratios and give angles . we need inverse trig functions ! the inverse trigonometric functions we already know about inverse operations . for example , addition and subtraction are inverse operations , and multiplication and division are inverse operations . each operation does the opposite of its inverse . the idea is the same in trigonometry . inverse trig functions do the opposite of the “ regular ” trig functions . for example : inverse sine $ ( \sin^ { -1 } ) $ does the opposite of the sine . inverse cosine $ ( \cos^ { -1 } ) $ does the opposite of the cosine . inverse tangent $ ( \tan^ { -1 } ) $ does the opposite of the tangent . in general , if you know the trig ratio but not the angle , you can use the corresponding inverse trig function to find the angle . this is expressed mathematically in the statements below . trigonometric functions input angles and output side ratios | |inverse trigonometric functions input side ratios and output angles : - : | : -| : - : $ \sin ( \theta ) =\dfrac { \text { opposite } } { \text { hypotenuse } } $ | $ \rightarrow $ | $ \sin^ { -1 } \left ( \dfrac { \text { opposite } } { \text { hypotenuse } } \right ) =\theta $ $ \cos ( \theta ) =\dfrac { \text { adjacent } } { \text { hypotenuse } } $ | $ \rightarrow $ | $ \cos^ { -1 } \left ( \dfrac { \text { adjacent } } { \text { hypotenuse } } \right ) =\theta $ $ \tan ( \theta ) =\dfrac { \text { opposite } } { \text { adjacent } } $ | $ \rightarrow $ | $ \tan^ { -1 } \left ( \dfrac { \text { opposite } } { \text { adjacent } } \right ) =\theta $ misconception alert ! the expression $ \sin^ { -1 } ( x ) $ is not the same as $ \dfrac { 1 } { \sin ( x ) } $ . in other words , the $ \small { -1 } $ is not an exponent . instead , it simply means inverse function . however , there is an alternate notation that avoids this pitfall ! the inverse sine can also be expressed as $ \arcsin $ , the inverse cosine as $ \arccos $ , and the inverse tangent as $ \arctan $ . this notation is common in computer programming languages , but not in mathematics . solving the introductory problem in the introductory problem , we were given the opposite and adjacent side lengths , so we can use inverse tangent to find the angle . $ \begin { align } { m\angle l } & amp ; =\tan^ { -1 } \left ( \dfrac { \text { } \blued { \text { opposite } } } { \text { } \maroonc { \text { adjacent } } \text { } } \right ) \quad\small { \gray { \text { define . } } } \\ m\angle l & amp ; =\tan^ { -1 } \left ( \dfrac { \blued { 35 } } { \maroonc { 65 } } \right ) \quad\small { \gray { \text { substitute values . } } } \\ m\angle l & amp ; \approx 28.30^\circ \quad\small { \gray { \text { evaluate with a calculator . } } } \end { align } $ now let 's try some practice problems .
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inverse trig functions do the opposite of the “ regular ” trig functions . for example : inverse sine $ ( \sin^ { -1 } ) $ does the opposite of the sine . inverse cosine $ ( \cos^ { -1 } ) $ does the opposite of the cosine .
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like sine is opposite/hypotenuse , but is n't there also hypotenuse/opposite ?
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let 's take a look at a new type of trigonometry problem . interestingly , these problems ca n't be solved with sine , cosine , or tangent . a problem : in the triangle below , what is the measure of angle $ l $ ? what we know : relative to $ \angle l $ , we know the lengths of the opposite and adjacent sides , so we can write : $ \tan ( l ) = \dfrac { \text { opposite } } { \text { adjacent } } = \dfrac { 35 } { 65 } $ but this does n't help us find the measure of $ \angle l $ . we 're stuck ! what we need : we need new mathematical tools to solve problems like these . our old friends sine , cosine , and tangent aren ’ t up to the task . they take angles and give side ratios , but we need functions that take side ratios and give angles . we need inverse trig functions ! the inverse trigonometric functions we already know about inverse operations . for example , addition and subtraction are inverse operations , and multiplication and division are inverse operations . each operation does the opposite of its inverse . the idea is the same in trigonometry . inverse trig functions do the opposite of the “ regular ” trig functions . for example : inverse sine $ ( \sin^ { -1 } ) $ does the opposite of the sine . inverse cosine $ ( \cos^ { -1 } ) $ does the opposite of the cosine . inverse tangent $ ( \tan^ { -1 } ) $ does the opposite of the tangent . in general , if you know the trig ratio but not the angle , you can use the corresponding inverse trig function to find the angle . this is expressed mathematically in the statements below . trigonometric functions input angles and output side ratios | |inverse trigonometric functions input side ratios and output angles : - : | : -| : - : $ \sin ( \theta ) =\dfrac { \text { opposite } } { \text { hypotenuse } } $ | $ \rightarrow $ | $ \sin^ { -1 } \left ( \dfrac { \text { opposite } } { \text { hypotenuse } } \right ) =\theta $ $ \cos ( \theta ) =\dfrac { \text { adjacent } } { \text { hypotenuse } } $ | $ \rightarrow $ | $ \cos^ { -1 } \left ( \dfrac { \text { adjacent } } { \text { hypotenuse } } \right ) =\theta $ $ \tan ( \theta ) =\dfrac { \text { opposite } } { \text { adjacent } } $ | $ \rightarrow $ | $ \tan^ { -1 } \left ( \dfrac { \text { opposite } } { \text { adjacent } } \right ) =\theta $ misconception alert ! the expression $ \sin^ { -1 } ( x ) $ is not the same as $ \dfrac { 1 } { \sin ( x ) } $ . in other words , the $ \small { -1 } $ is not an exponent . instead , it simply means inverse function . however , there is an alternate notation that avoids this pitfall ! the inverse sine can also be expressed as $ \arcsin $ , the inverse cosine as $ \arccos $ , and the inverse tangent as $ \arctan $ . this notation is common in computer programming languages , but not in mathematics . solving the introductory problem in the introductory problem , we were given the opposite and adjacent side lengths , so we can use inverse tangent to find the angle . $ \begin { align } { m\angle l } & amp ; =\tan^ { -1 } \left ( \dfrac { \text { } \blued { \text { opposite } } } { \text { } \maroonc { \text { adjacent } } \text { } } \right ) \quad\small { \gray { \text { define . } } } \\ m\angle l & amp ; =\tan^ { -1 } \left ( \dfrac { \blued { 35 } } { \maroonc { 65 } } \right ) \quad\small { \gray { \text { substitute values . } } } \\ m\angle l & amp ; \approx 28.30^\circ \quad\small { \gray { \text { evaluate with a calculator . } } } \end { align } $ now let 's try some practice problems .
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let 's take a look at a new type of trigonometry problem . interestingly , these problems ca n't be solved with sine , cosine , or tangent .
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how can i calculate such equations ?
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let 's take a look at a new type of trigonometry problem . interestingly , these problems ca n't be solved with sine , cosine , or tangent . a problem : in the triangle below , what is the measure of angle $ l $ ? what we know : relative to $ \angle l $ , we know the lengths of the opposite and adjacent sides , so we can write : $ \tan ( l ) = \dfrac { \text { opposite } } { \text { adjacent } } = \dfrac { 35 } { 65 } $ but this does n't help us find the measure of $ \angle l $ . we 're stuck ! what we need : we need new mathematical tools to solve problems like these . our old friends sine , cosine , and tangent aren ’ t up to the task . they take angles and give side ratios , but we need functions that take side ratios and give angles . we need inverse trig functions ! the inverse trigonometric functions we already know about inverse operations . for example , addition and subtraction are inverse operations , and multiplication and division are inverse operations . each operation does the opposite of its inverse . the idea is the same in trigonometry . inverse trig functions do the opposite of the “ regular ” trig functions . for example : inverse sine $ ( \sin^ { -1 } ) $ does the opposite of the sine . inverse cosine $ ( \cos^ { -1 } ) $ does the opposite of the cosine . inverse tangent $ ( \tan^ { -1 } ) $ does the opposite of the tangent . in general , if you know the trig ratio but not the angle , you can use the corresponding inverse trig function to find the angle . this is expressed mathematically in the statements below . trigonometric functions input angles and output side ratios | |inverse trigonometric functions input side ratios and output angles : - : | : -| : - : $ \sin ( \theta ) =\dfrac { \text { opposite } } { \text { hypotenuse } } $ | $ \rightarrow $ | $ \sin^ { -1 } \left ( \dfrac { \text { opposite } } { \text { hypotenuse } } \right ) =\theta $ $ \cos ( \theta ) =\dfrac { \text { adjacent } } { \text { hypotenuse } } $ | $ \rightarrow $ | $ \cos^ { -1 } \left ( \dfrac { \text { adjacent } } { \text { hypotenuse } } \right ) =\theta $ $ \tan ( \theta ) =\dfrac { \text { opposite } } { \text { adjacent } } $ | $ \rightarrow $ | $ \tan^ { -1 } \left ( \dfrac { \text { opposite } } { \text { adjacent } } \right ) =\theta $ misconception alert ! the expression $ \sin^ { -1 } ( x ) $ is not the same as $ \dfrac { 1 } { \sin ( x ) } $ . in other words , the $ \small { -1 } $ is not an exponent . instead , it simply means inverse function . however , there is an alternate notation that avoids this pitfall ! the inverse sine can also be expressed as $ \arcsin $ , the inverse cosine as $ \arccos $ , and the inverse tangent as $ \arctan $ . this notation is common in computer programming languages , but not in mathematics . solving the introductory problem in the introductory problem , we were given the opposite and adjacent side lengths , so we can use inverse tangent to find the angle . $ \begin { align } { m\angle l } & amp ; =\tan^ { -1 } \left ( \dfrac { \text { } \blued { \text { opposite } } } { \text { } \maroonc { \text { adjacent } } \text { } } \right ) \quad\small { \gray { \text { define . } } } \\ m\angle l & amp ; =\tan^ { -1 } \left ( \dfrac { \blued { 35 } } { \maroonc { 65 } } \right ) \quad\small { \gray { \text { substitute values . } } } \\ m\angle l & amp ; \approx 28.30^\circ \quad\small { \gray { \text { evaluate with a calculator . } } } \end { align } $ now let 's try some practice problems .
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they take angles and give side ratios , but we need functions that take side ratios and give angles . we need inverse trig functions ! the inverse trigonometric functions we already know about inverse operations .
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how to use inverse trig functions in calculator ?
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let 's take a look at a new type of trigonometry problem . interestingly , these problems ca n't be solved with sine , cosine , or tangent . a problem : in the triangle below , what is the measure of angle $ l $ ? what we know : relative to $ \angle l $ , we know the lengths of the opposite and adjacent sides , so we can write : $ \tan ( l ) = \dfrac { \text { opposite } } { \text { adjacent } } = \dfrac { 35 } { 65 } $ but this does n't help us find the measure of $ \angle l $ . we 're stuck ! what we need : we need new mathematical tools to solve problems like these . our old friends sine , cosine , and tangent aren ’ t up to the task . they take angles and give side ratios , but we need functions that take side ratios and give angles . we need inverse trig functions ! the inverse trigonometric functions we already know about inverse operations . for example , addition and subtraction are inverse operations , and multiplication and division are inverse operations . each operation does the opposite of its inverse . the idea is the same in trigonometry . inverse trig functions do the opposite of the “ regular ” trig functions . for example : inverse sine $ ( \sin^ { -1 } ) $ does the opposite of the sine . inverse cosine $ ( \cos^ { -1 } ) $ does the opposite of the cosine . inverse tangent $ ( \tan^ { -1 } ) $ does the opposite of the tangent . in general , if you know the trig ratio but not the angle , you can use the corresponding inverse trig function to find the angle . this is expressed mathematically in the statements below . trigonometric functions input angles and output side ratios | |inverse trigonometric functions input side ratios and output angles : - : | : -| : - : $ \sin ( \theta ) =\dfrac { \text { opposite } } { \text { hypotenuse } } $ | $ \rightarrow $ | $ \sin^ { -1 } \left ( \dfrac { \text { opposite } } { \text { hypotenuse } } \right ) =\theta $ $ \cos ( \theta ) =\dfrac { \text { adjacent } } { \text { hypotenuse } } $ | $ \rightarrow $ | $ \cos^ { -1 } \left ( \dfrac { \text { adjacent } } { \text { hypotenuse } } \right ) =\theta $ $ \tan ( \theta ) =\dfrac { \text { opposite } } { \text { adjacent } } $ | $ \rightarrow $ | $ \tan^ { -1 } \left ( \dfrac { \text { opposite } } { \text { adjacent } } \right ) =\theta $ misconception alert ! the expression $ \sin^ { -1 } ( x ) $ is not the same as $ \dfrac { 1 } { \sin ( x ) } $ . in other words , the $ \small { -1 } $ is not an exponent . instead , it simply means inverse function . however , there is an alternate notation that avoids this pitfall ! the inverse sine can also be expressed as $ \arcsin $ , the inverse cosine as $ \arccos $ , and the inverse tangent as $ \arctan $ . this notation is common in computer programming languages , but not in mathematics . solving the introductory problem in the introductory problem , we were given the opposite and adjacent side lengths , so we can use inverse tangent to find the angle . $ \begin { align } { m\angle l } & amp ; =\tan^ { -1 } \left ( \dfrac { \text { } \blued { \text { opposite } } } { \text { } \maroonc { \text { adjacent } } \text { } } \right ) \quad\small { \gray { \text { define . } } } \\ m\angle l & amp ; =\tan^ { -1 } \left ( \dfrac { \blued { 35 } } { \maroonc { 65 } } \right ) \quad\small { \gray { \text { substitute values . } } } \\ m\angle l & amp ; \approx 28.30^\circ \quad\small { \gray { \text { evaluate with a calculator . } } } \end { align } $ now let 's try some practice problems .
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let 's take a look at a new type of trigonometry problem . interestingly , these problems ca n't be solved with sine , cosine , or tangent .
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what is the velocity of an airborne swallow ?
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let 's take a look at a new type of trigonometry problem . interestingly , these problems ca n't be solved with sine , cosine , or tangent . a problem : in the triangle below , what is the measure of angle $ l $ ? what we know : relative to $ \angle l $ , we know the lengths of the opposite and adjacent sides , so we can write : $ \tan ( l ) = \dfrac { \text { opposite } } { \text { adjacent } } = \dfrac { 35 } { 65 } $ but this does n't help us find the measure of $ \angle l $ . we 're stuck ! what we need : we need new mathematical tools to solve problems like these . our old friends sine , cosine , and tangent aren ’ t up to the task . they take angles and give side ratios , but we need functions that take side ratios and give angles . we need inverse trig functions ! the inverse trigonometric functions we already know about inverse operations . for example , addition and subtraction are inverse operations , and multiplication and division are inverse operations . each operation does the opposite of its inverse . the idea is the same in trigonometry . inverse trig functions do the opposite of the “ regular ” trig functions . for example : inverse sine $ ( \sin^ { -1 } ) $ does the opposite of the sine . inverse cosine $ ( \cos^ { -1 } ) $ does the opposite of the cosine . inverse tangent $ ( \tan^ { -1 } ) $ does the opposite of the tangent . in general , if you know the trig ratio but not the angle , you can use the corresponding inverse trig function to find the angle . this is expressed mathematically in the statements below . trigonometric functions input angles and output side ratios | |inverse trigonometric functions input side ratios and output angles : - : | : -| : - : $ \sin ( \theta ) =\dfrac { \text { opposite } } { \text { hypotenuse } } $ | $ \rightarrow $ | $ \sin^ { -1 } \left ( \dfrac { \text { opposite } } { \text { hypotenuse } } \right ) =\theta $ $ \cos ( \theta ) =\dfrac { \text { adjacent } } { \text { hypotenuse } } $ | $ \rightarrow $ | $ \cos^ { -1 } \left ( \dfrac { \text { adjacent } } { \text { hypotenuse } } \right ) =\theta $ $ \tan ( \theta ) =\dfrac { \text { opposite } } { \text { adjacent } } $ | $ \rightarrow $ | $ \tan^ { -1 } \left ( \dfrac { \text { opposite } } { \text { adjacent } } \right ) =\theta $ misconception alert ! the expression $ \sin^ { -1 } ( x ) $ is not the same as $ \dfrac { 1 } { \sin ( x ) } $ . in other words , the $ \small { -1 } $ is not an exponent . instead , it simply means inverse function . however , there is an alternate notation that avoids this pitfall ! the inverse sine can also be expressed as $ \arcsin $ , the inverse cosine as $ \arccos $ , and the inverse tangent as $ \arctan $ . this notation is common in computer programming languages , but not in mathematics . solving the introductory problem in the introductory problem , we were given the opposite and adjacent side lengths , so we can use inverse tangent to find the angle . $ \begin { align } { m\angle l } & amp ; =\tan^ { -1 } \left ( \dfrac { \text { } \blued { \text { opposite } } } { \text { } \maroonc { \text { adjacent } } \text { } } \right ) \quad\small { \gray { \text { define . } } } \\ m\angle l & amp ; =\tan^ { -1 } \left ( \dfrac { \blued { 35 } } { \maroonc { 65 } } \right ) \quad\small { \gray { \text { substitute values . } } } \\ m\angle l & amp ; \approx 28.30^\circ \quad\small { \gray { \text { evaluate with a calculator . } } } \end { align } $ now let 's try some practice problems .
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they take angles and give side ratios , but we need functions that take side ratios and give angles . we need inverse trig functions ! the inverse trigonometric functions we already know about inverse operations .
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what is the difference between a reciprocal trig function and an inverse trig function ?
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let 's take a look at a new type of trigonometry problem . interestingly , these problems ca n't be solved with sine , cosine , or tangent . a problem : in the triangle below , what is the measure of angle $ l $ ? what we know : relative to $ \angle l $ , we know the lengths of the opposite and adjacent sides , so we can write : $ \tan ( l ) = \dfrac { \text { opposite } } { \text { adjacent } } = \dfrac { 35 } { 65 } $ but this does n't help us find the measure of $ \angle l $ . we 're stuck ! what we need : we need new mathematical tools to solve problems like these . our old friends sine , cosine , and tangent aren ’ t up to the task . they take angles and give side ratios , but we need functions that take side ratios and give angles . we need inverse trig functions ! the inverse trigonometric functions we already know about inverse operations . for example , addition and subtraction are inverse operations , and multiplication and division are inverse operations . each operation does the opposite of its inverse . the idea is the same in trigonometry . inverse trig functions do the opposite of the “ regular ” trig functions . for example : inverse sine $ ( \sin^ { -1 } ) $ does the opposite of the sine . inverse cosine $ ( \cos^ { -1 } ) $ does the opposite of the cosine . inverse tangent $ ( \tan^ { -1 } ) $ does the opposite of the tangent . in general , if you know the trig ratio but not the angle , you can use the corresponding inverse trig function to find the angle . this is expressed mathematically in the statements below . trigonometric functions input angles and output side ratios | |inverse trigonometric functions input side ratios and output angles : - : | : -| : - : $ \sin ( \theta ) =\dfrac { \text { opposite } } { \text { hypotenuse } } $ | $ \rightarrow $ | $ \sin^ { -1 } \left ( \dfrac { \text { opposite } } { \text { hypotenuse } } \right ) =\theta $ $ \cos ( \theta ) =\dfrac { \text { adjacent } } { \text { hypotenuse } } $ | $ \rightarrow $ | $ \cos^ { -1 } \left ( \dfrac { \text { adjacent } } { \text { hypotenuse } } \right ) =\theta $ $ \tan ( \theta ) =\dfrac { \text { opposite } } { \text { adjacent } } $ | $ \rightarrow $ | $ \tan^ { -1 } \left ( \dfrac { \text { opposite } } { \text { adjacent } } \right ) =\theta $ misconception alert ! the expression $ \sin^ { -1 } ( x ) $ is not the same as $ \dfrac { 1 } { \sin ( x ) } $ . in other words , the $ \small { -1 } $ is not an exponent . instead , it simply means inverse function . however , there is an alternate notation that avoids this pitfall ! the inverse sine can also be expressed as $ \arcsin $ , the inverse cosine as $ \arccos $ , and the inverse tangent as $ \arctan $ . this notation is common in computer programming languages , but not in mathematics . solving the introductory problem in the introductory problem , we were given the opposite and adjacent side lengths , so we can use inverse tangent to find the angle . $ \begin { align } { m\angle l } & amp ; =\tan^ { -1 } \left ( \dfrac { \text { } \blued { \text { opposite } } } { \text { } \maroonc { \text { adjacent } } \text { } } \right ) \quad\small { \gray { \text { define . } } } \\ m\angle l & amp ; =\tan^ { -1 } \left ( \dfrac { \blued { 35 } } { \maroonc { 65 } } \right ) \quad\small { \gray { \text { substitute values . } } } \\ m\angle l & amp ; \approx 28.30^\circ \quad\small { \gray { \text { evaluate with a calculator . } } } \end { align } $ now let 's try some practice problems .
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$ \begin { align } { m\angle l } & amp ; =\tan^ { -1 } \left ( \dfrac { \text { } \blued { \text { opposite } } } { \text { } \maroonc { \text { adjacent } } \text { } } \right ) \quad\small { \gray { \text { define . } } } \\ m\angle l & amp ; =\tan^ { -1 } \left ( \dfrac { \blued { 35 } } { \maroonc { 65 } } \right ) \quad\small { \gray { \text { substitute values . } } } \\ m\angle l & amp ; \approx 28.30^\circ \quad\small { \gray { \text { evaluate with a calculator .
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why are concert tickets sold for $ 35 when they would sell out at $ 50 ?
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how can we solve 2-dimensional collision problems ? in other articles , we have looked at how momentum is conserved in collisions . we have also looked at how kinetic energy is transferred between bodies and converted into other forms of energy . we have applied these principles to simple problems , often in which the motion is constrained in one dimension . if two objects make a head on collision , they can bounce and move along the same direction they approached from ( i.e . only a single dimension ) . however , if two objects make a glancing collision , they 'll move off in two dimensions after the collision ( like a glancing collision between two billiard balls ) . for a collision where objects will be moving in 2 dimensions ( e.g . x and y ) , the momentum will be conserved in each direction independently ( as long as there 's no external impulse in that direction ) . in other words , the total momentum in the x direction will be the same before and after the collision . $ \large \sigma p_ { xi } =\sigma p_ { xf } $ also , the total momentum in the y direction will be the same before and after the collision . $ \large \sigma p_ { yi } =\sigma p_ { yf } $ in solving 2 dimensional collision problems , a good approach usually follows a general procedure : identify all the bodies in the system . assign clear symbols to each and draw a simple diagram if necessary . write down all the values you know and decide exactly what you need to find out to solve the problem . select a coordinate system . if many of the forces and velocities fall along a particular direction , it is advisable to use this direction as your x or y axis to simplify calculation ; even if it makes your axes not parallel to the page in your diagram . identify all the forces acting on each of the bodies in the system . make sure that all impulse is accounted for , or that you understand where external impulses can be neglected . remember that conservation of momentum only applies in cases where there is no external impulse . however , conservation of momentum can be applied separately to horizontal and vertical components . sometimes it is possible to neglect an external impulse if it is not in the direction of interest . write down equations which equate the momentum of the system before and after the collision . separate equations can be written down for momentum in the x and y directions . solve the resulting equations to determine an expression for the variable ( s ) you need . substitute in the numbers you know to find the final value . should this require adding vectors , it is often useful to do this graphically . a vector diagram can be drawn and the method of adding vectors head-to-tail used . trigonometry can then be used to find the magnitude and direction of all the vectors you need to know . billiard ball problem figure 1 describes the geometry of a collision of a white and a yellow billiard ball . the yellow ball is initially at rest . the white ball is played in the positive-x direction such that it collides with the yellow ball . the collision causes the yellow ball to move off towards the lower-right pocket at an angle of 28° from the x-axis . the mass of the yellow ball is 0.15 kg and the white ball is 0.18 kg . a sound recording reveals that the collision happens 0.25 s after the player has struck the white ball . the yellow ball falls into the pocket 0.35 s after the collision . exercise 1a : what is the velocity of the white ball after the collision ? exercise 1b : is the white ball likely to fall into any of the pockets ? if so , how could that be avoided ? exercise 1c : how much energy is lost to the environment in this collision ? bouncing baseball consider the situation in which a baseball is delivered toward a stationary wooden back-board by a pitching machine . the machine is set to deliver the 0.145 kg ball at 10 m/s . the ball hits the board , making an angle of 45° from the surface of the board . the board is slightly flexible and the collision is inelastic . the ball bounces back at an angle of 40° from the surface of the board as shown in figure 3 . hint : when a ball bounces off a surface , the impulse responsible for the bounce is always directed normal to the surface . exercise 2a : what is the velocity of the ball after the collision ? exercise 2b : if the ball is in contact with the wall for 0.5 ms , what is the magnitude of the force on the ball due to the wall ? exercise 2c : how much work was done on the wall by the ball ?
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billiard ball problem figure 1 describes the geometry of a collision of a white and a yellow billiard ball . the yellow ball is initially at rest . the white ball is played in the positive-x direction such that it collides with the yellow ball . the collision causes the yellow ball to move off towards the lower-right pocket at an angle of 28° from the x-axis .
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1a : why is the yellow ball at a 62 angle ?
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how can we solve 2-dimensional collision problems ? in other articles , we have looked at how momentum is conserved in collisions . we have also looked at how kinetic energy is transferred between bodies and converted into other forms of energy . we have applied these principles to simple problems , often in which the motion is constrained in one dimension . if two objects make a head on collision , they can bounce and move along the same direction they approached from ( i.e . only a single dimension ) . however , if two objects make a glancing collision , they 'll move off in two dimensions after the collision ( like a glancing collision between two billiard balls ) . for a collision where objects will be moving in 2 dimensions ( e.g . x and y ) , the momentum will be conserved in each direction independently ( as long as there 's no external impulse in that direction ) . in other words , the total momentum in the x direction will be the same before and after the collision . $ \large \sigma p_ { xi } =\sigma p_ { xf } $ also , the total momentum in the y direction will be the same before and after the collision . $ \large \sigma p_ { yi } =\sigma p_ { yf } $ in solving 2 dimensional collision problems , a good approach usually follows a general procedure : identify all the bodies in the system . assign clear symbols to each and draw a simple diagram if necessary . write down all the values you know and decide exactly what you need to find out to solve the problem . select a coordinate system . if many of the forces and velocities fall along a particular direction , it is advisable to use this direction as your x or y axis to simplify calculation ; even if it makes your axes not parallel to the page in your diagram . identify all the forces acting on each of the bodies in the system . make sure that all impulse is accounted for , or that you understand where external impulses can be neglected . remember that conservation of momentum only applies in cases where there is no external impulse . however , conservation of momentum can be applied separately to horizontal and vertical components . sometimes it is possible to neglect an external impulse if it is not in the direction of interest . write down equations which equate the momentum of the system before and after the collision . separate equations can be written down for momentum in the x and y directions . solve the resulting equations to determine an expression for the variable ( s ) you need . substitute in the numbers you know to find the final value . should this require adding vectors , it is often useful to do this graphically . a vector diagram can be drawn and the method of adding vectors head-to-tail used . trigonometry can then be used to find the magnitude and direction of all the vectors you need to know . billiard ball problem figure 1 describes the geometry of a collision of a white and a yellow billiard ball . the yellow ball is initially at rest . the white ball is played in the positive-x direction such that it collides with the yellow ball . the collision causes the yellow ball to move off towards the lower-right pocket at an angle of 28° from the x-axis . the mass of the yellow ball is 0.15 kg and the white ball is 0.18 kg . a sound recording reveals that the collision happens 0.25 s after the player has struck the white ball . the yellow ball falls into the pocket 0.35 s after the collision . exercise 1a : what is the velocity of the white ball after the collision ? exercise 1b : is the white ball likely to fall into any of the pockets ? if so , how could that be avoided ? exercise 1c : how much energy is lost to the environment in this collision ? bouncing baseball consider the situation in which a baseball is delivered toward a stationary wooden back-board by a pitching machine . the machine is set to deliver the 0.145 kg ball at 10 m/s . the ball hits the board , making an angle of 45° from the surface of the board . the board is slightly flexible and the collision is inelastic . the ball bounces back at an angle of 40° from the surface of the board as shown in figure 3 . hint : when a ball bounces off a surface , the impulse responsible for the bounce is always directed normal to the surface . exercise 2a : what is the velocity of the ball after the collision ? exercise 2b : if the ball is in contact with the wall for 0.5 ms , what is the magnitude of the force on the ball due to the wall ? exercise 2c : how much work was done on the wall by the ball ?
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in other articles , we have looked at how momentum is conserved in collisions . we have also looked at how kinetic energy is transferred between bodies and converted into other forms of energy . we have applied these principles to simple problems , often in which the motion is constrained in one dimension .
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in exercise 2c : should n't the work be equal to the final kinetic energy minus the initial kinetic energy in stead of otherwise ?
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how can we solve 2-dimensional collision problems ? in other articles , we have looked at how momentum is conserved in collisions . we have also looked at how kinetic energy is transferred between bodies and converted into other forms of energy . we have applied these principles to simple problems , often in which the motion is constrained in one dimension . if two objects make a head on collision , they can bounce and move along the same direction they approached from ( i.e . only a single dimension ) . however , if two objects make a glancing collision , they 'll move off in two dimensions after the collision ( like a glancing collision between two billiard balls ) . for a collision where objects will be moving in 2 dimensions ( e.g . x and y ) , the momentum will be conserved in each direction independently ( as long as there 's no external impulse in that direction ) . in other words , the total momentum in the x direction will be the same before and after the collision . $ \large \sigma p_ { xi } =\sigma p_ { xf } $ also , the total momentum in the y direction will be the same before and after the collision . $ \large \sigma p_ { yi } =\sigma p_ { yf } $ in solving 2 dimensional collision problems , a good approach usually follows a general procedure : identify all the bodies in the system . assign clear symbols to each and draw a simple diagram if necessary . write down all the values you know and decide exactly what you need to find out to solve the problem . select a coordinate system . if many of the forces and velocities fall along a particular direction , it is advisable to use this direction as your x or y axis to simplify calculation ; even if it makes your axes not parallel to the page in your diagram . identify all the forces acting on each of the bodies in the system . make sure that all impulse is accounted for , or that you understand where external impulses can be neglected . remember that conservation of momentum only applies in cases where there is no external impulse . however , conservation of momentum can be applied separately to horizontal and vertical components . sometimes it is possible to neglect an external impulse if it is not in the direction of interest . write down equations which equate the momentum of the system before and after the collision . separate equations can be written down for momentum in the x and y directions . solve the resulting equations to determine an expression for the variable ( s ) you need . substitute in the numbers you know to find the final value . should this require adding vectors , it is often useful to do this graphically . a vector diagram can be drawn and the method of adding vectors head-to-tail used . trigonometry can then be used to find the magnitude and direction of all the vectors you need to know . billiard ball problem figure 1 describes the geometry of a collision of a white and a yellow billiard ball . the yellow ball is initially at rest . the white ball is played in the positive-x direction such that it collides with the yellow ball . the collision causes the yellow ball to move off towards the lower-right pocket at an angle of 28° from the x-axis . the mass of the yellow ball is 0.15 kg and the white ball is 0.18 kg . a sound recording reveals that the collision happens 0.25 s after the player has struck the white ball . the yellow ball falls into the pocket 0.35 s after the collision . exercise 1a : what is the velocity of the white ball after the collision ? exercise 1b : is the white ball likely to fall into any of the pockets ? if so , how could that be avoided ? exercise 1c : how much energy is lost to the environment in this collision ? bouncing baseball consider the situation in which a baseball is delivered toward a stationary wooden back-board by a pitching machine . the machine is set to deliver the 0.145 kg ball at 10 m/s . the ball hits the board , making an angle of 45° from the surface of the board . the board is slightly flexible and the collision is inelastic . the ball bounces back at an angle of 40° from the surface of the board as shown in figure 3 . hint : when a ball bounces off a surface , the impulse responsible for the bounce is always directed normal to the surface . exercise 2a : what is the velocity of the ball after the collision ? exercise 2b : if the ball is in contact with the wall for 0.5 ms , what is the magnitude of the force on the ball due to the wall ? exercise 2c : how much work was done on the wall by the ball ?
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the collision causes the yellow ball to move off towards the lower-right pocket at an angle of 28° from the x-axis . the mass of the yellow ball is 0.15 kg and the white ball is 0.18 kg . a sound recording reveals that the collision happens 0.25 s after the player has struck the white ball .
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in the bouncing ball example , why was 1.45 used in the calculations instead of the given 0.145 kg in the problem ?
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how can we solve 2-dimensional collision problems ? in other articles , we have looked at how momentum is conserved in collisions . we have also looked at how kinetic energy is transferred between bodies and converted into other forms of energy . we have applied these principles to simple problems , often in which the motion is constrained in one dimension . if two objects make a head on collision , they can bounce and move along the same direction they approached from ( i.e . only a single dimension ) . however , if two objects make a glancing collision , they 'll move off in two dimensions after the collision ( like a glancing collision between two billiard balls ) . for a collision where objects will be moving in 2 dimensions ( e.g . x and y ) , the momentum will be conserved in each direction independently ( as long as there 's no external impulse in that direction ) . in other words , the total momentum in the x direction will be the same before and after the collision . $ \large \sigma p_ { xi } =\sigma p_ { xf } $ also , the total momentum in the y direction will be the same before and after the collision . $ \large \sigma p_ { yi } =\sigma p_ { yf } $ in solving 2 dimensional collision problems , a good approach usually follows a general procedure : identify all the bodies in the system . assign clear symbols to each and draw a simple diagram if necessary . write down all the values you know and decide exactly what you need to find out to solve the problem . select a coordinate system . if many of the forces and velocities fall along a particular direction , it is advisable to use this direction as your x or y axis to simplify calculation ; even if it makes your axes not parallel to the page in your diagram . identify all the forces acting on each of the bodies in the system . make sure that all impulse is accounted for , or that you understand where external impulses can be neglected . remember that conservation of momentum only applies in cases where there is no external impulse . however , conservation of momentum can be applied separately to horizontal and vertical components . sometimes it is possible to neglect an external impulse if it is not in the direction of interest . write down equations which equate the momentum of the system before and after the collision . separate equations can be written down for momentum in the x and y directions . solve the resulting equations to determine an expression for the variable ( s ) you need . substitute in the numbers you know to find the final value . should this require adding vectors , it is often useful to do this graphically . a vector diagram can be drawn and the method of adding vectors head-to-tail used . trigonometry can then be used to find the magnitude and direction of all the vectors you need to know . billiard ball problem figure 1 describes the geometry of a collision of a white and a yellow billiard ball . the yellow ball is initially at rest . the white ball is played in the positive-x direction such that it collides with the yellow ball . the collision causes the yellow ball to move off towards the lower-right pocket at an angle of 28° from the x-axis . the mass of the yellow ball is 0.15 kg and the white ball is 0.18 kg . a sound recording reveals that the collision happens 0.25 s after the player has struck the white ball . the yellow ball falls into the pocket 0.35 s after the collision . exercise 1a : what is the velocity of the white ball after the collision ? exercise 1b : is the white ball likely to fall into any of the pockets ? if so , how could that be avoided ? exercise 1c : how much energy is lost to the environment in this collision ? bouncing baseball consider the situation in which a baseball is delivered toward a stationary wooden back-board by a pitching machine . the machine is set to deliver the 0.145 kg ball at 10 m/s . the ball hits the board , making an angle of 45° from the surface of the board . the board is slightly flexible and the collision is inelastic . the ball bounces back at an angle of 40° from the surface of the board as shown in figure 3 . hint : when a ball bounces off a surface , the impulse responsible for the bounce is always directed normal to the surface . exercise 2a : what is the velocity of the ball after the collision ? exercise 2b : if the ball is in contact with the wall for 0.5 ms , what is the magnitude of the force on the ball due to the wall ? exercise 2c : how much work was done on the wall by the ball ?
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exercise 2b : if the ball is in contact with the wall for 0.5 ms , what is the magnitude of the force on the ball due to the wall ? exercise 2c : how much work was done on the wall by the ball ?
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regarding 2c , will not the collision result in heat that will be distributed to both the ball and the wall ?
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how can we solve 2-dimensional collision problems ? in other articles , we have looked at how momentum is conserved in collisions . we have also looked at how kinetic energy is transferred between bodies and converted into other forms of energy . we have applied these principles to simple problems , often in which the motion is constrained in one dimension . if two objects make a head on collision , they can bounce and move along the same direction they approached from ( i.e . only a single dimension ) . however , if two objects make a glancing collision , they 'll move off in two dimensions after the collision ( like a glancing collision between two billiard balls ) . for a collision where objects will be moving in 2 dimensions ( e.g . x and y ) , the momentum will be conserved in each direction independently ( as long as there 's no external impulse in that direction ) . in other words , the total momentum in the x direction will be the same before and after the collision . $ \large \sigma p_ { xi } =\sigma p_ { xf } $ also , the total momentum in the y direction will be the same before and after the collision . $ \large \sigma p_ { yi } =\sigma p_ { yf } $ in solving 2 dimensional collision problems , a good approach usually follows a general procedure : identify all the bodies in the system . assign clear symbols to each and draw a simple diagram if necessary . write down all the values you know and decide exactly what you need to find out to solve the problem . select a coordinate system . if many of the forces and velocities fall along a particular direction , it is advisable to use this direction as your x or y axis to simplify calculation ; even if it makes your axes not parallel to the page in your diagram . identify all the forces acting on each of the bodies in the system . make sure that all impulse is accounted for , or that you understand where external impulses can be neglected . remember that conservation of momentum only applies in cases where there is no external impulse . however , conservation of momentum can be applied separately to horizontal and vertical components . sometimes it is possible to neglect an external impulse if it is not in the direction of interest . write down equations which equate the momentum of the system before and after the collision . separate equations can be written down for momentum in the x and y directions . solve the resulting equations to determine an expression for the variable ( s ) you need . substitute in the numbers you know to find the final value . should this require adding vectors , it is often useful to do this graphically . a vector diagram can be drawn and the method of adding vectors head-to-tail used . trigonometry can then be used to find the magnitude and direction of all the vectors you need to know . billiard ball problem figure 1 describes the geometry of a collision of a white and a yellow billiard ball . the yellow ball is initially at rest . the white ball is played in the positive-x direction such that it collides with the yellow ball . the collision causes the yellow ball to move off towards the lower-right pocket at an angle of 28° from the x-axis . the mass of the yellow ball is 0.15 kg and the white ball is 0.18 kg . a sound recording reveals that the collision happens 0.25 s after the player has struck the white ball . the yellow ball falls into the pocket 0.35 s after the collision . exercise 1a : what is the velocity of the white ball after the collision ? exercise 1b : is the white ball likely to fall into any of the pockets ? if so , how could that be avoided ? exercise 1c : how much energy is lost to the environment in this collision ? bouncing baseball consider the situation in which a baseball is delivered toward a stationary wooden back-board by a pitching machine . the machine is set to deliver the 0.145 kg ball at 10 m/s . the ball hits the board , making an angle of 45° from the surface of the board . the board is slightly flexible and the collision is inelastic . the ball bounces back at an angle of 40° from the surface of the board as shown in figure 3 . hint : when a ball bounces off a surface , the impulse responsible for the bounce is always directed normal to the surface . exercise 2a : what is the velocity of the ball after the collision ? exercise 2b : if the ball is in contact with the wall for 0.5 ms , what is the magnitude of the force on the ball due to the wall ? exercise 2c : how much work was done on the wall by the ball ?
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exercise 2b : if the ball is in contact with the wall for 0.5 ms , what is the magnitude of the force on the ball due to the wall ? exercise 2c : how much work was done on the wall by the ball ?
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without knowing how much heat the ball accumulates , how can one calculate how much work went into the wall ?
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how can we solve 2-dimensional collision problems ? in other articles , we have looked at how momentum is conserved in collisions . we have also looked at how kinetic energy is transferred between bodies and converted into other forms of energy . we have applied these principles to simple problems , often in which the motion is constrained in one dimension . if two objects make a head on collision , they can bounce and move along the same direction they approached from ( i.e . only a single dimension ) . however , if two objects make a glancing collision , they 'll move off in two dimensions after the collision ( like a glancing collision between two billiard balls ) . for a collision where objects will be moving in 2 dimensions ( e.g . x and y ) , the momentum will be conserved in each direction independently ( as long as there 's no external impulse in that direction ) . in other words , the total momentum in the x direction will be the same before and after the collision . $ \large \sigma p_ { xi } =\sigma p_ { xf } $ also , the total momentum in the y direction will be the same before and after the collision . $ \large \sigma p_ { yi } =\sigma p_ { yf } $ in solving 2 dimensional collision problems , a good approach usually follows a general procedure : identify all the bodies in the system . assign clear symbols to each and draw a simple diagram if necessary . write down all the values you know and decide exactly what you need to find out to solve the problem . select a coordinate system . if many of the forces and velocities fall along a particular direction , it is advisable to use this direction as your x or y axis to simplify calculation ; even if it makes your axes not parallel to the page in your diagram . identify all the forces acting on each of the bodies in the system . make sure that all impulse is accounted for , or that you understand where external impulses can be neglected . remember that conservation of momentum only applies in cases where there is no external impulse . however , conservation of momentum can be applied separately to horizontal and vertical components . sometimes it is possible to neglect an external impulse if it is not in the direction of interest . write down equations which equate the momentum of the system before and after the collision . separate equations can be written down for momentum in the x and y directions . solve the resulting equations to determine an expression for the variable ( s ) you need . substitute in the numbers you know to find the final value . should this require adding vectors , it is often useful to do this graphically . a vector diagram can be drawn and the method of adding vectors head-to-tail used . trigonometry can then be used to find the magnitude and direction of all the vectors you need to know . billiard ball problem figure 1 describes the geometry of a collision of a white and a yellow billiard ball . the yellow ball is initially at rest . the white ball is played in the positive-x direction such that it collides with the yellow ball . the collision causes the yellow ball to move off towards the lower-right pocket at an angle of 28° from the x-axis . the mass of the yellow ball is 0.15 kg and the white ball is 0.18 kg . a sound recording reveals that the collision happens 0.25 s after the player has struck the white ball . the yellow ball falls into the pocket 0.35 s after the collision . exercise 1a : what is the velocity of the white ball after the collision ? exercise 1b : is the white ball likely to fall into any of the pockets ? if so , how could that be avoided ? exercise 1c : how much energy is lost to the environment in this collision ? bouncing baseball consider the situation in which a baseball is delivered toward a stationary wooden back-board by a pitching machine . the machine is set to deliver the 0.145 kg ball at 10 m/s . the ball hits the board , making an angle of 45° from the surface of the board . the board is slightly flexible and the collision is inelastic . the ball bounces back at an angle of 40° from the surface of the board as shown in figure 3 . hint : when a ball bounces off a surface , the impulse responsible for the bounce is always directed normal to the surface . exercise 2a : what is the velocity of the ball after the collision ? exercise 2b : if the ball is in contact with the wall for 0.5 ms , what is the magnitude of the force on the ball due to the wall ? exercise 2c : how much work was done on the wall by the ball ?
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the board is slightly flexible and the collision is inelastic . the ball bounces back at an angle of 40° from the surface of the board as shown in figure 3 . hint : when a ball bounces off a surface , the impulse responsible for the bounce is always directed normal to the surface .
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is n't the 40 degrees outside of the triangle , not inside as shown in the solution ?
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how can we solve 2-dimensional collision problems ? in other articles , we have looked at how momentum is conserved in collisions . we have also looked at how kinetic energy is transferred between bodies and converted into other forms of energy . we have applied these principles to simple problems , often in which the motion is constrained in one dimension . if two objects make a head on collision , they can bounce and move along the same direction they approached from ( i.e . only a single dimension ) . however , if two objects make a glancing collision , they 'll move off in two dimensions after the collision ( like a glancing collision between two billiard balls ) . for a collision where objects will be moving in 2 dimensions ( e.g . x and y ) , the momentum will be conserved in each direction independently ( as long as there 's no external impulse in that direction ) . in other words , the total momentum in the x direction will be the same before and after the collision . $ \large \sigma p_ { xi } =\sigma p_ { xf } $ also , the total momentum in the y direction will be the same before and after the collision . $ \large \sigma p_ { yi } =\sigma p_ { yf } $ in solving 2 dimensional collision problems , a good approach usually follows a general procedure : identify all the bodies in the system . assign clear symbols to each and draw a simple diagram if necessary . write down all the values you know and decide exactly what you need to find out to solve the problem . select a coordinate system . if many of the forces and velocities fall along a particular direction , it is advisable to use this direction as your x or y axis to simplify calculation ; even if it makes your axes not parallel to the page in your diagram . identify all the forces acting on each of the bodies in the system . make sure that all impulse is accounted for , or that you understand where external impulses can be neglected . remember that conservation of momentum only applies in cases where there is no external impulse . however , conservation of momentum can be applied separately to horizontal and vertical components . sometimes it is possible to neglect an external impulse if it is not in the direction of interest . write down equations which equate the momentum of the system before and after the collision . separate equations can be written down for momentum in the x and y directions . solve the resulting equations to determine an expression for the variable ( s ) you need . substitute in the numbers you know to find the final value . should this require adding vectors , it is often useful to do this graphically . a vector diagram can be drawn and the method of adding vectors head-to-tail used . trigonometry can then be used to find the magnitude and direction of all the vectors you need to know . billiard ball problem figure 1 describes the geometry of a collision of a white and a yellow billiard ball . the yellow ball is initially at rest . the white ball is played in the positive-x direction such that it collides with the yellow ball . the collision causes the yellow ball to move off towards the lower-right pocket at an angle of 28° from the x-axis . the mass of the yellow ball is 0.15 kg and the white ball is 0.18 kg . a sound recording reveals that the collision happens 0.25 s after the player has struck the white ball . the yellow ball falls into the pocket 0.35 s after the collision . exercise 1a : what is the velocity of the white ball after the collision ? exercise 1b : is the white ball likely to fall into any of the pockets ? if so , how could that be avoided ? exercise 1c : how much energy is lost to the environment in this collision ? bouncing baseball consider the situation in which a baseball is delivered toward a stationary wooden back-board by a pitching machine . the machine is set to deliver the 0.145 kg ball at 10 m/s . the ball hits the board , making an angle of 45° from the surface of the board . the board is slightly flexible and the collision is inelastic . the ball bounces back at an angle of 40° from the surface of the board as shown in figure 3 . hint : when a ball bounces off a surface , the impulse responsible for the bounce is always directed normal to the surface . exercise 2a : what is the velocity of the ball after the collision ? exercise 2b : if the ball is in contact with the wall for 0.5 ms , what is the magnitude of the force on the ball due to the wall ? exercise 2c : how much work was done on the wall by the ball ?
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trigonometry can then be used to find the magnitude and direction of all the vectors you need to know . billiard ball problem figure 1 describes the geometry of a collision of a white and a yellow billiard ball . the yellow ball is initially at rest .
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would n't it be 1.025 / cos50 or 1.025 / sin40 , rather than 1.025 / cos40 ?
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how can we solve 2-dimensional collision problems ? in other articles , we have looked at how momentum is conserved in collisions . we have also looked at how kinetic energy is transferred between bodies and converted into other forms of energy . we have applied these principles to simple problems , often in which the motion is constrained in one dimension . if two objects make a head on collision , they can bounce and move along the same direction they approached from ( i.e . only a single dimension ) . however , if two objects make a glancing collision , they 'll move off in two dimensions after the collision ( like a glancing collision between two billiard balls ) . for a collision where objects will be moving in 2 dimensions ( e.g . x and y ) , the momentum will be conserved in each direction independently ( as long as there 's no external impulse in that direction ) . in other words , the total momentum in the x direction will be the same before and after the collision . $ \large \sigma p_ { xi } =\sigma p_ { xf } $ also , the total momentum in the y direction will be the same before and after the collision . $ \large \sigma p_ { yi } =\sigma p_ { yf } $ in solving 2 dimensional collision problems , a good approach usually follows a general procedure : identify all the bodies in the system . assign clear symbols to each and draw a simple diagram if necessary . write down all the values you know and decide exactly what you need to find out to solve the problem . select a coordinate system . if many of the forces and velocities fall along a particular direction , it is advisable to use this direction as your x or y axis to simplify calculation ; even if it makes your axes not parallel to the page in your diagram . identify all the forces acting on each of the bodies in the system . make sure that all impulse is accounted for , or that you understand where external impulses can be neglected . remember that conservation of momentum only applies in cases where there is no external impulse . however , conservation of momentum can be applied separately to horizontal and vertical components . sometimes it is possible to neglect an external impulse if it is not in the direction of interest . write down equations which equate the momentum of the system before and after the collision . separate equations can be written down for momentum in the x and y directions . solve the resulting equations to determine an expression for the variable ( s ) you need . substitute in the numbers you know to find the final value . should this require adding vectors , it is often useful to do this graphically . a vector diagram can be drawn and the method of adding vectors head-to-tail used . trigonometry can then be used to find the magnitude and direction of all the vectors you need to know . billiard ball problem figure 1 describes the geometry of a collision of a white and a yellow billiard ball . the yellow ball is initially at rest . the white ball is played in the positive-x direction such that it collides with the yellow ball . the collision causes the yellow ball to move off towards the lower-right pocket at an angle of 28° from the x-axis . the mass of the yellow ball is 0.15 kg and the white ball is 0.18 kg . a sound recording reveals that the collision happens 0.25 s after the player has struck the white ball . the yellow ball falls into the pocket 0.35 s after the collision . exercise 1a : what is the velocity of the white ball after the collision ? exercise 1b : is the white ball likely to fall into any of the pockets ? if so , how could that be avoided ? exercise 1c : how much energy is lost to the environment in this collision ? bouncing baseball consider the situation in which a baseball is delivered toward a stationary wooden back-board by a pitching machine . the machine is set to deliver the 0.145 kg ball at 10 m/s . the ball hits the board , making an angle of 45° from the surface of the board . the board is slightly flexible and the collision is inelastic . the ball bounces back at an angle of 40° from the surface of the board as shown in figure 3 . hint : when a ball bounces off a surface , the impulse responsible for the bounce is always directed normal to the surface . exercise 2a : what is the velocity of the ball after the collision ? exercise 2b : if the ball is in contact with the wall for 0.5 ms , what is the magnitude of the force on the ball due to the wall ? exercise 2c : how much work was done on the wall by the ball ?
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for a collision where objects will be moving in 2 dimensions ( e.g . x and y ) , the momentum will be conserved in each direction independently ( as long as there 's no external impulse in that direction ) . in other words , the total momentum in the x direction will be the same before and after the collision .
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for questions 2b and 2c , do we only use the velocity ( and momentum ) in the x direction for our calculations because the y direction was affected by the momentum lost to the wall in the inelastic collision ?
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how can we solve 2-dimensional collision problems ? in other articles , we have looked at how momentum is conserved in collisions . we have also looked at how kinetic energy is transferred between bodies and converted into other forms of energy . we have applied these principles to simple problems , often in which the motion is constrained in one dimension . if two objects make a head on collision , they can bounce and move along the same direction they approached from ( i.e . only a single dimension ) . however , if two objects make a glancing collision , they 'll move off in two dimensions after the collision ( like a glancing collision between two billiard balls ) . for a collision where objects will be moving in 2 dimensions ( e.g . x and y ) , the momentum will be conserved in each direction independently ( as long as there 's no external impulse in that direction ) . in other words , the total momentum in the x direction will be the same before and after the collision . $ \large \sigma p_ { xi } =\sigma p_ { xf } $ also , the total momentum in the y direction will be the same before and after the collision . $ \large \sigma p_ { yi } =\sigma p_ { yf } $ in solving 2 dimensional collision problems , a good approach usually follows a general procedure : identify all the bodies in the system . assign clear symbols to each and draw a simple diagram if necessary . write down all the values you know and decide exactly what you need to find out to solve the problem . select a coordinate system . if many of the forces and velocities fall along a particular direction , it is advisable to use this direction as your x or y axis to simplify calculation ; even if it makes your axes not parallel to the page in your diagram . identify all the forces acting on each of the bodies in the system . make sure that all impulse is accounted for , or that you understand where external impulses can be neglected . remember that conservation of momentum only applies in cases where there is no external impulse . however , conservation of momentum can be applied separately to horizontal and vertical components . sometimes it is possible to neglect an external impulse if it is not in the direction of interest . write down equations which equate the momentum of the system before and after the collision . separate equations can be written down for momentum in the x and y directions . solve the resulting equations to determine an expression for the variable ( s ) you need . substitute in the numbers you know to find the final value . should this require adding vectors , it is often useful to do this graphically . a vector diagram can be drawn and the method of adding vectors head-to-tail used . trigonometry can then be used to find the magnitude and direction of all the vectors you need to know . billiard ball problem figure 1 describes the geometry of a collision of a white and a yellow billiard ball . the yellow ball is initially at rest . the white ball is played in the positive-x direction such that it collides with the yellow ball . the collision causes the yellow ball to move off towards the lower-right pocket at an angle of 28° from the x-axis . the mass of the yellow ball is 0.15 kg and the white ball is 0.18 kg . a sound recording reveals that the collision happens 0.25 s after the player has struck the white ball . the yellow ball falls into the pocket 0.35 s after the collision . exercise 1a : what is the velocity of the white ball after the collision ? exercise 1b : is the white ball likely to fall into any of the pockets ? if so , how could that be avoided ? exercise 1c : how much energy is lost to the environment in this collision ? bouncing baseball consider the situation in which a baseball is delivered toward a stationary wooden back-board by a pitching machine . the machine is set to deliver the 0.145 kg ball at 10 m/s . the ball hits the board , making an angle of 45° from the surface of the board . the board is slightly flexible and the collision is inelastic . the ball bounces back at an angle of 40° from the surface of the board as shown in figure 3 . hint : when a ball bounces off a surface , the impulse responsible for the bounce is always directed normal to the surface . exercise 2a : what is the velocity of the ball after the collision ? exercise 2b : if the ball is in contact with the wall for 0.5 ms , what is the magnitude of the force on the ball due to the wall ? exercise 2c : how much work was done on the wall by the ball ?
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write down equations which equate the momentum of the system before and after the collision . separate equations can be written down for momentum in the x and y directions . solve the resulting equations to determine an expression for the variable ( s ) you need .
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if this were an elastic collision and no momentum were lost in the y direction , would we do this using both the x and y directions ?
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this content is provided by the 49ers museum education program . physics is the study of matter and its motion through time and space , as well as its interaction with energy and the forces created by this interaction . so , what is a force ? a force is a push or a pull exerted on one object from another . forces make things move . you can make something start or stop when you push or pull an object . there are many different types of forces in action in football . a player kicking a football is a force that makes the football fly through the air . a quarterback throwing a football is another example of a force that makes the football fly in a game . when studying the concept of force , we can look to history to find mathematical principles that guide the laws of motion . sir isaac newton was one of the most famous scientists of the 17th century to study the laws of forces and motion . through careful study of how objects react to various forces , newton developed the three laws of motion . below are explanations of each law and how these laws can be applied to football . newton 's first law of motion : the law of inertia now , imagine a football sitting on levi ’ s® stadium ’ s 50 yard line . what do you expect to happen to the football ? based on your experience , you would probably expect the ball to just sit there unless someone picks it up or kicks it . the law of inertia tells us that the football will remain at rest unless someone or something moves it by a specific force . the law of inertia also states that an object ( like a football ) in motion remains in motion unless acted on by an external force . inertia is defined as the force that keeps an object at rest or in motion . the law of inertia also tells us that an object at rest , like the football at the 50 yard line , will remain at rest unless another force acts upon it . if our quarterback picks up the ball and throws it , it will fly in the direction he threw it , at a specific speed based upon how much force he applies in his throw . once the ball leaves the quarterback 's hands , the first law tells us that if there are no other forces on the ball , the ball would continue to travel in the same direction and with the same speed until other forces affect its flight . based on our real life experience throwing things , we know that the ball does not continue to travel in the air with the same velocity indefinitely . that is because there are other forces that are acting on the ball , including gravity . newton 's second law of motion : force and acceleration as a football player , you might want to throw or kick the ball as far as possible , or you might want to throw it high enough that the player on the other team ca n't catch it . we can use newton ’ s second law of motion to figure out the best angle to apply the force on the ball so that it does what the football player wants it to . in fact , the football player has probably been doing physics in their head without even realizing it ! the second law of motion states that force on an object is equal to the mass of the object multiplied by its acceleration . if we apply this law to a football , it tells us that the amount that the ball accelerates depends on the force applied by the quarterback and the mass of the ball . $ \text { force } = \text { mass } \times \text { acceleration } $ we use the word mass to talk about how much matter there is in something . matter is anything that takes up space . acceleration is the name we give to any process where the speed and direction of an object changes . there are only two ways for you to accelerate : change your speed ( speed up or slow down ) or change your direction—or change both . to understand this physics law , let ’ s see it in practice through another football example . when a quarterback throws the ball , the player is applying a force on the object . this causes the football to accelerate . we know the football accelerates because it starts from a resting point in the quarterback ’ s hand and then speeds up after the appropriate force is applied to the football to reach the targeted receivers on the field . the second law also tells us that acceleration of the football also depends on the mass of the football . if the player is throwing an extra heavy football , the acceleration of the football would decrease compared to a lighter football that is thrown with the same force . newton 's third law of motion the third law of motion states that for every force applied there is an equal and opposite reaction force . an illustration of this might be when a player is trying to catch a football from a very high kick . the football , coming down from above , exerts a force ( a push ) on the player as he catches it . the player then exerts a force that is equal in magnitude ( the size of a force ) and opposite in direction . this slows down the ball so the player can catch the football and bring it to rest . another example of newton ’ s third law can be witnessed in the 49ers locker room . picture a football helmet sitting on top of a player ’ s locker . even at rest , the force of gravity is pulling on the helmet . the locker is between the helmet and the locker room floor . based on the third law , we know that the locker must be pushing up on the helmet with a force equal in magnitude . this keeps the helmet from falling due to the force of gravity .
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physics is the study of matter and its motion through time and space , as well as its interaction with energy and the forces created by this interaction . so , what is a force ? a force is a push or a pull exerted on one object from another .
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what is the force of 200pounds of pressure in your chest ?
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this content is provided by the 49ers museum education program . physics is the study of matter and its motion through time and space , as well as its interaction with energy and the forces created by this interaction . so , what is a force ? a force is a push or a pull exerted on one object from another . forces make things move . you can make something start or stop when you push or pull an object . there are many different types of forces in action in football . a player kicking a football is a force that makes the football fly through the air . a quarterback throwing a football is another example of a force that makes the football fly in a game . when studying the concept of force , we can look to history to find mathematical principles that guide the laws of motion . sir isaac newton was one of the most famous scientists of the 17th century to study the laws of forces and motion . through careful study of how objects react to various forces , newton developed the three laws of motion . below are explanations of each law and how these laws can be applied to football . newton 's first law of motion : the law of inertia now , imagine a football sitting on levi ’ s® stadium ’ s 50 yard line . what do you expect to happen to the football ? based on your experience , you would probably expect the ball to just sit there unless someone picks it up or kicks it . the law of inertia tells us that the football will remain at rest unless someone or something moves it by a specific force . the law of inertia also states that an object ( like a football ) in motion remains in motion unless acted on by an external force . inertia is defined as the force that keeps an object at rest or in motion . the law of inertia also tells us that an object at rest , like the football at the 50 yard line , will remain at rest unless another force acts upon it . if our quarterback picks up the ball and throws it , it will fly in the direction he threw it , at a specific speed based upon how much force he applies in his throw . once the ball leaves the quarterback 's hands , the first law tells us that if there are no other forces on the ball , the ball would continue to travel in the same direction and with the same speed until other forces affect its flight . based on our real life experience throwing things , we know that the ball does not continue to travel in the air with the same velocity indefinitely . that is because there are other forces that are acting on the ball , including gravity . newton 's second law of motion : force and acceleration as a football player , you might want to throw or kick the ball as far as possible , or you might want to throw it high enough that the player on the other team ca n't catch it . we can use newton ’ s second law of motion to figure out the best angle to apply the force on the ball so that it does what the football player wants it to . in fact , the football player has probably been doing physics in their head without even realizing it ! the second law of motion states that force on an object is equal to the mass of the object multiplied by its acceleration . if we apply this law to a football , it tells us that the amount that the ball accelerates depends on the force applied by the quarterback and the mass of the ball . $ \text { force } = \text { mass } \times \text { acceleration } $ we use the word mass to talk about how much matter there is in something . matter is anything that takes up space . acceleration is the name we give to any process where the speed and direction of an object changes . there are only two ways for you to accelerate : change your speed ( speed up or slow down ) or change your direction—or change both . to understand this physics law , let ’ s see it in practice through another football example . when a quarterback throws the ball , the player is applying a force on the object . this causes the football to accelerate . we know the football accelerates because it starts from a resting point in the quarterback ’ s hand and then speeds up after the appropriate force is applied to the football to reach the targeted receivers on the field . the second law also tells us that acceleration of the football also depends on the mass of the football . if the player is throwing an extra heavy football , the acceleration of the football would decrease compared to a lighter football that is thrown with the same force . newton 's third law of motion the third law of motion states that for every force applied there is an equal and opposite reaction force . an illustration of this might be when a player is trying to catch a football from a very high kick . the football , coming down from above , exerts a force ( a push ) on the player as he catches it . the player then exerts a force that is equal in magnitude ( the size of a force ) and opposite in direction . this slows down the ball so the player can catch the football and bring it to rest . another example of newton ’ s third law can be witnessed in the 49ers locker room . picture a football helmet sitting on top of a player ’ s locker . even at rest , the force of gravity is pulling on the helmet . the locker is between the helmet and the locker room floor . based on the third law , we know that the locker must be pushing up on the helmet with a force equal in magnitude . this keeps the helmet from falling due to the force of gravity .
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based on the third law , we know that the locker must be pushing up on the helmet with a force equal in magnitude . this keeps the helmet from falling due to the force of gravity .
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this keeps the helmet from falling due to the force of gravity '' , i already know that , so why would they put that there ?
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this content is provided by the 49ers museum education program . physics is the study of matter and its motion through time and space , as well as its interaction with energy and the forces created by this interaction . so , what is a force ? a force is a push or a pull exerted on one object from another . forces make things move . you can make something start or stop when you push or pull an object . there are many different types of forces in action in football . a player kicking a football is a force that makes the football fly through the air . a quarterback throwing a football is another example of a force that makes the football fly in a game . when studying the concept of force , we can look to history to find mathematical principles that guide the laws of motion . sir isaac newton was one of the most famous scientists of the 17th century to study the laws of forces and motion . through careful study of how objects react to various forces , newton developed the three laws of motion . below are explanations of each law and how these laws can be applied to football . newton 's first law of motion : the law of inertia now , imagine a football sitting on levi ’ s® stadium ’ s 50 yard line . what do you expect to happen to the football ? based on your experience , you would probably expect the ball to just sit there unless someone picks it up or kicks it . the law of inertia tells us that the football will remain at rest unless someone or something moves it by a specific force . the law of inertia also states that an object ( like a football ) in motion remains in motion unless acted on by an external force . inertia is defined as the force that keeps an object at rest or in motion . the law of inertia also tells us that an object at rest , like the football at the 50 yard line , will remain at rest unless another force acts upon it . if our quarterback picks up the ball and throws it , it will fly in the direction he threw it , at a specific speed based upon how much force he applies in his throw . once the ball leaves the quarterback 's hands , the first law tells us that if there are no other forces on the ball , the ball would continue to travel in the same direction and with the same speed until other forces affect its flight . based on our real life experience throwing things , we know that the ball does not continue to travel in the air with the same velocity indefinitely . that is because there are other forces that are acting on the ball , including gravity . newton 's second law of motion : force and acceleration as a football player , you might want to throw or kick the ball as far as possible , or you might want to throw it high enough that the player on the other team ca n't catch it . we can use newton ’ s second law of motion to figure out the best angle to apply the force on the ball so that it does what the football player wants it to . in fact , the football player has probably been doing physics in their head without even realizing it ! the second law of motion states that force on an object is equal to the mass of the object multiplied by its acceleration . if we apply this law to a football , it tells us that the amount that the ball accelerates depends on the force applied by the quarterback and the mass of the ball . $ \text { force } = \text { mass } \times \text { acceleration } $ we use the word mass to talk about how much matter there is in something . matter is anything that takes up space . acceleration is the name we give to any process where the speed and direction of an object changes . there are only two ways for you to accelerate : change your speed ( speed up or slow down ) or change your direction—or change both . to understand this physics law , let ’ s see it in practice through another football example . when a quarterback throws the ball , the player is applying a force on the object . this causes the football to accelerate . we know the football accelerates because it starts from a resting point in the quarterback ’ s hand and then speeds up after the appropriate force is applied to the football to reach the targeted receivers on the field . the second law also tells us that acceleration of the football also depends on the mass of the football . if the player is throwing an extra heavy football , the acceleration of the football would decrease compared to a lighter football that is thrown with the same force . newton 's third law of motion the third law of motion states that for every force applied there is an equal and opposite reaction force . an illustration of this might be when a player is trying to catch a football from a very high kick . the football , coming down from above , exerts a force ( a push ) on the player as he catches it . the player then exerts a force that is equal in magnitude ( the size of a force ) and opposite in direction . this slows down the ball so the player can catch the football and bring it to rest . another example of newton ’ s third law can be witnessed in the 49ers locker room . picture a football helmet sitting on top of a player ’ s locker . even at rest , the force of gravity is pulling on the helmet . the locker is between the helmet and the locker room floor . based on the third law , we know that the locker must be pushing up on the helmet with a force equal in magnitude . this keeps the helmet from falling due to the force of gravity .
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based on the third law , we know that the locker must be pushing up on the helmet with a force equal in magnitude . this keeps the helmet from falling due to the force of gravity .
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how can the helmet not be falling just because if the force of gravity in the last paragraph ?
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this content is provided by the 49ers museum education program . physics is the study of matter and its motion through time and space , as well as its interaction with energy and the forces created by this interaction . so , what is a force ? a force is a push or a pull exerted on one object from another . forces make things move . you can make something start or stop when you push or pull an object . there are many different types of forces in action in football . a player kicking a football is a force that makes the football fly through the air . a quarterback throwing a football is another example of a force that makes the football fly in a game . when studying the concept of force , we can look to history to find mathematical principles that guide the laws of motion . sir isaac newton was one of the most famous scientists of the 17th century to study the laws of forces and motion . through careful study of how objects react to various forces , newton developed the three laws of motion . below are explanations of each law and how these laws can be applied to football . newton 's first law of motion : the law of inertia now , imagine a football sitting on levi ’ s® stadium ’ s 50 yard line . what do you expect to happen to the football ? based on your experience , you would probably expect the ball to just sit there unless someone picks it up or kicks it . the law of inertia tells us that the football will remain at rest unless someone or something moves it by a specific force . the law of inertia also states that an object ( like a football ) in motion remains in motion unless acted on by an external force . inertia is defined as the force that keeps an object at rest or in motion . the law of inertia also tells us that an object at rest , like the football at the 50 yard line , will remain at rest unless another force acts upon it . if our quarterback picks up the ball and throws it , it will fly in the direction he threw it , at a specific speed based upon how much force he applies in his throw . once the ball leaves the quarterback 's hands , the first law tells us that if there are no other forces on the ball , the ball would continue to travel in the same direction and with the same speed until other forces affect its flight . based on our real life experience throwing things , we know that the ball does not continue to travel in the air with the same velocity indefinitely . that is because there are other forces that are acting on the ball , including gravity . newton 's second law of motion : force and acceleration as a football player , you might want to throw or kick the ball as far as possible , or you might want to throw it high enough that the player on the other team ca n't catch it . we can use newton ’ s second law of motion to figure out the best angle to apply the force on the ball so that it does what the football player wants it to . in fact , the football player has probably been doing physics in their head without even realizing it ! the second law of motion states that force on an object is equal to the mass of the object multiplied by its acceleration . if we apply this law to a football , it tells us that the amount that the ball accelerates depends on the force applied by the quarterback and the mass of the ball . $ \text { force } = \text { mass } \times \text { acceleration } $ we use the word mass to talk about how much matter there is in something . matter is anything that takes up space . acceleration is the name we give to any process where the speed and direction of an object changes . there are only two ways for you to accelerate : change your speed ( speed up or slow down ) or change your direction—or change both . to understand this physics law , let ’ s see it in practice through another football example . when a quarterback throws the ball , the player is applying a force on the object . this causes the football to accelerate . we know the football accelerates because it starts from a resting point in the quarterback ’ s hand and then speeds up after the appropriate force is applied to the football to reach the targeted receivers on the field . the second law also tells us that acceleration of the football also depends on the mass of the football . if the player is throwing an extra heavy football , the acceleration of the football would decrease compared to a lighter football that is thrown with the same force . newton 's third law of motion the third law of motion states that for every force applied there is an equal and opposite reaction force . an illustration of this might be when a player is trying to catch a football from a very high kick . the football , coming down from above , exerts a force ( a push ) on the player as he catches it . the player then exerts a force that is equal in magnitude ( the size of a force ) and opposite in direction . this slows down the ball so the player can catch the football and bring it to rest . another example of newton ’ s third law can be witnessed in the 49ers locker room . picture a football helmet sitting on top of a player ’ s locker . even at rest , the force of gravity is pulling on the helmet . the locker is between the helmet and the locker room floor . based on the third law , we know that the locker must be pushing up on the helmet with a force equal in magnitude . this keeps the helmet from falling due to the force of gravity .
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we know the football accelerates because it starts from a resting point in the quarterback ’ s hand and then speeds up after the appropriate force is applied to the football to reach the targeted receivers on the field . the second law also tells us that acceleration of the football also depends on the mass of the football . if the player is throwing an extra heavy football , the acceleration of the football would decrease compared to a lighter football that is thrown with the same force .
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so if you 're a football kicker is it more efficient to work on achieving a faster acceleration or building towards heavier mass ?
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what does energy and work mean ? energy is a word which tends to be used a lot in everyday life . though it is often used quite loosely , it does have a very specific physical meaning . energy is a measurement of the ability of something to do work . it is not a material substance . energy can be stored and measured in many forms . although we often hear people talking about energy consumption , energy is never really destroyed . it is just transferred from one form to another , doing work in the process . some forms of energy are less useful to us than others—for example , low level heat energy . it is better to talk about the consumption or extraction of energy resources , for example coal , oil , or wind , than consumption of energy itself . a speeding bullet has a measurable amount of energy associated with it ; this is known as kinetic energy . the bullet gained this energy because work was done on it by a charge of gunpowder which lost some chemical potential energy in the process . a hot cup of coffee has a measurable amount of thermal energy which it acquired via work done by a microwave oven , which in turn took electrical energy from the electrical grid . in practice , whenever work is done to move energy from one form to another , there is always some loss to other forms of energy such as heat and sound . for example , a traditional light bulb is only about 3 % efficient at converting electrical energy to visible light , while a human being is about 25 % efficient at converting chemical energy from food into work . how do we measure energy and work ? the standard unit used to measure energy and work done in physics is the joule , which has the symbol j . in mechanics , 1 joule is the energy transferred when a force of 1 newton is applied to an object and moves it through a distance of 1 meter . another unit of energy you may have come across is the calorie . the amount of energy in an item of food is often written in calories on the back of the packet . a typical 60 gram chocolate bar for example contains about 280 calories of energy . one calorie is the amount of energy required to raise 1 kg of water by 1 $ ^ { \circ } $ celsius . this is equal to 4184 joules , so one chocolate bar has 1.17 million joules or 1.17 mj of stored energy . that 's a lot of joules ! how long do i have to push a heavy box around to burn off one chocolate bar ? suppose we 're feeling guilty about eating a chocolate bar ; we want to find how much exercise we need to do to offset those extra 280 calories . let 's consider a simple form of exercise : pushing a heavy box around a room , see figure 1 below . using a bathroom scale between ourselves and the box , we find that we can push with a force of 500 n. meanwhile , we use a stopwatch and measuring tape to measure our speed . this comes out to be 0.25 meters per second . so how much work do we need to do to the box to burn off the candy bar ? the definition of work , $ w $ , is below : $ \large w = f\cdot \delta x $ the work we need to do to burn the energy in the candy bar is $ e=280 \mathrm { cal } \cdot 4184 \mathrm { j/cal } =1.17 \mathrm { mj } $ . therefore , the distance , $ δx $ , we need to move the box through is : $ \begin { align } w & amp ; = f\cdot \delta x \ 1.17 \text { mj } & amp ; = ( 500 \text { n } ) \cdot \delta x \ \dfrac { 1.17\times 10^6 \text { j } } { 500 \text { n } } & amp ; = \delta x \ 2,340 \text { m } & amp ; = \delta x \end { align } $ remember , however , that our bodies are about 25 % efficient at transferring stored energy from food into work . the actual energy we will offset is then four times as high as the work done to the box . so , we only need to push the box through a distance of 585 m , which is still over five football fields long . given the known speed of 0.25 m/s that will take us : $ \frac { 585 \mathrm { m } } { 0.25 \mathrm { m/s } } =2340 \mathrm { s } $ exercise : suppose that the force that we apply to the box , see figure 1 above , is initially reduced but increases to a constant value as we warm up . for instance , in the graph below we see that as the box is displaced further—i.e. , $ x $ gets larger—the force , $ f $ , increases for the first 30 m , see figure 2 below . how could we find the work done during the period where the force is changing ? if the force is not constant , one way to determine the work done is to divide the problem up into small sections over which the change is negligible and sum up the work done in each section . just as we have learned when looking at velocity time graphs , this can be done by calculating the area under the curve using geometry . the work done by a force is equal to the area under a force vs. position graph . in the case of figure 2 , it would be : $ ( 200\text { n } \cdot 30 \text { m } ) + \frac { 1 } { 2 } \left ( ( 500\text { n } -200\text { n } ) \cdot 30 \text { m } \right ) =10500 \mathrm { j } $ for the initial $ 30 \text { m } $ of displacement . similarly , the work done for the final 40 m of displacement would be : $ 500\text { n } \cdot 40 \text { m } =20,000 \mathrm { j } $ what if we are n't pushing straight on ? there is one thing we need to watch out for when doing these problems . the previous equation , $ w = f \cdot \delta x $ , does n't take into account situations where the force we are applying is not in the same direction as the motion . for instance , imagine we use a rope to pull on the box . in that case there will be an angle between the rope and the ground . to untangle this situation , we begin by drawing a triangle to separate out the horizontal and vertical components of the applied force . the key point here is that it is only the component of the force , $ f_ { || } $ , that lies parallel to the displacement that does work on an object . in the case of the box shown above , only the horizontal component of the applied force , $ f \text { cos } ( \theta ) $ , is doing work on the box since the box is being displaced horizontally . this means that a more general equation for the work done on the box by a force at an angle θ could be written as : $ w=f_ { || } \cdot\delta x $ $ w= ( f \cos { \theta } ) \cdot\delta x $ which is more often written as , $ \large w=f \delta x\cos { \theta } $ exercise : suppose we use a rope to pull the box , and the angle between the rope and the ground is 30º . this time we pull along the rope with a force of 500 n. how much of a chocolate bar can we eat this time if we pull the box through the same 585 m ? what about lifting weights instead ? in the previous example , we were doing work on a box which we were pushing around a floor . in doing so , we were working against a frictional force . another common form of exercise is lifting weights . in this case we are working against the force of gravity rather than friction . using newton 's laws we can find the force , $ f $ , required to lift a weight with mass $ m $ straight up , placing it on a rack which is at a height $ h $ above us : $ f=mg $ the change in position—previously $ \delta x $ —is simply the height , so the work , $ w $ , that we have done in lifting the weight is then $ w=m g h $ the exercise we have done in lifting the weight has resulted in energy being stored in the form of gravitational potential energy . it is called potential energy because it has the potential to be released at any moment with a crash as the weight falls back to the ground . we did positive work on the weight since we exerted our force in the same direction as the displacement of the weight , i.e. , upward . the work done by gravity on the weight while it was lifted was negative since the force of gravity is directed in the opposite direction to the displacement . also , since the weight is stationary after the lift , we know that the work that we have done is exactly canceled out by the work done by gravity . the work done by us is $ mgh $ , and the work done by gravity is $ -mgh $ . we will talk more about this when we look into kinetic energy . ok , let 's put some numbers in and find how much of that chocolate bar we would offset by lifting a weight of 50 kg up to a height of 0.5 m. the work done to the weight is $ w= ( 50\mathrm { kg } ) ( 9.81\mathrm { m/s^2 } ) ( 0.5\mathrm { m } ) = 245.25 \mathrm { j } $ ok , so how many 280 calorie—i.e. , $ 1.17 \times 10^6 $ joule—chocolate bars is this ? well , 245.25 j is about $ \dfrac { 1 } { 4770 } $ of a chocolate bar . but remember , our bodies are only about 25 % efficient , so the work done by the person is actually four times larger , about 981.8 j , which is $ \dfrac { 1 } { 1190 } $ chocolate bars . so , if we can lift this weight once every 2 seconds , it will take us about 2380 seconds or 40 minutes of hard work to burn off this chocolate bar ! what about simply holding a weight stationary ? one frequent source of confusion people have with the concept of work comes about when thinking about holding a heavy weight stationary above our heads , against the force of gravity . we are not moving the weight through any distance , so no work is being done to the weight . we could also achieve this by placing the weight on a table ; it is clear that the table is not doing any work to keep the weight in position . yet , we know from our experience that we get tired when doing the same job . so what is going on here ? it turns out that what is actually happening here is that our bodies are doing work on our muscles to maintain the necessary tension to hold the weight up . the body does this by sending a cascade of nerve impulses to each muscle . each impulse causes the muscle to momentarily contract and release . this all happens so fast that we might only notice a slight twitching at first . eventually though , not enough chemical energy is available in the muscle and it can no longer keep up . we then begin to shake and eventually must rest for a while . so work is being done , it is just not being done on the weight .
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what does energy and work mean ? energy is a word which tends to be used a lot in everyday life . though it is often used quite loosely , it does have a very specific physical meaning .
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how could i calculate the energy i used to attempt to cause motion ?
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what does energy and work mean ? energy is a word which tends to be used a lot in everyday life . though it is often used quite loosely , it does have a very specific physical meaning . energy is a measurement of the ability of something to do work . it is not a material substance . energy can be stored and measured in many forms . although we often hear people talking about energy consumption , energy is never really destroyed . it is just transferred from one form to another , doing work in the process . some forms of energy are less useful to us than others—for example , low level heat energy . it is better to talk about the consumption or extraction of energy resources , for example coal , oil , or wind , than consumption of energy itself . a speeding bullet has a measurable amount of energy associated with it ; this is known as kinetic energy . the bullet gained this energy because work was done on it by a charge of gunpowder which lost some chemical potential energy in the process . a hot cup of coffee has a measurable amount of thermal energy which it acquired via work done by a microwave oven , which in turn took electrical energy from the electrical grid . in practice , whenever work is done to move energy from one form to another , there is always some loss to other forms of energy such as heat and sound . for example , a traditional light bulb is only about 3 % efficient at converting electrical energy to visible light , while a human being is about 25 % efficient at converting chemical energy from food into work . how do we measure energy and work ? the standard unit used to measure energy and work done in physics is the joule , which has the symbol j . in mechanics , 1 joule is the energy transferred when a force of 1 newton is applied to an object and moves it through a distance of 1 meter . another unit of energy you may have come across is the calorie . the amount of energy in an item of food is often written in calories on the back of the packet . a typical 60 gram chocolate bar for example contains about 280 calories of energy . one calorie is the amount of energy required to raise 1 kg of water by 1 $ ^ { \circ } $ celsius . this is equal to 4184 joules , so one chocolate bar has 1.17 million joules or 1.17 mj of stored energy . that 's a lot of joules ! how long do i have to push a heavy box around to burn off one chocolate bar ? suppose we 're feeling guilty about eating a chocolate bar ; we want to find how much exercise we need to do to offset those extra 280 calories . let 's consider a simple form of exercise : pushing a heavy box around a room , see figure 1 below . using a bathroom scale between ourselves and the box , we find that we can push with a force of 500 n. meanwhile , we use a stopwatch and measuring tape to measure our speed . this comes out to be 0.25 meters per second . so how much work do we need to do to the box to burn off the candy bar ? the definition of work , $ w $ , is below : $ \large w = f\cdot \delta x $ the work we need to do to burn the energy in the candy bar is $ e=280 \mathrm { cal } \cdot 4184 \mathrm { j/cal } =1.17 \mathrm { mj } $ . therefore , the distance , $ δx $ , we need to move the box through is : $ \begin { align } w & amp ; = f\cdot \delta x \ 1.17 \text { mj } & amp ; = ( 500 \text { n } ) \cdot \delta x \ \dfrac { 1.17\times 10^6 \text { j } } { 500 \text { n } } & amp ; = \delta x \ 2,340 \text { m } & amp ; = \delta x \end { align } $ remember , however , that our bodies are about 25 % efficient at transferring stored energy from food into work . the actual energy we will offset is then four times as high as the work done to the box . so , we only need to push the box through a distance of 585 m , which is still over five football fields long . given the known speed of 0.25 m/s that will take us : $ \frac { 585 \mathrm { m } } { 0.25 \mathrm { m/s } } =2340 \mathrm { s } $ exercise : suppose that the force that we apply to the box , see figure 1 above , is initially reduced but increases to a constant value as we warm up . for instance , in the graph below we see that as the box is displaced further—i.e. , $ x $ gets larger—the force , $ f $ , increases for the first 30 m , see figure 2 below . how could we find the work done during the period where the force is changing ? if the force is not constant , one way to determine the work done is to divide the problem up into small sections over which the change is negligible and sum up the work done in each section . just as we have learned when looking at velocity time graphs , this can be done by calculating the area under the curve using geometry . the work done by a force is equal to the area under a force vs. position graph . in the case of figure 2 , it would be : $ ( 200\text { n } \cdot 30 \text { m } ) + \frac { 1 } { 2 } \left ( ( 500\text { n } -200\text { n } ) \cdot 30 \text { m } \right ) =10500 \mathrm { j } $ for the initial $ 30 \text { m } $ of displacement . similarly , the work done for the final 40 m of displacement would be : $ 500\text { n } \cdot 40 \text { m } =20,000 \mathrm { j } $ what if we are n't pushing straight on ? there is one thing we need to watch out for when doing these problems . the previous equation , $ w = f \cdot \delta x $ , does n't take into account situations where the force we are applying is not in the same direction as the motion . for instance , imagine we use a rope to pull on the box . in that case there will be an angle between the rope and the ground . to untangle this situation , we begin by drawing a triangle to separate out the horizontal and vertical components of the applied force . the key point here is that it is only the component of the force , $ f_ { || } $ , that lies parallel to the displacement that does work on an object . in the case of the box shown above , only the horizontal component of the applied force , $ f \text { cos } ( \theta ) $ , is doing work on the box since the box is being displaced horizontally . this means that a more general equation for the work done on the box by a force at an angle θ could be written as : $ w=f_ { || } \cdot\delta x $ $ w= ( f \cos { \theta } ) \cdot\delta x $ which is more often written as , $ \large w=f \delta x\cos { \theta } $ exercise : suppose we use a rope to pull the box , and the angle between the rope and the ground is 30º . this time we pull along the rope with a force of 500 n. how much of a chocolate bar can we eat this time if we pull the box through the same 585 m ? what about lifting weights instead ? in the previous example , we were doing work on a box which we were pushing around a floor . in doing so , we were working against a frictional force . another common form of exercise is lifting weights . in this case we are working against the force of gravity rather than friction . using newton 's laws we can find the force , $ f $ , required to lift a weight with mass $ m $ straight up , placing it on a rack which is at a height $ h $ above us : $ f=mg $ the change in position—previously $ \delta x $ —is simply the height , so the work , $ w $ , that we have done in lifting the weight is then $ w=m g h $ the exercise we have done in lifting the weight has resulted in energy being stored in the form of gravitational potential energy . it is called potential energy because it has the potential to be released at any moment with a crash as the weight falls back to the ground . we did positive work on the weight since we exerted our force in the same direction as the displacement of the weight , i.e. , upward . the work done by gravity on the weight while it was lifted was negative since the force of gravity is directed in the opposite direction to the displacement . also , since the weight is stationary after the lift , we know that the work that we have done is exactly canceled out by the work done by gravity . the work done by us is $ mgh $ , and the work done by gravity is $ -mgh $ . we will talk more about this when we look into kinetic energy . ok , let 's put some numbers in and find how much of that chocolate bar we would offset by lifting a weight of 50 kg up to a height of 0.5 m. the work done to the weight is $ w= ( 50\mathrm { kg } ) ( 9.81\mathrm { m/s^2 } ) ( 0.5\mathrm { m } ) = 245.25 \mathrm { j } $ ok , so how many 280 calorie—i.e. , $ 1.17 \times 10^6 $ joule—chocolate bars is this ? well , 245.25 j is about $ \dfrac { 1 } { 4770 } $ of a chocolate bar . but remember , our bodies are only about 25 % efficient , so the work done by the person is actually four times larger , about 981.8 j , which is $ \dfrac { 1 } { 1190 } $ chocolate bars . so , if we can lift this weight once every 2 seconds , it will take us about 2380 seconds or 40 minutes of hard work to burn off this chocolate bar ! what about simply holding a weight stationary ? one frequent source of confusion people have with the concept of work comes about when thinking about holding a heavy weight stationary above our heads , against the force of gravity . we are not moving the weight through any distance , so no work is being done to the weight . we could also achieve this by placing the weight on a table ; it is clear that the table is not doing any work to keep the weight in position . yet , we know from our experience that we get tired when doing the same job . so what is going on here ? it turns out that what is actually happening here is that our bodies are doing work on our muscles to maintain the necessary tension to hold the weight up . the body does this by sending a cascade of nerve impulses to each muscle . each impulse causes the muscle to momentarily contract and release . this all happens so fast that we might only notice a slight twitching at first . eventually though , not enough chemical energy is available in the muscle and it can no longer keep up . we then begin to shake and eventually must rest for a while . so work is being done , it is just not being done on the weight .
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a speeding bullet has a measurable amount of energy associated with it ; this is known as kinetic energy . the bullet gained this energy because work was done on it by a charge of gunpowder which lost some chemical potential energy in the process . a hot cup of coffee has a measurable amount of thermal energy which it acquired via work done by a microwave oven , which in turn took electrical energy from the electrical grid .
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what is chemical potential energy ?
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what does energy and work mean ? energy is a word which tends to be used a lot in everyday life . though it is often used quite loosely , it does have a very specific physical meaning . energy is a measurement of the ability of something to do work . it is not a material substance . energy can be stored and measured in many forms . although we often hear people talking about energy consumption , energy is never really destroyed . it is just transferred from one form to another , doing work in the process . some forms of energy are less useful to us than others—for example , low level heat energy . it is better to talk about the consumption or extraction of energy resources , for example coal , oil , or wind , than consumption of energy itself . a speeding bullet has a measurable amount of energy associated with it ; this is known as kinetic energy . the bullet gained this energy because work was done on it by a charge of gunpowder which lost some chemical potential energy in the process . a hot cup of coffee has a measurable amount of thermal energy which it acquired via work done by a microwave oven , which in turn took electrical energy from the electrical grid . in practice , whenever work is done to move energy from one form to another , there is always some loss to other forms of energy such as heat and sound . for example , a traditional light bulb is only about 3 % efficient at converting electrical energy to visible light , while a human being is about 25 % efficient at converting chemical energy from food into work . how do we measure energy and work ? the standard unit used to measure energy and work done in physics is the joule , which has the symbol j . in mechanics , 1 joule is the energy transferred when a force of 1 newton is applied to an object and moves it through a distance of 1 meter . another unit of energy you may have come across is the calorie . the amount of energy in an item of food is often written in calories on the back of the packet . a typical 60 gram chocolate bar for example contains about 280 calories of energy . one calorie is the amount of energy required to raise 1 kg of water by 1 $ ^ { \circ } $ celsius . this is equal to 4184 joules , so one chocolate bar has 1.17 million joules or 1.17 mj of stored energy . that 's a lot of joules ! how long do i have to push a heavy box around to burn off one chocolate bar ? suppose we 're feeling guilty about eating a chocolate bar ; we want to find how much exercise we need to do to offset those extra 280 calories . let 's consider a simple form of exercise : pushing a heavy box around a room , see figure 1 below . using a bathroom scale between ourselves and the box , we find that we can push with a force of 500 n. meanwhile , we use a stopwatch and measuring tape to measure our speed . this comes out to be 0.25 meters per second . so how much work do we need to do to the box to burn off the candy bar ? the definition of work , $ w $ , is below : $ \large w = f\cdot \delta x $ the work we need to do to burn the energy in the candy bar is $ e=280 \mathrm { cal } \cdot 4184 \mathrm { j/cal } =1.17 \mathrm { mj } $ . therefore , the distance , $ δx $ , we need to move the box through is : $ \begin { align } w & amp ; = f\cdot \delta x \ 1.17 \text { mj } & amp ; = ( 500 \text { n } ) \cdot \delta x \ \dfrac { 1.17\times 10^6 \text { j } } { 500 \text { n } } & amp ; = \delta x \ 2,340 \text { m } & amp ; = \delta x \end { align } $ remember , however , that our bodies are about 25 % efficient at transferring stored energy from food into work . the actual energy we will offset is then four times as high as the work done to the box . so , we only need to push the box through a distance of 585 m , which is still over five football fields long . given the known speed of 0.25 m/s that will take us : $ \frac { 585 \mathrm { m } } { 0.25 \mathrm { m/s } } =2340 \mathrm { s } $ exercise : suppose that the force that we apply to the box , see figure 1 above , is initially reduced but increases to a constant value as we warm up . for instance , in the graph below we see that as the box is displaced further—i.e. , $ x $ gets larger—the force , $ f $ , increases for the first 30 m , see figure 2 below . how could we find the work done during the period where the force is changing ? if the force is not constant , one way to determine the work done is to divide the problem up into small sections over which the change is negligible and sum up the work done in each section . just as we have learned when looking at velocity time graphs , this can be done by calculating the area under the curve using geometry . the work done by a force is equal to the area under a force vs. position graph . in the case of figure 2 , it would be : $ ( 200\text { n } \cdot 30 \text { m } ) + \frac { 1 } { 2 } \left ( ( 500\text { n } -200\text { n } ) \cdot 30 \text { m } \right ) =10500 \mathrm { j } $ for the initial $ 30 \text { m } $ of displacement . similarly , the work done for the final 40 m of displacement would be : $ 500\text { n } \cdot 40 \text { m } =20,000 \mathrm { j } $ what if we are n't pushing straight on ? there is one thing we need to watch out for when doing these problems . the previous equation , $ w = f \cdot \delta x $ , does n't take into account situations where the force we are applying is not in the same direction as the motion . for instance , imagine we use a rope to pull on the box . in that case there will be an angle between the rope and the ground . to untangle this situation , we begin by drawing a triangle to separate out the horizontal and vertical components of the applied force . the key point here is that it is only the component of the force , $ f_ { || } $ , that lies parallel to the displacement that does work on an object . in the case of the box shown above , only the horizontal component of the applied force , $ f \text { cos } ( \theta ) $ , is doing work on the box since the box is being displaced horizontally . this means that a more general equation for the work done on the box by a force at an angle θ could be written as : $ w=f_ { || } \cdot\delta x $ $ w= ( f \cos { \theta } ) \cdot\delta x $ which is more often written as , $ \large w=f \delta x\cos { \theta } $ exercise : suppose we use a rope to pull the box , and the angle between the rope and the ground is 30º . this time we pull along the rope with a force of 500 n. how much of a chocolate bar can we eat this time if we pull the box through the same 585 m ? what about lifting weights instead ? in the previous example , we were doing work on a box which we were pushing around a floor . in doing so , we were working against a frictional force . another common form of exercise is lifting weights . in this case we are working against the force of gravity rather than friction . using newton 's laws we can find the force , $ f $ , required to lift a weight with mass $ m $ straight up , placing it on a rack which is at a height $ h $ above us : $ f=mg $ the change in position—previously $ \delta x $ —is simply the height , so the work , $ w $ , that we have done in lifting the weight is then $ w=m g h $ the exercise we have done in lifting the weight has resulted in energy being stored in the form of gravitational potential energy . it is called potential energy because it has the potential to be released at any moment with a crash as the weight falls back to the ground . we did positive work on the weight since we exerted our force in the same direction as the displacement of the weight , i.e. , upward . the work done by gravity on the weight while it was lifted was negative since the force of gravity is directed in the opposite direction to the displacement . also , since the weight is stationary after the lift , we know that the work that we have done is exactly canceled out by the work done by gravity . the work done by us is $ mgh $ , and the work done by gravity is $ -mgh $ . we will talk more about this when we look into kinetic energy . ok , let 's put some numbers in and find how much of that chocolate bar we would offset by lifting a weight of 50 kg up to a height of 0.5 m. the work done to the weight is $ w= ( 50\mathrm { kg } ) ( 9.81\mathrm { m/s^2 } ) ( 0.5\mathrm { m } ) = 245.25 \mathrm { j } $ ok , so how many 280 calorie—i.e. , $ 1.17 \times 10^6 $ joule—chocolate bars is this ? well , 245.25 j is about $ \dfrac { 1 } { 4770 } $ of a chocolate bar . but remember , our bodies are only about 25 % efficient , so the work done by the person is actually four times larger , about 981.8 j , which is $ \dfrac { 1 } { 1190 } $ chocolate bars . so , if we can lift this weight once every 2 seconds , it will take us about 2380 seconds or 40 minutes of hard work to burn off this chocolate bar ! what about simply holding a weight stationary ? one frequent source of confusion people have with the concept of work comes about when thinking about holding a heavy weight stationary above our heads , against the force of gravity . we are not moving the weight through any distance , so no work is being done to the weight . we could also achieve this by placing the weight on a table ; it is clear that the table is not doing any work to keep the weight in position . yet , we know from our experience that we get tired when doing the same job . so what is going on here ? it turns out that what is actually happening here is that our bodies are doing work on our muscles to maintain the necessary tension to hold the weight up . the body does this by sending a cascade of nerve impulses to each muscle . each impulse causes the muscle to momentarily contract and release . this all happens so fast that we might only notice a slight twitching at first . eventually though , not enough chemical energy is available in the muscle and it can no longer keep up . we then begin to shake and eventually must rest for a while . so work is being done , it is just not being done on the weight .
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one frequent source of confusion people have with the concept of work comes about when thinking about holding a heavy weight stationary above our heads , against the force of gravity . we are not moving the weight through any distance , so no work is being done to the weight . we could also achieve this by placing the weight on a table ; it is clear that the table is not doing any work to keep the weight in position .
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for example , if we lower it at constant velocity , are n't we doing negative work which is equal in magnitude to the work we did lifting the weight ?
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what does energy and work mean ? energy is a word which tends to be used a lot in everyday life . though it is often used quite loosely , it does have a very specific physical meaning . energy is a measurement of the ability of something to do work . it is not a material substance . energy can be stored and measured in many forms . although we often hear people talking about energy consumption , energy is never really destroyed . it is just transferred from one form to another , doing work in the process . some forms of energy are less useful to us than others—for example , low level heat energy . it is better to talk about the consumption or extraction of energy resources , for example coal , oil , or wind , than consumption of energy itself . a speeding bullet has a measurable amount of energy associated with it ; this is known as kinetic energy . the bullet gained this energy because work was done on it by a charge of gunpowder which lost some chemical potential energy in the process . a hot cup of coffee has a measurable amount of thermal energy which it acquired via work done by a microwave oven , which in turn took electrical energy from the electrical grid . in practice , whenever work is done to move energy from one form to another , there is always some loss to other forms of energy such as heat and sound . for example , a traditional light bulb is only about 3 % efficient at converting electrical energy to visible light , while a human being is about 25 % efficient at converting chemical energy from food into work . how do we measure energy and work ? the standard unit used to measure energy and work done in physics is the joule , which has the symbol j . in mechanics , 1 joule is the energy transferred when a force of 1 newton is applied to an object and moves it through a distance of 1 meter . another unit of energy you may have come across is the calorie . the amount of energy in an item of food is often written in calories on the back of the packet . a typical 60 gram chocolate bar for example contains about 280 calories of energy . one calorie is the amount of energy required to raise 1 kg of water by 1 $ ^ { \circ } $ celsius . this is equal to 4184 joules , so one chocolate bar has 1.17 million joules or 1.17 mj of stored energy . that 's a lot of joules ! how long do i have to push a heavy box around to burn off one chocolate bar ? suppose we 're feeling guilty about eating a chocolate bar ; we want to find how much exercise we need to do to offset those extra 280 calories . let 's consider a simple form of exercise : pushing a heavy box around a room , see figure 1 below . using a bathroom scale between ourselves and the box , we find that we can push with a force of 500 n. meanwhile , we use a stopwatch and measuring tape to measure our speed . this comes out to be 0.25 meters per second . so how much work do we need to do to the box to burn off the candy bar ? the definition of work , $ w $ , is below : $ \large w = f\cdot \delta x $ the work we need to do to burn the energy in the candy bar is $ e=280 \mathrm { cal } \cdot 4184 \mathrm { j/cal } =1.17 \mathrm { mj } $ . therefore , the distance , $ δx $ , we need to move the box through is : $ \begin { align } w & amp ; = f\cdot \delta x \ 1.17 \text { mj } & amp ; = ( 500 \text { n } ) \cdot \delta x \ \dfrac { 1.17\times 10^6 \text { j } } { 500 \text { n } } & amp ; = \delta x \ 2,340 \text { m } & amp ; = \delta x \end { align } $ remember , however , that our bodies are about 25 % efficient at transferring stored energy from food into work . the actual energy we will offset is then four times as high as the work done to the box . so , we only need to push the box through a distance of 585 m , which is still over five football fields long . given the known speed of 0.25 m/s that will take us : $ \frac { 585 \mathrm { m } } { 0.25 \mathrm { m/s } } =2340 \mathrm { s } $ exercise : suppose that the force that we apply to the box , see figure 1 above , is initially reduced but increases to a constant value as we warm up . for instance , in the graph below we see that as the box is displaced further—i.e. , $ x $ gets larger—the force , $ f $ , increases for the first 30 m , see figure 2 below . how could we find the work done during the period where the force is changing ? if the force is not constant , one way to determine the work done is to divide the problem up into small sections over which the change is negligible and sum up the work done in each section . just as we have learned when looking at velocity time graphs , this can be done by calculating the area under the curve using geometry . the work done by a force is equal to the area under a force vs. position graph . in the case of figure 2 , it would be : $ ( 200\text { n } \cdot 30 \text { m } ) + \frac { 1 } { 2 } \left ( ( 500\text { n } -200\text { n } ) \cdot 30 \text { m } \right ) =10500 \mathrm { j } $ for the initial $ 30 \text { m } $ of displacement . similarly , the work done for the final 40 m of displacement would be : $ 500\text { n } \cdot 40 \text { m } =20,000 \mathrm { j } $ what if we are n't pushing straight on ? there is one thing we need to watch out for when doing these problems . the previous equation , $ w = f \cdot \delta x $ , does n't take into account situations where the force we are applying is not in the same direction as the motion . for instance , imagine we use a rope to pull on the box . in that case there will be an angle between the rope and the ground . to untangle this situation , we begin by drawing a triangle to separate out the horizontal and vertical components of the applied force . the key point here is that it is only the component of the force , $ f_ { || } $ , that lies parallel to the displacement that does work on an object . in the case of the box shown above , only the horizontal component of the applied force , $ f \text { cos } ( \theta ) $ , is doing work on the box since the box is being displaced horizontally . this means that a more general equation for the work done on the box by a force at an angle θ could be written as : $ w=f_ { || } \cdot\delta x $ $ w= ( f \cos { \theta } ) \cdot\delta x $ which is more often written as , $ \large w=f \delta x\cos { \theta } $ exercise : suppose we use a rope to pull the box , and the angle between the rope and the ground is 30º . this time we pull along the rope with a force of 500 n. how much of a chocolate bar can we eat this time if we pull the box through the same 585 m ? what about lifting weights instead ? in the previous example , we were doing work on a box which we were pushing around a floor . in doing so , we were working against a frictional force . another common form of exercise is lifting weights . in this case we are working against the force of gravity rather than friction . using newton 's laws we can find the force , $ f $ , required to lift a weight with mass $ m $ straight up , placing it on a rack which is at a height $ h $ above us : $ f=mg $ the change in position—previously $ \delta x $ —is simply the height , so the work , $ w $ , that we have done in lifting the weight is then $ w=m g h $ the exercise we have done in lifting the weight has resulted in energy being stored in the form of gravitational potential energy . it is called potential energy because it has the potential to be released at any moment with a crash as the weight falls back to the ground . we did positive work on the weight since we exerted our force in the same direction as the displacement of the weight , i.e. , upward . the work done by gravity on the weight while it was lifted was negative since the force of gravity is directed in the opposite direction to the displacement . also , since the weight is stationary after the lift , we know that the work that we have done is exactly canceled out by the work done by gravity . the work done by us is $ mgh $ , and the work done by gravity is $ -mgh $ . we will talk more about this when we look into kinetic energy . ok , let 's put some numbers in and find how much of that chocolate bar we would offset by lifting a weight of 50 kg up to a height of 0.5 m. the work done to the weight is $ w= ( 50\mathrm { kg } ) ( 9.81\mathrm { m/s^2 } ) ( 0.5\mathrm { m } ) = 245.25 \mathrm { j } $ ok , so how many 280 calorie—i.e. , $ 1.17 \times 10^6 $ joule—chocolate bars is this ? well , 245.25 j is about $ \dfrac { 1 } { 4770 } $ of a chocolate bar . but remember , our bodies are only about 25 % efficient , so the work done by the person is actually four times larger , about 981.8 j , which is $ \dfrac { 1 } { 1190 } $ chocolate bars . so , if we can lift this weight once every 2 seconds , it will take us about 2380 seconds or 40 minutes of hard work to burn off this chocolate bar ! what about simply holding a weight stationary ? one frequent source of confusion people have with the concept of work comes about when thinking about holding a heavy weight stationary above our heads , against the force of gravity . we are not moving the weight through any distance , so no work is being done to the weight . we could also achieve this by placing the weight on a table ; it is clear that the table is not doing any work to keep the weight in position . yet , we know from our experience that we get tired when doing the same job . so what is going on here ? it turns out that what is actually happening here is that our bodies are doing work on our muscles to maintain the necessary tension to hold the weight up . the body does this by sending a cascade of nerve impulses to each muscle . each impulse causes the muscle to momentarily contract and release . this all happens so fast that we might only notice a slight twitching at first . eventually though , not enough chemical energy is available in the muscle and it can no longer keep up . we then begin to shake and eventually must rest for a while . so work is being done , it is just not being done on the weight .
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what does energy and work mean ? energy is a word which tends to be used a lot in everyday life .
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does n't it make more sense to say that because of our body 's 25 % efficiency the available energy for work is 4 times smaller , instead of saying the work done is 4 times larger ?
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what does energy and work mean ? energy is a word which tends to be used a lot in everyday life . though it is often used quite loosely , it does have a very specific physical meaning . energy is a measurement of the ability of something to do work . it is not a material substance . energy can be stored and measured in many forms . although we often hear people talking about energy consumption , energy is never really destroyed . it is just transferred from one form to another , doing work in the process . some forms of energy are less useful to us than others—for example , low level heat energy . it is better to talk about the consumption or extraction of energy resources , for example coal , oil , or wind , than consumption of energy itself . a speeding bullet has a measurable amount of energy associated with it ; this is known as kinetic energy . the bullet gained this energy because work was done on it by a charge of gunpowder which lost some chemical potential energy in the process . a hot cup of coffee has a measurable amount of thermal energy which it acquired via work done by a microwave oven , which in turn took electrical energy from the electrical grid . in practice , whenever work is done to move energy from one form to another , there is always some loss to other forms of energy such as heat and sound . for example , a traditional light bulb is only about 3 % efficient at converting electrical energy to visible light , while a human being is about 25 % efficient at converting chemical energy from food into work . how do we measure energy and work ? the standard unit used to measure energy and work done in physics is the joule , which has the symbol j . in mechanics , 1 joule is the energy transferred when a force of 1 newton is applied to an object and moves it through a distance of 1 meter . another unit of energy you may have come across is the calorie . the amount of energy in an item of food is often written in calories on the back of the packet . a typical 60 gram chocolate bar for example contains about 280 calories of energy . one calorie is the amount of energy required to raise 1 kg of water by 1 $ ^ { \circ } $ celsius . this is equal to 4184 joules , so one chocolate bar has 1.17 million joules or 1.17 mj of stored energy . that 's a lot of joules ! how long do i have to push a heavy box around to burn off one chocolate bar ? suppose we 're feeling guilty about eating a chocolate bar ; we want to find how much exercise we need to do to offset those extra 280 calories . let 's consider a simple form of exercise : pushing a heavy box around a room , see figure 1 below . using a bathroom scale between ourselves and the box , we find that we can push with a force of 500 n. meanwhile , we use a stopwatch and measuring tape to measure our speed . this comes out to be 0.25 meters per second . so how much work do we need to do to the box to burn off the candy bar ? the definition of work , $ w $ , is below : $ \large w = f\cdot \delta x $ the work we need to do to burn the energy in the candy bar is $ e=280 \mathrm { cal } \cdot 4184 \mathrm { j/cal } =1.17 \mathrm { mj } $ . therefore , the distance , $ δx $ , we need to move the box through is : $ \begin { align } w & amp ; = f\cdot \delta x \ 1.17 \text { mj } & amp ; = ( 500 \text { n } ) \cdot \delta x \ \dfrac { 1.17\times 10^6 \text { j } } { 500 \text { n } } & amp ; = \delta x \ 2,340 \text { m } & amp ; = \delta x \end { align } $ remember , however , that our bodies are about 25 % efficient at transferring stored energy from food into work . the actual energy we will offset is then four times as high as the work done to the box . so , we only need to push the box through a distance of 585 m , which is still over five football fields long . given the known speed of 0.25 m/s that will take us : $ \frac { 585 \mathrm { m } } { 0.25 \mathrm { m/s } } =2340 \mathrm { s } $ exercise : suppose that the force that we apply to the box , see figure 1 above , is initially reduced but increases to a constant value as we warm up . for instance , in the graph below we see that as the box is displaced further—i.e. , $ x $ gets larger—the force , $ f $ , increases for the first 30 m , see figure 2 below . how could we find the work done during the period where the force is changing ? if the force is not constant , one way to determine the work done is to divide the problem up into small sections over which the change is negligible and sum up the work done in each section . just as we have learned when looking at velocity time graphs , this can be done by calculating the area under the curve using geometry . the work done by a force is equal to the area under a force vs. position graph . in the case of figure 2 , it would be : $ ( 200\text { n } \cdot 30 \text { m } ) + \frac { 1 } { 2 } \left ( ( 500\text { n } -200\text { n } ) \cdot 30 \text { m } \right ) =10500 \mathrm { j } $ for the initial $ 30 \text { m } $ of displacement . similarly , the work done for the final 40 m of displacement would be : $ 500\text { n } \cdot 40 \text { m } =20,000 \mathrm { j } $ what if we are n't pushing straight on ? there is one thing we need to watch out for when doing these problems . the previous equation , $ w = f \cdot \delta x $ , does n't take into account situations where the force we are applying is not in the same direction as the motion . for instance , imagine we use a rope to pull on the box . in that case there will be an angle between the rope and the ground . to untangle this situation , we begin by drawing a triangle to separate out the horizontal and vertical components of the applied force . the key point here is that it is only the component of the force , $ f_ { || } $ , that lies parallel to the displacement that does work on an object . in the case of the box shown above , only the horizontal component of the applied force , $ f \text { cos } ( \theta ) $ , is doing work on the box since the box is being displaced horizontally . this means that a more general equation for the work done on the box by a force at an angle θ could be written as : $ w=f_ { || } \cdot\delta x $ $ w= ( f \cos { \theta } ) \cdot\delta x $ which is more often written as , $ \large w=f \delta x\cos { \theta } $ exercise : suppose we use a rope to pull the box , and the angle between the rope and the ground is 30º . this time we pull along the rope with a force of 500 n. how much of a chocolate bar can we eat this time if we pull the box through the same 585 m ? what about lifting weights instead ? in the previous example , we were doing work on a box which we were pushing around a floor . in doing so , we were working against a frictional force . another common form of exercise is lifting weights . in this case we are working against the force of gravity rather than friction . using newton 's laws we can find the force , $ f $ , required to lift a weight with mass $ m $ straight up , placing it on a rack which is at a height $ h $ above us : $ f=mg $ the change in position—previously $ \delta x $ —is simply the height , so the work , $ w $ , that we have done in lifting the weight is then $ w=m g h $ the exercise we have done in lifting the weight has resulted in energy being stored in the form of gravitational potential energy . it is called potential energy because it has the potential to be released at any moment with a crash as the weight falls back to the ground . we did positive work on the weight since we exerted our force in the same direction as the displacement of the weight , i.e. , upward . the work done by gravity on the weight while it was lifted was negative since the force of gravity is directed in the opposite direction to the displacement . also , since the weight is stationary after the lift , we know that the work that we have done is exactly canceled out by the work done by gravity . the work done by us is $ mgh $ , and the work done by gravity is $ -mgh $ . we will talk more about this when we look into kinetic energy . ok , let 's put some numbers in and find how much of that chocolate bar we would offset by lifting a weight of 50 kg up to a height of 0.5 m. the work done to the weight is $ w= ( 50\mathrm { kg } ) ( 9.81\mathrm { m/s^2 } ) ( 0.5\mathrm { m } ) = 245.25 \mathrm { j } $ ok , so how many 280 calorie—i.e. , $ 1.17 \times 10^6 $ joule—chocolate bars is this ? well , 245.25 j is about $ \dfrac { 1 } { 4770 } $ of a chocolate bar . but remember , our bodies are only about 25 % efficient , so the work done by the person is actually four times larger , about 981.8 j , which is $ \dfrac { 1 } { 1190 } $ chocolate bars . so , if we can lift this weight once every 2 seconds , it will take us about 2380 seconds or 40 minutes of hard work to burn off this chocolate bar ! what about simply holding a weight stationary ? one frequent source of confusion people have with the concept of work comes about when thinking about holding a heavy weight stationary above our heads , against the force of gravity . we are not moving the weight through any distance , so no work is being done to the weight . we could also achieve this by placing the weight on a table ; it is clear that the table is not doing any work to keep the weight in position . yet , we know from our experience that we get tired when doing the same job . so what is going on here ? it turns out that what is actually happening here is that our bodies are doing work on our muscles to maintain the necessary tension to hold the weight up . the body does this by sending a cascade of nerve impulses to each muscle . each impulse causes the muscle to momentarily contract and release . this all happens so fast that we might only notice a slight twitching at first . eventually though , not enough chemical energy is available in the muscle and it can no longer keep up . we then begin to shake and eventually must rest for a while . so work is being done , it is just not being done on the weight .
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another common form of exercise is lifting weights . in this case we are working against the force of gravity rather than friction . using newton 's laws we can find the force , $ f $ , required to lift a weight with mass $ m $ straight up , placing it on a rack which is at a height $ h $ above us : $ f=mg $ the change in position—previously $ \delta x $ —is simply the height , so the work , $ w $ , that we have done in lifting the weight is then $ w=m g h $ the exercise we have done in lifting the weight has resulted in energy being stored in the form of gravitational potential energy .
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if we take friction into account , is the force we use in the work equation with or without the friction ?
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what does energy and work mean ? energy is a word which tends to be used a lot in everyday life . though it is often used quite loosely , it does have a very specific physical meaning . energy is a measurement of the ability of something to do work . it is not a material substance . energy can be stored and measured in many forms . although we often hear people talking about energy consumption , energy is never really destroyed . it is just transferred from one form to another , doing work in the process . some forms of energy are less useful to us than others—for example , low level heat energy . it is better to talk about the consumption or extraction of energy resources , for example coal , oil , or wind , than consumption of energy itself . a speeding bullet has a measurable amount of energy associated with it ; this is known as kinetic energy . the bullet gained this energy because work was done on it by a charge of gunpowder which lost some chemical potential energy in the process . a hot cup of coffee has a measurable amount of thermal energy which it acquired via work done by a microwave oven , which in turn took electrical energy from the electrical grid . in practice , whenever work is done to move energy from one form to another , there is always some loss to other forms of energy such as heat and sound . for example , a traditional light bulb is only about 3 % efficient at converting electrical energy to visible light , while a human being is about 25 % efficient at converting chemical energy from food into work . how do we measure energy and work ? the standard unit used to measure energy and work done in physics is the joule , which has the symbol j . in mechanics , 1 joule is the energy transferred when a force of 1 newton is applied to an object and moves it through a distance of 1 meter . another unit of energy you may have come across is the calorie . the amount of energy in an item of food is often written in calories on the back of the packet . a typical 60 gram chocolate bar for example contains about 280 calories of energy . one calorie is the amount of energy required to raise 1 kg of water by 1 $ ^ { \circ } $ celsius . this is equal to 4184 joules , so one chocolate bar has 1.17 million joules or 1.17 mj of stored energy . that 's a lot of joules ! how long do i have to push a heavy box around to burn off one chocolate bar ? suppose we 're feeling guilty about eating a chocolate bar ; we want to find how much exercise we need to do to offset those extra 280 calories . let 's consider a simple form of exercise : pushing a heavy box around a room , see figure 1 below . using a bathroom scale between ourselves and the box , we find that we can push with a force of 500 n. meanwhile , we use a stopwatch and measuring tape to measure our speed . this comes out to be 0.25 meters per second . so how much work do we need to do to the box to burn off the candy bar ? the definition of work , $ w $ , is below : $ \large w = f\cdot \delta x $ the work we need to do to burn the energy in the candy bar is $ e=280 \mathrm { cal } \cdot 4184 \mathrm { j/cal } =1.17 \mathrm { mj } $ . therefore , the distance , $ δx $ , we need to move the box through is : $ \begin { align } w & amp ; = f\cdot \delta x \ 1.17 \text { mj } & amp ; = ( 500 \text { n } ) \cdot \delta x \ \dfrac { 1.17\times 10^6 \text { j } } { 500 \text { n } } & amp ; = \delta x \ 2,340 \text { m } & amp ; = \delta x \end { align } $ remember , however , that our bodies are about 25 % efficient at transferring stored energy from food into work . the actual energy we will offset is then four times as high as the work done to the box . so , we only need to push the box through a distance of 585 m , which is still over five football fields long . given the known speed of 0.25 m/s that will take us : $ \frac { 585 \mathrm { m } } { 0.25 \mathrm { m/s } } =2340 \mathrm { s } $ exercise : suppose that the force that we apply to the box , see figure 1 above , is initially reduced but increases to a constant value as we warm up . for instance , in the graph below we see that as the box is displaced further—i.e. , $ x $ gets larger—the force , $ f $ , increases for the first 30 m , see figure 2 below . how could we find the work done during the period where the force is changing ? if the force is not constant , one way to determine the work done is to divide the problem up into small sections over which the change is negligible and sum up the work done in each section . just as we have learned when looking at velocity time graphs , this can be done by calculating the area under the curve using geometry . the work done by a force is equal to the area under a force vs. position graph . in the case of figure 2 , it would be : $ ( 200\text { n } \cdot 30 \text { m } ) + \frac { 1 } { 2 } \left ( ( 500\text { n } -200\text { n } ) \cdot 30 \text { m } \right ) =10500 \mathrm { j } $ for the initial $ 30 \text { m } $ of displacement . similarly , the work done for the final 40 m of displacement would be : $ 500\text { n } \cdot 40 \text { m } =20,000 \mathrm { j } $ what if we are n't pushing straight on ? there is one thing we need to watch out for when doing these problems . the previous equation , $ w = f \cdot \delta x $ , does n't take into account situations where the force we are applying is not in the same direction as the motion . for instance , imagine we use a rope to pull on the box . in that case there will be an angle between the rope and the ground . to untangle this situation , we begin by drawing a triangle to separate out the horizontal and vertical components of the applied force . the key point here is that it is only the component of the force , $ f_ { || } $ , that lies parallel to the displacement that does work on an object . in the case of the box shown above , only the horizontal component of the applied force , $ f \text { cos } ( \theta ) $ , is doing work on the box since the box is being displaced horizontally . this means that a more general equation for the work done on the box by a force at an angle θ could be written as : $ w=f_ { || } \cdot\delta x $ $ w= ( f \cos { \theta } ) \cdot\delta x $ which is more often written as , $ \large w=f \delta x\cos { \theta } $ exercise : suppose we use a rope to pull the box , and the angle between the rope and the ground is 30º . this time we pull along the rope with a force of 500 n. how much of a chocolate bar can we eat this time if we pull the box through the same 585 m ? what about lifting weights instead ? in the previous example , we were doing work on a box which we were pushing around a floor . in doing so , we were working against a frictional force . another common form of exercise is lifting weights . in this case we are working against the force of gravity rather than friction . using newton 's laws we can find the force , $ f $ , required to lift a weight with mass $ m $ straight up , placing it on a rack which is at a height $ h $ above us : $ f=mg $ the change in position—previously $ \delta x $ —is simply the height , so the work , $ w $ , that we have done in lifting the weight is then $ w=m g h $ the exercise we have done in lifting the weight has resulted in energy being stored in the form of gravitational potential energy . it is called potential energy because it has the potential to be released at any moment with a crash as the weight falls back to the ground . we did positive work on the weight since we exerted our force in the same direction as the displacement of the weight , i.e. , upward . the work done by gravity on the weight while it was lifted was negative since the force of gravity is directed in the opposite direction to the displacement . also , since the weight is stationary after the lift , we know that the work that we have done is exactly canceled out by the work done by gravity . the work done by us is $ mgh $ , and the work done by gravity is $ -mgh $ . we will talk more about this when we look into kinetic energy . ok , let 's put some numbers in and find how much of that chocolate bar we would offset by lifting a weight of 50 kg up to a height of 0.5 m. the work done to the weight is $ w= ( 50\mathrm { kg } ) ( 9.81\mathrm { m/s^2 } ) ( 0.5\mathrm { m } ) = 245.25 \mathrm { j } $ ok , so how many 280 calorie—i.e. , $ 1.17 \times 10^6 $ joule—chocolate bars is this ? well , 245.25 j is about $ \dfrac { 1 } { 4770 } $ of a chocolate bar . but remember , our bodies are only about 25 % efficient , so the work done by the person is actually four times larger , about 981.8 j , which is $ \dfrac { 1 } { 1190 } $ chocolate bars . so , if we can lift this weight once every 2 seconds , it will take us about 2380 seconds or 40 minutes of hard work to burn off this chocolate bar ! what about simply holding a weight stationary ? one frequent source of confusion people have with the concept of work comes about when thinking about holding a heavy weight stationary above our heads , against the force of gravity . we are not moving the weight through any distance , so no work is being done to the weight . we could also achieve this by placing the weight on a table ; it is clear that the table is not doing any work to keep the weight in position . yet , we know from our experience that we get tired when doing the same job . so what is going on here ? it turns out that what is actually happening here is that our bodies are doing work on our muscles to maintain the necessary tension to hold the weight up . the body does this by sending a cascade of nerve impulses to each muscle . each impulse causes the muscle to momentarily contract and release . this all happens so fast that we might only notice a slight twitching at first . eventually though , not enough chemical energy is available in the muscle and it can no longer keep up . we then begin to shake and eventually must rest for a while . so work is being done , it is just not being done on the weight .
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we did positive work on the weight since we exerted our force in the same direction as the displacement of the weight , i.e. , upward . the work done by gravity on the weight while it was lifted was negative since the force of gravity is directed in the opposite direction to the displacement . also , since the weight is stationary after the lift , we know that the work that we have done is exactly canceled out by the work done by gravity .
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`` w= ( 50kg ) ( 9.81m/s2 ) ( 0.5m ) =245.25j '' in this equation , should't the acceleration due to gravity be negative since it 's accelerating in the opposite direction ?
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what does energy and work mean ? energy is a word which tends to be used a lot in everyday life . though it is often used quite loosely , it does have a very specific physical meaning . energy is a measurement of the ability of something to do work . it is not a material substance . energy can be stored and measured in many forms . although we often hear people talking about energy consumption , energy is never really destroyed . it is just transferred from one form to another , doing work in the process . some forms of energy are less useful to us than others—for example , low level heat energy . it is better to talk about the consumption or extraction of energy resources , for example coal , oil , or wind , than consumption of energy itself . a speeding bullet has a measurable amount of energy associated with it ; this is known as kinetic energy . the bullet gained this energy because work was done on it by a charge of gunpowder which lost some chemical potential energy in the process . a hot cup of coffee has a measurable amount of thermal energy which it acquired via work done by a microwave oven , which in turn took electrical energy from the electrical grid . in practice , whenever work is done to move energy from one form to another , there is always some loss to other forms of energy such as heat and sound . for example , a traditional light bulb is only about 3 % efficient at converting electrical energy to visible light , while a human being is about 25 % efficient at converting chemical energy from food into work . how do we measure energy and work ? the standard unit used to measure energy and work done in physics is the joule , which has the symbol j . in mechanics , 1 joule is the energy transferred when a force of 1 newton is applied to an object and moves it through a distance of 1 meter . another unit of energy you may have come across is the calorie . the amount of energy in an item of food is often written in calories on the back of the packet . a typical 60 gram chocolate bar for example contains about 280 calories of energy . one calorie is the amount of energy required to raise 1 kg of water by 1 $ ^ { \circ } $ celsius . this is equal to 4184 joules , so one chocolate bar has 1.17 million joules or 1.17 mj of stored energy . that 's a lot of joules ! how long do i have to push a heavy box around to burn off one chocolate bar ? suppose we 're feeling guilty about eating a chocolate bar ; we want to find how much exercise we need to do to offset those extra 280 calories . let 's consider a simple form of exercise : pushing a heavy box around a room , see figure 1 below . using a bathroom scale between ourselves and the box , we find that we can push with a force of 500 n. meanwhile , we use a stopwatch and measuring tape to measure our speed . this comes out to be 0.25 meters per second . so how much work do we need to do to the box to burn off the candy bar ? the definition of work , $ w $ , is below : $ \large w = f\cdot \delta x $ the work we need to do to burn the energy in the candy bar is $ e=280 \mathrm { cal } \cdot 4184 \mathrm { j/cal } =1.17 \mathrm { mj } $ . therefore , the distance , $ δx $ , we need to move the box through is : $ \begin { align } w & amp ; = f\cdot \delta x \ 1.17 \text { mj } & amp ; = ( 500 \text { n } ) \cdot \delta x \ \dfrac { 1.17\times 10^6 \text { j } } { 500 \text { n } } & amp ; = \delta x \ 2,340 \text { m } & amp ; = \delta x \end { align } $ remember , however , that our bodies are about 25 % efficient at transferring stored energy from food into work . the actual energy we will offset is then four times as high as the work done to the box . so , we only need to push the box through a distance of 585 m , which is still over five football fields long . given the known speed of 0.25 m/s that will take us : $ \frac { 585 \mathrm { m } } { 0.25 \mathrm { m/s } } =2340 \mathrm { s } $ exercise : suppose that the force that we apply to the box , see figure 1 above , is initially reduced but increases to a constant value as we warm up . for instance , in the graph below we see that as the box is displaced further—i.e. , $ x $ gets larger—the force , $ f $ , increases for the first 30 m , see figure 2 below . how could we find the work done during the period where the force is changing ? if the force is not constant , one way to determine the work done is to divide the problem up into small sections over which the change is negligible and sum up the work done in each section . just as we have learned when looking at velocity time graphs , this can be done by calculating the area under the curve using geometry . the work done by a force is equal to the area under a force vs. position graph . in the case of figure 2 , it would be : $ ( 200\text { n } \cdot 30 \text { m } ) + \frac { 1 } { 2 } \left ( ( 500\text { n } -200\text { n } ) \cdot 30 \text { m } \right ) =10500 \mathrm { j } $ for the initial $ 30 \text { m } $ of displacement . similarly , the work done for the final 40 m of displacement would be : $ 500\text { n } \cdot 40 \text { m } =20,000 \mathrm { j } $ what if we are n't pushing straight on ? there is one thing we need to watch out for when doing these problems . the previous equation , $ w = f \cdot \delta x $ , does n't take into account situations where the force we are applying is not in the same direction as the motion . for instance , imagine we use a rope to pull on the box . in that case there will be an angle between the rope and the ground . to untangle this situation , we begin by drawing a triangle to separate out the horizontal and vertical components of the applied force . the key point here is that it is only the component of the force , $ f_ { || } $ , that lies parallel to the displacement that does work on an object . in the case of the box shown above , only the horizontal component of the applied force , $ f \text { cos } ( \theta ) $ , is doing work on the box since the box is being displaced horizontally . this means that a more general equation for the work done on the box by a force at an angle θ could be written as : $ w=f_ { || } \cdot\delta x $ $ w= ( f \cos { \theta } ) \cdot\delta x $ which is more often written as , $ \large w=f \delta x\cos { \theta } $ exercise : suppose we use a rope to pull the box , and the angle between the rope and the ground is 30º . this time we pull along the rope with a force of 500 n. how much of a chocolate bar can we eat this time if we pull the box through the same 585 m ? what about lifting weights instead ? in the previous example , we were doing work on a box which we were pushing around a floor . in doing so , we were working against a frictional force . another common form of exercise is lifting weights . in this case we are working against the force of gravity rather than friction . using newton 's laws we can find the force , $ f $ , required to lift a weight with mass $ m $ straight up , placing it on a rack which is at a height $ h $ above us : $ f=mg $ the change in position—previously $ \delta x $ —is simply the height , so the work , $ w $ , that we have done in lifting the weight is then $ w=m g h $ the exercise we have done in lifting the weight has resulted in energy being stored in the form of gravitational potential energy . it is called potential energy because it has the potential to be released at any moment with a crash as the weight falls back to the ground . we did positive work on the weight since we exerted our force in the same direction as the displacement of the weight , i.e. , upward . the work done by gravity on the weight while it was lifted was negative since the force of gravity is directed in the opposite direction to the displacement . also , since the weight is stationary after the lift , we know that the work that we have done is exactly canceled out by the work done by gravity . the work done by us is $ mgh $ , and the work done by gravity is $ -mgh $ . we will talk more about this when we look into kinetic energy . ok , let 's put some numbers in and find how much of that chocolate bar we would offset by lifting a weight of 50 kg up to a height of 0.5 m. the work done to the weight is $ w= ( 50\mathrm { kg } ) ( 9.81\mathrm { m/s^2 } ) ( 0.5\mathrm { m } ) = 245.25 \mathrm { j } $ ok , so how many 280 calorie—i.e. , $ 1.17 \times 10^6 $ joule—chocolate bars is this ? well , 245.25 j is about $ \dfrac { 1 } { 4770 } $ of a chocolate bar . but remember , our bodies are only about 25 % efficient , so the work done by the person is actually four times larger , about 981.8 j , which is $ \dfrac { 1 } { 1190 } $ chocolate bars . so , if we can lift this weight once every 2 seconds , it will take us about 2380 seconds or 40 minutes of hard work to burn off this chocolate bar ! what about simply holding a weight stationary ? one frequent source of confusion people have with the concept of work comes about when thinking about holding a heavy weight stationary above our heads , against the force of gravity . we are not moving the weight through any distance , so no work is being done to the weight . we could also achieve this by placing the weight on a table ; it is clear that the table is not doing any work to keep the weight in position . yet , we know from our experience that we get tired when doing the same job . so what is going on here ? it turns out that what is actually happening here is that our bodies are doing work on our muscles to maintain the necessary tension to hold the weight up . the body does this by sending a cascade of nerve impulses to each muscle . each impulse causes the muscle to momentarily contract and release . this all happens so fast that we might only notice a slight twitching at first . eventually though , not enough chemical energy is available in the muscle and it can no longer keep up . we then begin to shake and eventually must rest for a while . so work is being done , it is just not being done on the weight .
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it is called potential energy because it has the potential to be released at any moment with a crash as the weight falls back to the ground . we did positive work on the weight since we exerted our force in the same direction as the displacement of the weight , i.e. , upward . the work done by gravity on the weight while it was lifted was negative since the force of gravity is directed in the opposite direction to the displacement .
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wait ... why is the force to lift the weight equals mg ?
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what does energy and work mean ? energy is a word which tends to be used a lot in everyday life . though it is often used quite loosely , it does have a very specific physical meaning . energy is a measurement of the ability of something to do work . it is not a material substance . energy can be stored and measured in many forms . although we often hear people talking about energy consumption , energy is never really destroyed . it is just transferred from one form to another , doing work in the process . some forms of energy are less useful to us than others—for example , low level heat energy . it is better to talk about the consumption or extraction of energy resources , for example coal , oil , or wind , than consumption of energy itself . a speeding bullet has a measurable amount of energy associated with it ; this is known as kinetic energy . the bullet gained this energy because work was done on it by a charge of gunpowder which lost some chemical potential energy in the process . a hot cup of coffee has a measurable amount of thermal energy which it acquired via work done by a microwave oven , which in turn took electrical energy from the electrical grid . in practice , whenever work is done to move energy from one form to another , there is always some loss to other forms of energy such as heat and sound . for example , a traditional light bulb is only about 3 % efficient at converting electrical energy to visible light , while a human being is about 25 % efficient at converting chemical energy from food into work . how do we measure energy and work ? the standard unit used to measure energy and work done in physics is the joule , which has the symbol j . in mechanics , 1 joule is the energy transferred when a force of 1 newton is applied to an object and moves it through a distance of 1 meter . another unit of energy you may have come across is the calorie . the amount of energy in an item of food is often written in calories on the back of the packet . a typical 60 gram chocolate bar for example contains about 280 calories of energy . one calorie is the amount of energy required to raise 1 kg of water by 1 $ ^ { \circ } $ celsius . this is equal to 4184 joules , so one chocolate bar has 1.17 million joules or 1.17 mj of stored energy . that 's a lot of joules ! how long do i have to push a heavy box around to burn off one chocolate bar ? suppose we 're feeling guilty about eating a chocolate bar ; we want to find how much exercise we need to do to offset those extra 280 calories . let 's consider a simple form of exercise : pushing a heavy box around a room , see figure 1 below . using a bathroom scale between ourselves and the box , we find that we can push with a force of 500 n. meanwhile , we use a stopwatch and measuring tape to measure our speed . this comes out to be 0.25 meters per second . so how much work do we need to do to the box to burn off the candy bar ? the definition of work , $ w $ , is below : $ \large w = f\cdot \delta x $ the work we need to do to burn the energy in the candy bar is $ e=280 \mathrm { cal } \cdot 4184 \mathrm { j/cal } =1.17 \mathrm { mj } $ . therefore , the distance , $ δx $ , we need to move the box through is : $ \begin { align } w & amp ; = f\cdot \delta x \ 1.17 \text { mj } & amp ; = ( 500 \text { n } ) \cdot \delta x \ \dfrac { 1.17\times 10^6 \text { j } } { 500 \text { n } } & amp ; = \delta x \ 2,340 \text { m } & amp ; = \delta x \end { align } $ remember , however , that our bodies are about 25 % efficient at transferring stored energy from food into work . the actual energy we will offset is then four times as high as the work done to the box . so , we only need to push the box through a distance of 585 m , which is still over five football fields long . given the known speed of 0.25 m/s that will take us : $ \frac { 585 \mathrm { m } } { 0.25 \mathrm { m/s } } =2340 \mathrm { s } $ exercise : suppose that the force that we apply to the box , see figure 1 above , is initially reduced but increases to a constant value as we warm up . for instance , in the graph below we see that as the box is displaced further—i.e. , $ x $ gets larger—the force , $ f $ , increases for the first 30 m , see figure 2 below . how could we find the work done during the period where the force is changing ? if the force is not constant , one way to determine the work done is to divide the problem up into small sections over which the change is negligible and sum up the work done in each section . just as we have learned when looking at velocity time graphs , this can be done by calculating the area under the curve using geometry . the work done by a force is equal to the area under a force vs. position graph . in the case of figure 2 , it would be : $ ( 200\text { n } \cdot 30 \text { m } ) + \frac { 1 } { 2 } \left ( ( 500\text { n } -200\text { n } ) \cdot 30 \text { m } \right ) =10500 \mathrm { j } $ for the initial $ 30 \text { m } $ of displacement . similarly , the work done for the final 40 m of displacement would be : $ 500\text { n } \cdot 40 \text { m } =20,000 \mathrm { j } $ what if we are n't pushing straight on ? there is one thing we need to watch out for when doing these problems . the previous equation , $ w = f \cdot \delta x $ , does n't take into account situations where the force we are applying is not in the same direction as the motion . for instance , imagine we use a rope to pull on the box . in that case there will be an angle between the rope and the ground . to untangle this situation , we begin by drawing a triangle to separate out the horizontal and vertical components of the applied force . the key point here is that it is only the component of the force , $ f_ { || } $ , that lies parallel to the displacement that does work on an object . in the case of the box shown above , only the horizontal component of the applied force , $ f \text { cos } ( \theta ) $ , is doing work on the box since the box is being displaced horizontally . this means that a more general equation for the work done on the box by a force at an angle θ could be written as : $ w=f_ { || } \cdot\delta x $ $ w= ( f \cos { \theta } ) \cdot\delta x $ which is more often written as , $ \large w=f \delta x\cos { \theta } $ exercise : suppose we use a rope to pull the box , and the angle between the rope and the ground is 30º . this time we pull along the rope with a force of 500 n. how much of a chocolate bar can we eat this time if we pull the box through the same 585 m ? what about lifting weights instead ? in the previous example , we were doing work on a box which we were pushing around a floor . in doing so , we were working against a frictional force . another common form of exercise is lifting weights . in this case we are working against the force of gravity rather than friction . using newton 's laws we can find the force , $ f $ , required to lift a weight with mass $ m $ straight up , placing it on a rack which is at a height $ h $ above us : $ f=mg $ the change in position—previously $ \delta x $ —is simply the height , so the work , $ w $ , that we have done in lifting the weight is then $ w=m g h $ the exercise we have done in lifting the weight has resulted in energy being stored in the form of gravitational potential energy . it is called potential energy because it has the potential to be released at any moment with a crash as the weight falls back to the ground . we did positive work on the weight since we exerted our force in the same direction as the displacement of the weight , i.e. , upward . the work done by gravity on the weight while it was lifted was negative since the force of gravity is directed in the opposite direction to the displacement . also , since the weight is stationary after the lift , we know that the work that we have done is exactly canceled out by the work done by gravity . the work done by us is $ mgh $ , and the work done by gravity is $ -mgh $ . we will talk more about this when we look into kinetic energy . ok , let 's put some numbers in and find how much of that chocolate bar we would offset by lifting a weight of 50 kg up to a height of 0.5 m. the work done to the weight is $ w= ( 50\mathrm { kg } ) ( 9.81\mathrm { m/s^2 } ) ( 0.5\mathrm { m } ) = 245.25 \mathrm { j } $ ok , so how many 280 calorie—i.e. , $ 1.17 \times 10^6 $ joule—chocolate bars is this ? well , 245.25 j is about $ \dfrac { 1 } { 4770 } $ of a chocolate bar . but remember , our bodies are only about 25 % efficient , so the work done by the person is actually four times larger , about 981.8 j , which is $ \dfrac { 1 } { 1190 } $ chocolate bars . so , if we can lift this weight once every 2 seconds , it will take us about 2380 seconds or 40 minutes of hard work to burn off this chocolate bar ! what about simply holding a weight stationary ? one frequent source of confusion people have with the concept of work comes about when thinking about holding a heavy weight stationary above our heads , against the force of gravity . we are not moving the weight through any distance , so no work is being done to the weight . we could also achieve this by placing the weight on a table ; it is clear that the table is not doing any work to keep the weight in position . yet , we know from our experience that we get tired when doing the same job . so what is going on here ? it turns out that what is actually happening here is that our bodies are doing work on our muscles to maintain the necessary tension to hold the weight up . the body does this by sending a cascade of nerve impulses to each muscle . each impulse causes the muscle to momentarily contract and release . this all happens so fast that we might only notice a slight twitching at first . eventually though , not enough chemical energy is available in the muscle and it can no longer keep up . we then begin to shake and eventually must rest for a while . so work is being done , it is just not being done on the weight .
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one frequent source of confusion people have with the concept of work comes about when thinking about holding a heavy weight stationary above our heads , against the force of gravity . we are not moving the weight through any distance , so no work is being done to the weight . we could also achieve this by placing the weight on a table ; it is clear that the table is not doing any work to keep the weight in position .
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how is it that we can do less work over the same distance by pulling at an angle ?
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what does energy and work mean ? energy is a word which tends to be used a lot in everyday life . though it is often used quite loosely , it does have a very specific physical meaning . energy is a measurement of the ability of something to do work . it is not a material substance . energy can be stored and measured in many forms . although we often hear people talking about energy consumption , energy is never really destroyed . it is just transferred from one form to another , doing work in the process . some forms of energy are less useful to us than others—for example , low level heat energy . it is better to talk about the consumption or extraction of energy resources , for example coal , oil , or wind , than consumption of energy itself . a speeding bullet has a measurable amount of energy associated with it ; this is known as kinetic energy . the bullet gained this energy because work was done on it by a charge of gunpowder which lost some chemical potential energy in the process . a hot cup of coffee has a measurable amount of thermal energy which it acquired via work done by a microwave oven , which in turn took electrical energy from the electrical grid . in practice , whenever work is done to move energy from one form to another , there is always some loss to other forms of energy such as heat and sound . for example , a traditional light bulb is only about 3 % efficient at converting electrical energy to visible light , while a human being is about 25 % efficient at converting chemical energy from food into work . how do we measure energy and work ? the standard unit used to measure energy and work done in physics is the joule , which has the symbol j . in mechanics , 1 joule is the energy transferred when a force of 1 newton is applied to an object and moves it through a distance of 1 meter . another unit of energy you may have come across is the calorie . the amount of energy in an item of food is often written in calories on the back of the packet . a typical 60 gram chocolate bar for example contains about 280 calories of energy . one calorie is the amount of energy required to raise 1 kg of water by 1 $ ^ { \circ } $ celsius . this is equal to 4184 joules , so one chocolate bar has 1.17 million joules or 1.17 mj of stored energy . that 's a lot of joules ! how long do i have to push a heavy box around to burn off one chocolate bar ? suppose we 're feeling guilty about eating a chocolate bar ; we want to find how much exercise we need to do to offset those extra 280 calories . let 's consider a simple form of exercise : pushing a heavy box around a room , see figure 1 below . using a bathroom scale between ourselves and the box , we find that we can push with a force of 500 n. meanwhile , we use a stopwatch and measuring tape to measure our speed . this comes out to be 0.25 meters per second . so how much work do we need to do to the box to burn off the candy bar ? the definition of work , $ w $ , is below : $ \large w = f\cdot \delta x $ the work we need to do to burn the energy in the candy bar is $ e=280 \mathrm { cal } \cdot 4184 \mathrm { j/cal } =1.17 \mathrm { mj } $ . therefore , the distance , $ δx $ , we need to move the box through is : $ \begin { align } w & amp ; = f\cdot \delta x \ 1.17 \text { mj } & amp ; = ( 500 \text { n } ) \cdot \delta x \ \dfrac { 1.17\times 10^6 \text { j } } { 500 \text { n } } & amp ; = \delta x \ 2,340 \text { m } & amp ; = \delta x \end { align } $ remember , however , that our bodies are about 25 % efficient at transferring stored energy from food into work . the actual energy we will offset is then four times as high as the work done to the box . so , we only need to push the box through a distance of 585 m , which is still over five football fields long . given the known speed of 0.25 m/s that will take us : $ \frac { 585 \mathrm { m } } { 0.25 \mathrm { m/s } } =2340 \mathrm { s } $ exercise : suppose that the force that we apply to the box , see figure 1 above , is initially reduced but increases to a constant value as we warm up . for instance , in the graph below we see that as the box is displaced further—i.e. , $ x $ gets larger—the force , $ f $ , increases for the first 30 m , see figure 2 below . how could we find the work done during the period where the force is changing ? if the force is not constant , one way to determine the work done is to divide the problem up into small sections over which the change is negligible and sum up the work done in each section . just as we have learned when looking at velocity time graphs , this can be done by calculating the area under the curve using geometry . the work done by a force is equal to the area under a force vs. position graph . in the case of figure 2 , it would be : $ ( 200\text { n } \cdot 30 \text { m } ) + \frac { 1 } { 2 } \left ( ( 500\text { n } -200\text { n } ) \cdot 30 \text { m } \right ) =10500 \mathrm { j } $ for the initial $ 30 \text { m } $ of displacement . similarly , the work done for the final 40 m of displacement would be : $ 500\text { n } \cdot 40 \text { m } =20,000 \mathrm { j } $ what if we are n't pushing straight on ? there is one thing we need to watch out for when doing these problems . the previous equation , $ w = f \cdot \delta x $ , does n't take into account situations where the force we are applying is not in the same direction as the motion . for instance , imagine we use a rope to pull on the box . in that case there will be an angle between the rope and the ground . to untangle this situation , we begin by drawing a triangle to separate out the horizontal and vertical components of the applied force . the key point here is that it is only the component of the force , $ f_ { || } $ , that lies parallel to the displacement that does work on an object . in the case of the box shown above , only the horizontal component of the applied force , $ f \text { cos } ( \theta ) $ , is doing work on the box since the box is being displaced horizontally . this means that a more general equation for the work done on the box by a force at an angle θ could be written as : $ w=f_ { || } \cdot\delta x $ $ w= ( f \cos { \theta } ) \cdot\delta x $ which is more often written as , $ \large w=f \delta x\cos { \theta } $ exercise : suppose we use a rope to pull the box , and the angle between the rope and the ground is 30º . this time we pull along the rope with a force of 500 n. how much of a chocolate bar can we eat this time if we pull the box through the same 585 m ? what about lifting weights instead ? in the previous example , we were doing work on a box which we were pushing around a floor . in doing so , we were working against a frictional force . another common form of exercise is lifting weights . in this case we are working against the force of gravity rather than friction . using newton 's laws we can find the force , $ f $ , required to lift a weight with mass $ m $ straight up , placing it on a rack which is at a height $ h $ above us : $ f=mg $ the change in position—previously $ \delta x $ —is simply the height , so the work , $ w $ , that we have done in lifting the weight is then $ w=m g h $ the exercise we have done in lifting the weight has resulted in energy being stored in the form of gravitational potential energy . it is called potential energy because it has the potential to be released at any moment with a crash as the weight falls back to the ground . we did positive work on the weight since we exerted our force in the same direction as the displacement of the weight , i.e. , upward . the work done by gravity on the weight while it was lifted was negative since the force of gravity is directed in the opposite direction to the displacement . also , since the weight is stationary after the lift , we know that the work that we have done is exactly canceled out by the work done by gravity . the work done by us is $ mgh $ , and the work done by gravity is $ -mgh $ . we will talk more about this when we look into kinetic energy . ok , let 's put some numbers in and find how much of that chocolate bar we would offset by lifting a weight of 50 kg up to a height of 0.5 m. the work done to the weight is $ w= ( 50\mathrm { kg } ) ( 9.81\mathrm { m/s^2 } ) ( 0.5\mathrm { m } ) = 245.25 \mathrm { j } $ ok , so how many 280 calorie—i.e. , $ 1.17 \times 10^6 $ joule—chocolate bars is this ? well , 245.25 j is about $ \dfrac { 1 } { 4770 } $ of a chocolate bar . but remember , our bodies are only about 25 % efficient , so the work done by the person is actually four times larger , about 981.8 j , which is $ \dfrac { 1 } { 1190 } $ chocolate bars . so , if we can lift this weight once every 2 seconds , it will take us about 2380 seconds or 40 minutes of hard work to burn off this chocolate bar ! what about simply holding a weight stationary ? one frequent source of confusion people have with the concept of work comes about when thinking about holding a heavy weight stationary above our heads , against the force of gravity . we are not moving the weight through any distance , so no work is being done to the weight . we could also achieve this by placing the weight on a table ; it is clear that the table is not doing any work to keep the weight in position . yet , we know from our experience that we get tired when doing the same job . so what is going on here ? it turns out that what is actually happening here is that our bodies are doing work on our muscles to maintain the necessary tension to hold the weight up . the body does this by sending a cascade of nerve impulses to each muscle . each impulse causes the muscle to momentarily contract and release . this all happens so fast that we might only notice a slight twitching at first . eventually though , not enough chemical energy is available in the muscle and it can no longer keep up . we then begin to shake and eventually must rest for a while . so work is being done , it is just not being done on the weight .
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it is not a material substance . energy can be stored and measured in many forms . although we often hear people talking about energy consumption , energy is never really destroyed . it is just transferred from one form to another , doing work in the process .
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in the first question , why do we consider only the horizontal energy that got extracted , instead of the total force the person extracted ?
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what does energy and work mean ? energy is a word which tends to be used a lot in everyday life . though it is often used quite loosely , it does have a very specific physical meaning . energy is a measurement of the ability of something to do work . it is not a material substance . energy can be stored and measured in many forms . although we often hear people talking about energy consumption , energy is never really destroyed . it is just transferred from one form to another , doing work in the process . some forms of energy are less useful to us than others—for example , low level heat energy . it is better to talk about the consumption or extraction of energy resources , for example coal , oil , or wind , than consumption of energy itself . a speeding bullet has a measurable amount of energy associated with it ; this is known as kinetic energy . the bullet gained this energy because work was done on it by a charge of gunpowder which lost some chemical potential energy in the process . a hot cup of coffee has a measurable amount of thermal energy which it acquired via work done by a microwave oven , which in turn took electrical energy from the electrical grid . in practice , whenever work is done to move energy from one form to another , there is always some loss to other forms of energy such as heat and sound . for example , a traditional light bulb is only about 3 % efficient at converting electrical energy to visible light , while a human being is about 25 % efficient at converting chemical energy from food into work . how do we measure energy and work ? the standard unit used to measure energy and work done in physics is the joule , which has the symbol j . in mechanics , 1 joule is the energy transferred when a force of 1 newton is applied to an object and moves it through a distance of 1 meter . another unit of energy you may have come across is the calorie . the amount of energy in an item of food is often written in calories on the back of the packet . a typical 60 gram chocolate bar for example contains about 280 calories of energy . one calorie is the amount of energy required to raise 1 kg of water by 1 $ ^ { \circ } $ celsius . this is equal to 4184 joules , so one chocolate bar has 1.17 million joules or 1.17 mj of stored energy . that 's a lot of joules ! how long do i have to push a heavy box around to burn off one chocolate bar ? suppose we 're feeling guilty about eating a chocolate bar ; we want to find how much exercise we need to do to offset those extra 280 calories . let 's consider a simple form of exercise : pushing a heavy box around a room , see figure 1 below . using a bathroom scale between ourselves and the box , we find that we can push with a force of 500 n. meanwhile , we use a stopwatch and measuring tape to measure our speed . this comes out to be 0.25 meters per second . so how much work do we need to do to the box to burn off the candy bar ? the definition of work , $ w $ , is below : $ \large w = f\cdot \delta x $ the work we need to do to burn the energy in the candy bar is $ e=280 \mathrm { cal } \cdot 4184 \mathrm { j/cal } =1.17 \mathrm { mj } $ . therefore , the distance , $ δx $ , we need to move the box through is : $ \begin { align } w & amp ; = f\cdot \delta x \ 1.17 \text { mj } & amp ; = ( 500 \text { n } ) \cdot \delta x \ \dfrac { 1.17\times 10^6 \text { j } } { 500 \text { n } } & amp ; = \delta x \ 2,340 \text { m } & amp ; = \delta x \end { align } $ remember , however , that our bodies are about 25 % efficient at transferring stored energy from food into work . the actual energy we will offset is then four times as high as the work done to the box . so , we only need to push the box through a distance of 585 m , which is still over five football fields long . given the known speed of 0.25 m/s that will take us : $ \frac { 585 \mathrm { m } } { 0.25 \mathrm { m/s } } =2340 \mathrm { s } $ exercise : suppose that the force that we apply to the box , see figure 1 above , is initially reduced but increases to a constant value as we warm up . for instance , in the graph below we see that as the box is displaced further—i.e. , $ x $ gets larger—the force , $ f $ , increases for the first 30 m , see figure 2 below . how could we find the work done during the period where the force is changing ? if the force is not constant , one way to determine the work done is to divide the problem up into small sections over which the change is negligible and sum up the work done in each section . just as we have learned when looking at velocity time graphs , this can be done by calculating the area under the curve using geometry . the work done by a force is equal to the area under a force vs. position graph . in the case of figure 2 , it would be : $ ( 200\text { n } \cdot 30 \text { m } ) + \frac { 1 } { 2 } \left ( ( 500\text { n } -200\text { n } ) \cdot 30 \text { m } \right ) =10500 \mathrm { j } $ for the initial $ 30 \text { m } $ of displacement . similarly , the work done for the final 40 m of displacement would be : $ 500\text { n } \cdot 40 \text { m } =20,000 \mathrm { j } $ what if we are n't pushing straight on ? there is one thing we need to watch out for when doing these problems . the previous equation , $ w = f \cdot \delta x $ , does n't take into account situations where the force we are applying is not in the same direction as the motion . for instance , imagine we use a rope to pull on the box . in that case there will be an angle between the rope and the ground . to untangle this situation , we begin by drawing a triangle to separate out the horizontal and vertical components of the applied force . the key point here is that it is only the component of the force , $ f_ { || } $ , that lies parallel to the displacement that does work on an object . in the case of the box shown above , only the horizontal component of the applied force , $ f \text { cos } ( \theta ) $ , is doing work on the box since the box is being displaced horizontally . this means that a more general equation for the work done on the box by a force at an angle θ could be written as : $ w=f_ { || } \cdot\delta x $ $ w= ( f \cos { \theta } ) \cdot\delta x $ which is more often written as , $ \large w=f \delta x\cos { \theta } $ exercise : suppose we use a rope to pull the box , and the angle between the rope and the ground is 30º . this time we pull along the rope with a force of 500 n. how much of a chocolate bar can we eat this time if we pull the box through the same 585 m ? what about lifting weights instead ? in the previous example , we were doing work on a box which we were pushing around a floor . in doing so , we were working against a frictional force . another common form of exercise is lifting weights . in this case we are working against the force of gravity rather than friction . using newton 's laws we can find the force , $ f $ , required to lift a weight with mass $ m $ straight up , placing it on a rack which is at a height $ h $ above us : $ f=mg $ the change in position—previously $ \delta x $ —is simply the height , so the work , $ w $ , that we have done in lifting the weight is then $ w=m g h $ the exercise we have done in lifting the weight has resulted in energy being stored in the form of gravitational potential energy . it is called potential energy because it has the potential to be released at any moment with a crash as the weight falls back to the ground . we did positive work on the weight since we exerted our force in the same direction as the displacement of the weight , i.e. , upward . the work done by gravity on the weight while it was lifted was negative since the force of gravity is directed in the opposite direction to the displacement . also , since the weight is stationary after the lift , we know that the work that we have done is exactly canceled out by the work done by gravity . the work done by us is $ mgh $ , and the work done by gravity is $ -mgh $ . we will talk more about this when we look into kinetic energy . ok , let 's put some numbers in and find how much of that chocolate bar we would offset by lifting a weight of 50 kg up to a height of 0.5 m. the work done to the weight is $ w= ( 50\mathrm { kg } ) ( 9.81\mathrm { m/s^2 } ) ( 0.5\mathrm { m } ) = 245.25 \mathrm { j } $ ok , so how many 280 calorie—i.e. , $ 1.17 \times 10^6 $ joule—chocolate bars is this ? well , 245.25 j is about $ \dfrac { 1 } { 4770 } $ of a chocolate bar . but remember , our bodies are only about 25 % efficient , so the work done by the person is actually four times larger , about 981.8 j , which is $ \dfrac { 1 } { 1190 } $ chocolate bars . so , if we can lift this weight once every 2 seconds , it will take us about 2380 seconds or 40 minutes of hard work to burn off this chocolate bar ! what about simply holding a weight stationary ? one frequent source of confusion people have with the concept of work comes about when thinking about holding a heavy weight stationary above our heads , against the force of gravity . we are not moving the weight through any distance , so no work is being done to the weight . we could also achieve this by placing the weight on a table ; it is clear that the table is not doing any work to keep the weight in position . yet , we know from our experience that we get tired when doing the same job . so what is going on here ? it turns out that what is actually happening here is that our bodies are doing work on our muscles to maintain the necessary tension to hold the weight up . the body does this by sending a cascade of nerve impulses to each muscle . each impulse causes the muscle to momentarily contract and release . this all happens so fast that we might only notice a slight twitching at first . eventually though , not enough chemical energy is available in the muscle and it can no longer keep up . we then begin to shake and eventually must rest for a while . so work is being done , it is just not being done on the weight .
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there is one thing we need to watch out for when doing these problems . the previous equation , $ w = f \cdot \delta x $ , does n't take into account situations where the force we are applying is not in the same direction as the motion . for instance , imagine we use a rope to pull on the box .
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is n't 500n what we should take in to account ?
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what does energy and work mean ? energy is a word which tends to be used a lot in everyday life . though it is often used quite loosely , it does have a very specific physical meaning . energy is a measurement of the ability of something to do work . it is not a material substance . energy can be stored and measured in many forms . although we often hear people talking about energy consumption , energy is never really destroyed . it is just transferred from one form to another , doing work in the process . some forms of energy are less useful to us than others—for example , low level heat energy . it is better to talk about the consumption or extraction of energy resources , for example coal , oil , or wind , than consumption of energy itself . a speeding bullet has a measurable amount of energy associated with it ; this is known as kinetic energy . the bullet gained this energy because work was done on it by a charge of gunpowder which lost some chemical potential energy in the process . a hot cup of coffee has a measurable amount of thermal energy which it acquired via work done by a microwave oven , which in turn took electrical energy from the electrical grid . in practice , whenever work is done to move energy from one form to another , there is always some loss to other forms of energy such as heat and sound . for example , a traditional light bulb is only about 3 % efficient at converting electrical energy to visible light , while a human being is about 25 % efficient at converting chemical energy from food into work . how do we measure energy and work ? the standard unit used to measure energy and work done in physics is the joule , which has the symbol j . in mechanics , 1 joule is the energy transferred when a force of 1 newton is applied to an object and moves it through a distance of 1 meter . another unit of energy you may have come across is the calorie . the amount of energy in an item of food is often written in calories on the back of the packet . a typical 60 gram chocolate bar for example contains about 280 calories of energy . one calorie is the amount of energy required to raise 1 kg of water by 1 $ ^ { \circ } $ celsius . this is equal to 4184 joules , so one chocolate bar has 1.17 million joules or 1.17 mj of stored energy . that 's a lot of joules ! how long do i have to push a heavy box around to burn off one chocolate bar ? suppose we 're feeling guilty about eating a chocolate bar ; we want to find how much exercise we need to do to offset those extra 280 calories . let 's consider a simple form of exercise : pushing a heavy box around a room , see figure 1 below . using a bathroom scale between ourselves and the box , we find that we can push with a force of 500 n. meanwhile , we use a stopwatch and measuring tape to measure our speed . this comes out to be 0.25 meters per second . so how much work do we need to do to the box to burn off the candy bar ? the definition of work , $ w $ , is below : $ \large w = f\cdot \delta x $ the work we need to do to burn the energy in the candy bar is $ e=280 \mathrm { cal } \cdot 4184 \mathrm { j/cal } =1.17 \mathrm { mj } $ . therefore , the distance , $ δx $ , we need to move the box through is : $ \begin { align } w & amp ; = f\cdot \delta x \ 1.17 \text { mj } & amp ; = ( 500 \text { n } ) \cdot \delta x \ \dfrac { 1.17\times 10^6 \text { j } } { 500 \text { n } } & amp ; = \delta x \ 2,340 \text { m } & amp ; = \delta x \end { align } $ remember , however , that our bodies are about 25 % efficient at transferring stored energy from food into work . the actual energy we will offset is then four times as high as the work done to the box . so , we only need to push the box through a distance of 585 m , which is still over five football fields long . given the known speed of 0.25 m/s that will take us : $ \frac { 585 \mathrm { m } } { 0.25 \mathrm { m/s } } =2340 \mathrm { s } $ exercise : suppose that the force that we apply to the box , see figure 1 above , is initially reduced but increases to a constant value as we warm up . for instance , in the graph below we see that as the box is displaced further—i.e. , $ x $ gets larger—the force , $ f $ , increases for the first 30 m , see figure 2 below . how could we find the work done during the period where the force is changing ? if the force is not constant , one way to determine the work done is to divide the problem up into small sections over which the change is negligible and sum up the work done in each section . just as we have learned when looking at velocity time graphs , this can be done by calculating the area under the curve using geometry . the work done by a force is equal to the area under a force vs. position graph . in the case of figure 2 , it would be : $ ( 200\text { n } \cdot 30 \text { m } ) + \frac { 1 } { 2 } \left ( ( 500\text { n } -200\text { n } ) \cdot 30 \text { m } \right ) =10500 \mathrm { j } $ for the initial $ 30 \text { m } $ of displacement . similarly , the work done for the final 40 m of displacement would be : $ 500\text { n } \cdot 40 \text { m } =20,000 \mathrm { j } $ what if we are n't pushing straight on ? there is one thing we need to watch out for when doing these problems . the previous equation , $ w = f \cdot \delta x $ , does n't take into account situations where the force we are applying is not in the same direction as the motion . for instance , imagine we use a rope to pull on the box . in that case there will be an angle between the rope and the ground . to untangle this situation , we begin by drawing a triangle to separate out the horizontal and vertical components of the applied force . the key point here is that it is only the component of the force , $ f_ { || } $ , that lies parallel to the displacement that does work on an object . in the case of the box shown above , only the horizontal component of the applied force , $ f \text { cos } ( \theta ) $ , is doing work on the box since the box is being displaced horizontally . this means that a more general equation for the work done on the box by a force at an angle θ could be written as : $ w=f_ { || } \cdot\delta x $ $ w= ( f \cos { \theta } ) \cdot\delta x $ which is more often written as , $ \large w=f \delta x\cos { \theta } $ exercise : suppose we use a rope to pull the box , and the angle between the rope and the ground is 30º . this time we pull along the rope with a force of 500 n. how much of a chocolate bar can we eat this time if we pull the box through the same 585 m ? what about lifting weights instead ? in the previous example , we were doing work on a box which we were pushing around a floor . in doing so , we were working against a frictional force . another common form of exercise is lifting weights . in this case we are working against the force of gravity rather than friction . using newton 's laws we can find the force , $ f $ , required to lift a weight with mass $ m $ straight up , placing it on a rack which is at a height $ h $ above us : $ f=mg $ the change in position—previously $ \delta x $ —is simply the height , so the work , $ w $ , that we have done in lifting the weight is then $ w=m g h $ the exercise we have done in lifting the weight has resulted in energy being stored in the form of gravitational potential energy . it is called potential energy because it has the potential to be released at any moment with a crash as the weight falls back to the ground . we did positive work on the weight since we exerted our force in the same direction as the displacement of the weight , i.e. , upward . the work done by gravity on the weight while it was lifted was negative since the force of gravity is directed in the opposite direction to the displacement . also , since the weight is stationary after the lift , we know that the work that we have done is exactly canceled out by the work done by gravity . the work done by us is $ mgh $ , and the work done by gravity is $ -mgh $ . we will talk more about this when we look into kinetic energy . ok , let 's put some numbers in and find how much of that chocolate bar we would offset by lifting a weight of 50 kg up to a height of 0.5 m. the work done to the weight is $ w= ( 50\mathrm { kg } ) ( 9.81\mathrm { m/s^2 } ) ( 0.5\mathrm { m } ) = 245.25 \mathrm { j } $ ok , so how many 280 calorie—i.e. , $ 1.17 \times 10^6 $ joule—chocolate bars is this ? well , 245.25 j is about $ \dfrac { 1 } { 4770 } $ of a chocolate bar . but remember , our bodies are only about 25 % efficient , so the work done by the person is actually four times larger , about 981.8 j , which is $ \dfrac { 1 } { 1190 } $ chocolate bars . so , if we can lift this weight once every 2 seconds , it will take us about 2380 seconds or 40 minutes of hard work to burn off this chocolate bar ! what about simply holding a weight stationary ? one frequent source of confusion people have with the concept of work comes about when thinking about holding a heavy weight stationary above our heads , against the force of gravity . we are not moving the weight through any distance , so no work is being done to the weight . we could also achieve this by placing the weight on a table ; it is clear that the table is not doing any work to keep the weight in position . yet , we know from our experience that we get tired when doing the same job . so what is going on here ? it turns out that what is actually happening here is that our bodies are doing work on our muscles to maintain the necessary tension to hold the weight up . the body does this by sending a cascade of nerve impulses to each muscle . each impulse causes the muscle to momentarily contract and release . this all happens so fast that we might only notice a slight twitching at first . eventually though , not enough chemical energy is available in the muscle and it can no longer keep up . we then begin to shake and eventually must rest for a while . so work is being done , it is just not being done on the weight .
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energy can be stored and measured in many forms . although we often hear people talking about energy consumption , energy is never really destroyed . it is just transferred from one form to another , doing work in the process .
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however , is it mathematically impossible to create or destroy energy , or has it just never happened ?
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what does energy and work mean ? energy is a word which tends to be used a lot in everyday life . though it is often used quite loosely , it does have a very specific physical meaning . energy is a measurement of the ability of something to do work . it is not a material substance . energy can be stored and measured in many forms . although we often hear people talking about energy consumption , energy is never really destroyed . it is just transferred from one form to another , doing work in the process . some forms of energy are less useful to us than others—for example , low level heat energy . it is better to talk about the consumption or extraction of energy resources , for example coal , oil , or wind , than consumption of energy itself . a speeding bullet has a measurable amount of energy associated with it ; this is known as kinetic energy . the bullet gained this energy because work was done on it by a charge of gunpowder which lost some chemical potential energy in the process . a hot cup of coffee has a measurable amount of thermal energy which it acquired via work done by a microwave oven , which in turn took electrical energy from the electrical grid . in practice , whenever work is done to move energy from one form to another , there is always some loss to other forms of energy such as heat and sound . for example , a traditional light bulb is only about 3 % efficient at converting electrical energy to visible light , while a human being is about 25 % efficient at converting chemical energy from food into work . how do we measure energy and work ? the standard unit used to measure energy and work done in physics is the joule , which has the symbol j . in mechanics , 1 joule is the energy transferred when a force of 1 newton is applied to an object and moves it through a distance of 1 meter . another unit of energy you may have come across is the calorie . the amount of energy in an item of food is often written in calories on the back of the packet . a typical 60 gram chocolate bar for example contains about 280 calories of energy . one calorie is the amount of energy required to raise 1 kg of water by 1 $ ^ { \circ } $ celsius . this is equal to 4184 joules , so one chocolate bar has 1.17 million joules or 1.17 mj of stored energy . that 's a lot of joules ! how long do i have to push a heavy box around to burn off one chocolate bar ? suppose we 're feeling guilty about eating a chocolate bar ; we want to find how much exercise we need to do to offset those extra 280 calories . let 's consider a simple form of exercise : pushing a heavy box around a room , see figure 1 below . using a bathroom scale between ourselves and the box , we find that we can push with a force of 500 n. meanwhile , we use a stopwatch and measuring tape to measure our speed . this comes out to be 0.25 meters per second . so how much work do we need to do to the box to burn off the candy bar ? the definition of work , $ w $ , is below : $ \large w = f\cdot \delta x $ the work we need to do to burn the energy in the candy bar is $ e=280 \mathrm { cal } \cdot 4184 \mathrm { j/cal } =1.17 \mathrm { mj } $ . therefore , the distance , $ δx $ , we need to move the box through is : $ \begin { align } w & amp ; = f\cdot \delta x \ 1.17 \text { mj } & amp ; = ( 500 \text { n } ) \cdot \delta x \ \dfrac { 1.17\times 10^6 \text { j } } { 500 \text { n } } & amp ; = \delta x \ 2,340 \text { m } & amp ; = \delta x \end { align } $ remember , however , that our bodies are about 25 % efficient at transferring stored energy from food into work . the actual energy we will offset is then four times as high as the work done to the box . so , we only need to push the box through a distance of 585 m , which is still over five football fields long . given the known speed of 0.25 m/s that will take us : $ \frac { 585 \mathrm { m } } { 0.25 \mathrm { m/s } } =2340 \mathrm { s } $ exercise : suppose that the force that we apply to the box , see figure 1 above , is initially reduced but increases to a constant value as we warm up . for instance , in the graph below we see that as the box is displaced further—i.e. , $ x $ gets larger—the force , $ f $ , increases for the first 30 m , see figure 2 below . how could we find the work done during the period where the force is changing ? if the force is not constant , one way to determine the work done is to divide the problem up into small sections over which the change is negligible and sum up the work done in each section . just as we have learned when looking at velocity time graphs , this can be done by calculating the area under the curve using geometry . the work done by a force is equal to the area under a force vs. position graph . in the case of figure 2 , it would be : $ ( 200\text { n } \cdot 30 \text { m } ) + \frac { 1 } { 2 } \left ( ( 500\text { n } -200\text { n } ) \cdot 30 \text { m } \right ) =10500 \mathrm { j } $ for the initial $ 30 \text { m } $ of displacement . similarly , the work done for the final 40 m of displacement would be : $ 500\text { n } \cdot 40 \text { m } =20,000 \mathrm { j } $ what if we are n't pushing straight on ? there is one thing we need to watch out for when doing these problems . the previous equation , $ w = f \cdot \delta x $ , does n't take into account situations where the force we are applying is not in the same direction as the motion . for instance , imagine we use a rope to pull on the box . in that case there will be an angle between the rope and the ground . to untangle this situation , we begin by drawing a triangle to separate out the horizontal and vertical components of the applied force . the key point here is that it is only the component of the force , $ f_ { || } $ , that lies parallel to the displacement that does work on an object . in the case of the box shown above , only the horizontal component of the applied force , $ f \text { cos } ( \theta ) $ , is doing work on the box since the box is being displaced horizontally . this means that a more general equation for the work done on the box by a force at an angle θ could be written as : $ w=f_ { || } \cdot\delta x $ $ w= ( f \cos { \theta } ) \cdot\delta x $ which is more often written as , $ \large w=f \delta x\cos { \theta } $ exercise : suppose we use a rope to pull the box , and the angle between the rope and the ground is 30º . this time we pull along the rope with a force of 500 n. how much of a chocolate bar can we eat this time if we pull the box through the same 585 m ? what about lifting weights instead ? in the previous example , we were doing work on a box which we were pushing around a floor . in doing so , we were working against a frictional force . another common form of exercise is lifting weights . in this case we are working against the force of gravity rather than friction . using newton 's laws we can find the force , $ f $ , required to lift a weight with mass $ m $ straight up , placing it on a rack which is at a height $ h $ above us : $ f=mg $ the change in position—previously $ \delta x $ —is simply the height , so the work , $ w $ , that we have done in lifting the weight is then $ w=m g h $ the exercise we have done in lifting the weight has resulted in energy being stored in the form of gravitational potential energy . it is called potential energy because it has the potential to be released at any moment with a crash as the weight falls back to the ground . we did positive work on the weight since we exerted our force in the same direction as the displacement of the weight , i.e. , upward . the work done by gravity on the weight while it was lifted was negative since the force of gravity is directed in the opposite direction to the displacement . also , since the weight is stationary after the lift , we know that the work that we have done is exactly canceled out by the work done by gravity . the work done by us is $ mgh $ , and the work done by gravity is $ -mgh $ . we will talk more about this when we look into kinetic energy . ok , let 's put some numbers in and find how much of that chocolate bar we would offset by lifting a weight of 50 kg up to a height of 0.5 m. the work done to the weight is $ w= ( 50\mathrm { kg } ) ( 9.81\mathrm { m/s^2 } ) ( 0.5\mathrm { m } ) = 245.25 \mathrm { j } $ ok , so how many 280 calorie—i.e. , $ 1.17 \times 10^6 $ joule—chocolate bars is this ? well , 245.25 j is about $ \dfrac { 1 } { 4770 } $ of a chocolate bar . but remember , our bodies are only about 25 % efficient , so the work done by the person is actually four times larger , about 981.8 j , which is $ \dfrac { 1 } { 1190 } $ chocolate bars . so , if we can lift this weight once every 2 seconds , it will take us about 2380 seconds or 40 minutes of hard work to burn off this chocolate bar ! what about simply holding a weight stationary ? one frequent source of confusion people have with the concept of work comes about when thinking about holding a heavy weight stationary above our heads , against the force of gravity . we are not moving the weight through any distance , so no work is being done to the weight . we could also achieve this by placing the weight on a table ; it is clear that the table is not doing any work to keep the weight in position . yet , we know from our experience that we get tired when doing the same job . so what is going on here ? it turns out that what is actually happening here is that our bodies are doing work on our muscles to maintain the necessary tension to hold the weight up . the body does this by sending a cascade of nerve impulses to each muscle . each impulse causes the muscle to momentarily contract and release . this all happens so fast that we might only notice a slight twitching at first . eventually though , not enough chemical energy is available in the muscle and it can no longer keep up . we then begin to shake and eventually must rest for a while . so work is being done , it is just not being done on the weight .
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but remember , our bodies are only about 25 % efficient , so the work done by the person is actually four times larger , about 981.8 j , which is $ \dfrac { 1 } { 1190 } $ chocolate bars . so , if we can lift this weight once every 2 seconds , it will take us about 2380 seconds or 40 minutes of hard work to burn off this chocolate bar ! what about simply holding a weight stationary ?
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according to the potential energy problem , i do not understand how we found the number of seconds it takes to burn off a chocolate bar ?
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what does energy and work mean ? energy is a word which tends to be used a lot in everyday life . though it is often used quite loosely , it does have a very specific physical meaning . energy is a measurement of the ability of something to do work . it is not a material substance . energy can be stored and measured in many forms . although we often hear people talking about energy consumption , energy is never really destroyed . it is just transferred from one form to another , doing work in the process . some forms of energy are less useful to us than others—for example , low level heat energy . it is better to talk about the consumption or extraction of energy resources , for example coal , oil , or wind , than consumption of energy itself . a speeding bullet has a measurable amount of energy associated with it ; this is known as kinetic energy . the bullet gained this energy because work was done on it by a charge of gunpowder which lost some chemical potential energy in the process . a hot cup of coffee has a measurable amount of thermal energy which it acquired via work done by a microwave oven , which in turn took electrical energy from the electrical grid . in practice , whenever work is done to move energy from one form to another , there is always some loss to other forms of energy such as heat and sound . for example , a traditional light bulb is only about 3 % efficient at converting electrical energy to visible light , while a human being is about 25 % efficient at converting chemical energy from food into work . how do we measure energy and work ? the standard unit used to measure energy and work done in physics is the joule , which has the symbol j . in mechanics , 1 joule is the energy transferred when a force of 1 newton is applied to an object and moves it through a distance of 1 meter . another unit of energy you may have come across is the calorie . the amount of energy in an item of food is often written in calories on the back of the packet . a typical 60 gram chocolate bar for example contains about 280 calories of energy . one calorie is the amount of energy required to raise 1 kg of water by 1 $ ^ { \circ } $ celsius . this is equal to 4184 joules , so one chocolate bar has 1.17 million joules or 1.17 mj of stored energy . that 's a lot of joules ! how long do i have to push a heavy box around to burn off one chocolate bar ? suppose we 're feeling guilty about eating a chocolate bar ; we want to find how much exercise we need to do to offset those extra 280 calories . let 's consider a simple form of exercise : pushing a heavy box around a room , see figure 1 below . using a bathroom scale between ourselves and the box , we find that we can push with a force of 500 n. meanwhile , we use a stopwatch and measuring tape to measure our speed . this comes out to be 0.25 meters per second . so how much work do we need to do to the box to burn off the candy bar ? the definition of work , $ w $ , is below : $ \large w = f\cdot \delta x $ the work we need to do to burn the energy in the candy bar is $ e=280 \mathrm { cal } \cdot 4184 \mathrm { j/cal } =1.17 \mathrm { mj } $ . therefore , the distance , $ δx $ , we need to move the box through is : $ \begin { align } w & amp ; = f\cdot \delta x \ 1.17 \text { mj } & amp ; = ( 500 \text { n } ) \cdot \delta x \ \dfrac { 1.17\times 10^6 \text { j } } { 500 \text { n } } & amp ; = \delta x \ 2,340 \text { m } & amp ; = \delta x \end { align } $ remember , however , that our bodies are about 25 % efficient at transferring stored energy from food into work . the actual energy we will offset is then four times as high as the work done to the box . so , we only need to push the box through a distance of 585 m , which is still over five football fields long . given the known speed of 0.25 m/s that will take us : $ \frac { 585 \mathrm { m } } { 0.25 \mathrm { m/s } } =2340 \mathrm { s } $ exercise : suppose that the force that we apply to the box , see figure 1 above , is initially reduced but increases to a constant value as we warm up . for instance , in the graph below we see that as the box is displaced further—i.e. , $ x $ gets larger—the force , $ f $ , increases for the first 30 m , see figure 2 below . how could we find the work done during the period where the force is changing ? if the force is not constant , one way to determine the work done is to divide the problem up into small sections over which the change is negligible and sum up the work done in each section . just as we have learned when looking at velocity time graphs , this can be done by calculating the area under the curve using geometry . the work done by a force is equal to the area under a force vs. position graph . in the case of figure 2 , it would be : $ ( 200\text { n } \cdot 30 \text { m } ) + \frac { 1 } { 2 } \left ( ( 500\text { n } -200\text { n } ) \cdot 30 \text { m } \right ) =10500 \mathrm { j } $ for the initial $ 30 \text { m } $ of displacement . similarly , the work done for the final 40 m of displacement would be : $ 500\text { n } \cdot 40 \text { m } =20,000 \mathrm { j } $ what if we are n't pushing straight on ? there is one thing we need to watch out for when doing these problems . the previous equation , $ w = f \cdot \delta x $ , does n't take into account situations where the force we are applying is not in the same direction as the motion . for instance , imagine we use a rope to pull on the box . in that case there will be an angle between the rope and the ground . to untangle this situation , we begin by drawing a triangle to separate out the horizontal and vertical components of the applied force . the key point here is that it is only the component of the force , $ f_ { || } $ , that lies parallel to the displacement that does work on an object . in the case of the box shown above , only the horizontal component of the applied force , $ f \text { cos } ( \theta ) $ , is doing work on the box since the box is being displaced horizontally . this means that a more general equation for the work done on the box by a force at an angle θ could be written as : $ w=f_ { || } \cdot\delta x $ $ w= ( f \cos { \theta } ) \cdot\delta x $ which is more often written as , $ \large w=f \delta x\cos { \theta } $ exercise : suppose we use a rope to pull the box , and the angle between the rope and the ground is 30º . this time we pull along the rope with a force of 500 n. how much of a chocolate bar can we eat this time if we pull the box through the same 585 m ? what about lifting weights instead ? in the previous example , we were doing work on a box which we were pushing around a floor . in doing so , we were working against a frictional force . another common form of exercise is lifting weights . in this case we are working against the force of gravity rather than friction . using newton 's laws we can find the force , $ f $ , required to lift a weight with mass $ m $ straight up , placing it on a rack which is at a height $ h $ above us : $ f=mg $ the change in position—previously $ \delta x $ —is simply the height , so the work , $ w $ , that we have done in lifting the weight is then $ w=m g h $ the exercise we have done in lifting the weight has resulted in energy being stored in the form of gravitational potential energy . it is called potential energy because it has the potential to be released at any moment with a crash as the weight falls back to the ground . we did positive work on the weight since we exerted our force in the same direction as the displacement of the weight , i.e. , upward . the work done by gravity on the weight while it was lifted was negative since the force of gravity is directed in the opposite direction to the displacement . also , since the weight is stationary after the lift , we know that the work that we have done is exactly canceled out by the work done by gravity . the work done by us is $ mgh $ , and the work done by gravity is $ -mgh $ . we will talk more about this when we look into kinetic energy . ok , let 's put some numbers in and find how much of that chocolate bar we would offset by lifting a weight of 50 kg up to a height of 0.5 m. the work done to the weight is $ w= ( 50\mathrm { kg } ) ( 9.81\mathrm { m/s^2 } ) ( 0.5\mathrm { m } ) = 245.25 \mathrm { j } $ ok , so how many 280 calorie—i.e. , $ 1.17 \times 10^6 $ joule—chocolate bars is this ? well , 245.25 j is about $ \dfrac { 1 } { 4770 } $ of a chocolate bar . but remember , our bodies are only about 25 % efficient , so the work done by the person is actually four times larger , about 981.8 j , which is $ \dfrac { 1 } { 1190 } $ chocolate bars . so , if we can lift this weight once every 2 seconds , it will take us about 2380 seconds or 40 minutes of hard work to burn off this chocolate bar ! what about simply holding a weight stationary ? one frequent source of confusion people have with the concept of work comes about when thinking about holding a heavy weight stationary above our heads , against the force of gravity . we are not moving the weight through any distance , so no work is being done to the weight . we could also achieve this by placing the weight on a table ; it is clear that the table is not doing any work to keep the weight in position . yet , we know from our experience that we get tired when doing the same job . so what is going on here ? it turns out that what is actually happening here is that our bodies are doing work on our muscles to maintain the necessary tension to hold the weight up . the body does this by sending a cascade of nerve impulses to each muscle . each impulse causes the muscle to momentarily contract and release . this all happens so fast that we might only notice a slight twitching at first . eventually though , not enough chemical energy is available in the muscle and it can no longer keep up . we then begin to shake and eventually must rest for a while . so work is being done , it is just not being done on the weight .
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one calorie is the amount of energy required to raise 1 kg of water by 1 $ ^ { \circ } $ celsius . this is equal to 4184 joules , so one chocolate bar has 1.17 million joules or 1.17 mj of stored energy . that 's a lot of joules ! how long do i have to push a heavy box around to burn off one chocolate bar ?
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how did we find 2380 sec from 981.8 joules ?
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what does energy and work mean ? energy is a word which tends to be used a lot in everyday life . though it is often used quite loosely , it does have a very specific physical meaning . energy is a measurement of the ability of something to do work . it is not a material substance . energy can be stored and measured in many forms . although we often hear people talking about energy consumption , energy is never really destroyed . it is just transferred from one form to another , doing work in the process . some forms of energy are less useful to us than others—for example , low level heat energy . it is better to talk about the consumption or extraction of energy resources , for example coal , oil , or wind , than consumption of energy itself . a speeding bullet has a measurable amount of energy associated with it ; this is known as kinetic energy . the bullet gained this energy because work was done on it by a charge of gunpowder which lost some chemical potential energy in the process . a hot cup of coffee has a measurable amount of thermal energy which it acquired via work done by a microwave oven , which in turn took electrical energy from the electrical grid . in practice , whenever work is done to move energy from one form to another , there is always some loss to other forms of energy such as heat and sound . for example , a traditional light bulb is only about 3 % efficient at converting electrical energy to visible light , while a human being is about 25 % efficient at converting chemical energy from food into work . how do we measure energy and work ? the standard unit used to measure energy and work done in physics is the joule , which has the symbol j . in mechanics , 1 joule is the energy transferred when a force of 1 newton is applied to an object and moves it through a distance of 1 meter . another unit of energy you may have come across is the calorie . the amount of energy in an item of food is often written in calories on the back of the packet . a typical 60 gram chocolate bar for example contains about 280 calories of energy . one calorie is the amount of energy required to raise 1 kg of water by 1 $ ^ { \circ } $ celsius . this is equal to 4184 joules , so one chocolate bar has 1.17 million joules or 1.17 mj of stored energy . that 's a lot of joules ! how long do i have to push a heavy box around to burn off one chocolate bar ? suppose we 're feeling guilty about eating a chocolate bar ; we want to find how much exercise we need to do to offset those extra 280 calories . let 's consider a simple form of exercise : pushing a heavy box around a room , see figure 1 below . using a bathroom scale between ourselves and the box , we find that we can push with a force of 500 n. meanwhile , we use a stopwatch and measuring tape to measure our speed . this comes out to be 0.25 meters per second . so how much work do we need to do to the box to burn off the candy bar ? the definition of work , $ w $ , is below : $ \large w = f\cdot \delta x $ the work we need to do to burn the energy in the candy bar is $ e=280 \mathrm { cal } \cdot 4184 \mathrm { j/cal } =1.17 \mathrm { mj } $ . therefore , the distance , $ δx $ , we need to move the box through is : $ \begin { align } w & amp ; = f\cdot \delta x \ 1.17 \text { mj } & amp ; = ( 500 \text { n } ) \cdot \delta x \ \dfrac { 1.17\times 10^6 \text { j } } { 500 \text { n } } & amp ; = \delta x \ 2,340 \text { m } & amp ; = \delta x \end { align } $ remember , however , that our bodies are about 25 % efficient at transferring stored energy from food into work . the actual energy we will offset is then four times as high as the work done to the box . so , we only need to push the box through a distance of 585 m , which is still over five football fields long . given the known speed of 0.25 m/s that will take us : $ \frac { 585 \mathrm { m } } { 0.25 \mathrm { m/s } } =2340 \mathrm { s } $ exercise : suppose that the force that we apply to the box , see figure 1 above , is initially reduced but increases to a constant value as we warm up . for instance , in the graph below we see that as the box is displaced further—i.e. , $ x $ gets larger—the force , $ f $ , increases for the first 30 m , see figure 2 below . how could we find the work done during the period where the force is changing ? if the force is not constant , one way to determine the work done is to divide the problem up into small sections over which the change is negligible and sum up the work done in each section . just as we have learned when looking at velocity time graphs , this can be done by calculating the area under the curve using geometry . the work done by a force is equal to the area under a force vs. position graph . in the case of figure 2 , it would be : $ ( 200\text { n } \cdot 30 \text { m } ) + \frac { 1 } { 2 } \left ( ( 500\text { n } -200\text { n } ) \cdot 30 \text { m } \right ) =10500 \mathrm { j } $ for the initial $ 30 \text { m } $ of displacement . similarly , the work done for the final 40 m of displacement would be : $ 500\text { n } \cdot 40 \text { m } =20,000 \mathrm { j } $ what if we are n't pushing straight on ? there is one thing we need to watch out for when doing these problems . the previous equation , $ w = f \cdot \delta x $ , does n't take into account situations where the force we are applying is not in the same direction as the motion . for instance , imagine we use a rope to pull on the box . in that case there will be an angle between the rope and the ground . to untangle this situation , we begin by drawing a triangle to separate out the horizontal and vertical components of the applied force . the key point here is that it is only the component of the force , $ f_ { || } $ , that lies parallel to the displacement that does work on an object . in the case of the box shown above , only the horizontal component of the applied force , $ f \text { cos } ( \theta ) $ , is doing work on the box since the box is being displaced horizontally . this means that a more general equation for the work done on the box by a force at an angle θ could be written as : $ w=f_ { || } \cdot\delta x $ $ w= ( f \cos { \theta } ) \cdot\delta x $ which is more often written as , $ \large w=f \delta x\cos { \theta } $ exercise : suppose we use a rope to pull the box , and the angle between the rope and the ground is 30º . this time we pull along the rope with a force of 500 n. how much of a chocolate bar can we eat this time if we pull the box through the same 585 m ? what about lifting weights instead ? in the previous example , we were doing work on a box which we were pushing around a floor . in doing so , we were working against a frictional force . another common form of exercise is lifting weights . in this case we are working against the force of gravity rather than friction . using newton 's laws we can find the force , $ f $ , required to lift a weight with mass $ m $ straight up , placing it on a rack which is at a height $ h $ above us : $ f=mg $ the change in position—previously $ \delta x $ —is simply the height , so the work , $ w $ , that we have done in lifting the weight is then $ w=m g h $ the exercise we have done in lifting the weight has resulted in energy being stored in the form of gravitational potential energy . it is called potential energy because it has the potential to be released at any moment with a crash as the weight falls back to the ground . we did positive work on the weight since we exerted our force in the same direction as the displacement of the weight , i.e. , upward . the work done by gravity on the weight while it was lifted was negative since the force of gravity is directed in the opposite direction to the displacement . also , since the weight is stationary after the lift , we know that the work that we have done is exactly canceled out by the work done by gravity . the work done by us is $ mgh $ , and the work done by gravity is $ -mgh $ . we will talk more about this when we look into kinetic energy . ok , let 's put some numbers in and find how much of that chocolate bar we would offset by lifting a weight of 50 kg up to a height of 0.5 m. the work done to the weight is $ w= ( 50\mathrm { kg } ) ( 9.81\mathrm { m/s^2 } ) ( 0.5\mathrm { m } ) = 245.25 \mathrm { j } $ ok , so how many 280 calorie—i.e. , $ 1.17 \times 10^6 $ joule—chocolate bars is this ? well , 245.25 j is about $ \dfrac { 1 } { 4770 } $ of a chocolate bar . but remember , our bodies are only about 25 % efficient , so the work done by the person is actually four times larger , about 981.8 j , which is $ \dfrac { 1 } { 1190 } $ chocolate bars . so , if we can lift this weight once every 2 seconds , it will take us about 2380 seconds or 40 minutes of hard work to burn off this chocolate bar ! what about simply holding a weight stationary ? one frequent source of confusion people have with the concept of work comes about when thinking about holding a heavy weight stationary above our heads , against the force of gravity . we are not moving the weight through any distance , so no work is being done to the weight . we could also achieve this by placing the weight on a table ; it is clear that the table is not doing any work to keep the weight in position . yet , we know from our experience that we get tired when doing the same job . so what is going on here ? it turns out that what is actually happening here is that our bodies are doing work on our muscles to maintain the necessary tension to hold the weight up . the body does this by sending a cascade of nerve impulses to each muscle . each impulse causes the muscle to momentarily contract and release . this all happens so fast that we might only notice a slight twitching at first . eventually though , not enough chemical energy is available in the muscle and it can no longer keep up . we then begin to shake and eventually must rest for a while . so work is being done , it is just not being done on the weight .
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the previous equation , $ w = f \cdot \delta x $ , does n't take into account situations where the force we are applying is not in the same direction as the motion . for instance , imagine we use a rope to pull on the box . in that case there will be an angle between the rope and the ground .
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when they use the word efficiency for example 25 % efficient how will i use that in an equation should i multiply or divide or does it depend on the situation ?
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what does energy and work mean ? energy is a word which tends to be used a lot in everyday life . though it is often used quite loosely , it does have a very specific physical meaning . energy is a measurement of the ability of something to do work . it is not a material substance . energy can be stored and measured in many forms . although we often hear people talking about energy consumption , energy is never really destroyed . it is just transferred from one form to another , doing work in the process . some forms of energy are less useful to us than others—for example , low level heat energy . it is better to talk about the consumption or extraction of energy resources , for example coal , oil , or wind , than consumption of energy itself . a speeding bullet has a measurable amount of energy associated with it ; this is known as kinetic energy . the bullet gained this energy because work was done on it by a charge of gunpowder which lost some chemical potential energy in the process . a hot cup of coffee has a measurable amount of thermal energy which it acquired via work done by a microwave oven , which in turn took electrical energy from the electrical grid . in practice , whenever work is done to move energy from one form to another , there is always some loss to other forms of energy such as heat and sound . for example , a traditional light bulb is only about 3 % efficient at converting electrical energy to visible light , while a human being is about 25 % efficient at converting chemical energy from food into work . how do we measure energy and work ? the standard unit used to measure energy and work done in physics is the joule , which has the symbol j . in mechanics , 1 joule is the energy transferred when a force of 1 newton is applied to an object and moves it through a distance of 1 meter . another unit of energy you may have come across is the calorie . the amount of energy in an item of food is often written in calories on the back of the packet . a typical 60 gram chocolate bar for example contains about 280 calories of energy . one calorie is the amount of energy required to raise 1 kg of water by 1 $ ^ { \circ } $ celsius . this is equal to 4184 joules , so one chocolate bar has 1.17 million joules or 1.17 mj of stored energy . that 's a lot of joules ! how long do i have to push a heavy box around to burn off one chocolate bar ? suppose we 're feeling guilty about eating a chocolate bar ; we want to find how much exercise we need to do to offset those extra 280 calories . let 's consider a simple form of exercise : pushing a heavy box around a room , see figure 1 below . using a bathroom scale between ourselves and the box , we find that we can push with a force of 500 n. meanwhile , we use a stopwatch and measuring tape to measure our speed . this comes out to be 0.25 meters per second . so how much work do we need to do to the box to burn off the candy bar ? the definition of work , $ w $ , is below : $ \large w = f\cdot \delta x $ the work we need to do to burn the energy in the candy bar is $ e=280 \mathrm { cal } \cdot 4184 \mathrm { j/cal } =1.17 \mathrm { mj } $ . therefore , the distance , $ δx $ , we need to move the box through is : $ \begin { align } w & amp ; = f\cdot \delta x \ 1.17 \text { mj } & amp ; = ( 500 \text { n } ) \cdot \delta x \ \dfrac { 1.17\times 10^6 \text { j } } { 500 \text { n } } & amp ; = \delta x \ 2,340 \text { m } & amp ; = \delta x \end { align } $ remember , however , that our bodies are about 25 % efficient at transferring stored energy from food into work . the actual energy we will offset is then four times as high as the work done to the box . so , we only need to push the box through a distance of 585 m , which is still over five football fields long . given the known speed of 0.25 m/s that will take us : $ \frac { 585 \mathrm { m } } { 0.25 \mathrm { m/s } } =2340 \mathrm { s } $ exercise : suppose that the force that we apply to the box , see figure 1 above , is initially reduced but increases to a constant value as we warm up . for instance , in the graph below we see that as the box is displaced further—i.e. , $ x $ gets larger—the force , $ f $ , increases for the first 30 m , see figure 2 below . how could we find the work done during the period where the force is changing ? if the force is not constant , one way to determine the work done is to divide the problem up into small sections over which the change is negligible and sum up the work done in each section . just as we have learned when looking at velocity time graphs , this can be done by calculating the area under the curve using geometry . the work done by a force is equal to the area under a force vs. position graph . in the case of figure 2 , it would be : $ ( 200\text { n } \cdot 30 \text { m } ) + \frac { 1 } { 2 } \left ( ( 500\text { n } -200\text { n } ) \cdot 30 \text { m } \right ) =10500 \mathrm { j } $ for the initial $ 30 \text { m } $ of displacement . similarly , the work done for the final 40 m of displacement would be : $ 500\text { n } \cdot 40 \text { m } =20,000 \mathrm { j } $ what if we are n't pushing straight on ? there is one thing we need to watch out for when doing these problems . the previous equation , $ w = f \cdot \delta x $ , does n't take into account situations where the force we are applying is not in the same direction as the motion . for instance , imagine we use a rope to pull on the box . in that case there will be an angle between the rope and the ground . to untangle this situation , we begin by drawing a triangle to separate out the horizontal and vertical components of the applied force . the key point here is that it is only the component of the force , $ f_ { || } $ , that lies parallel to the displacement that does work on an object . in the case of the box shown above , only the horizontal component of the applied force , $ f \text { cos } ( \theta ) $ , is doing work on the box since the box is being displaced horizontally . this means that a more general equation for the work done on the box by a force at an angle θ could be written as : $ w=f_ { || } \cdot\delta x $ $ w= ( f \cos { \theta } ) \cdot\delta x $ which is more often written as , $ \large w=f \delta x\cos { \theta } $ exercise : suppose we use a rope to pull the box , and the angle between the rope and the ground is 30º . this time we pull along the rope with a force of 500 n. how much of a chocolate bar can we eat this time if we pull the box through the same 585 m ? what about lifting weights instead ? in the previous example , we were doing work on a box which we were pushing around a floor . in doing so , we were working against a frictional force . another common form of exercise is lifting weights . in this case we are working against the force of gravity rather than friction . using newton 's laws we can find the force , $ f $ , required to lift a weight with mass $ m $ straight up , placing it on a rack which is at a height $ h $ above us : $ f=mg $ the change in position—previously $ \delta x $ —is simply the height , so the work , $ w $ , that we have done in lifting the weight is then $ w=m g h $ the exercise we have done in lifting the weight has resulted in energy being stored in the form of gravitational potential energy . it is called potential energy because it has the potential to be released at any moment with a crash as the weight falls back to the ground . we did positive work on the weight since we exerted our force in the same direction as the displacement of the weight , i.e. , upward . the work done by gravity on the weight while it was lifted was negative since the force of gravity is directed in the opposite direction to the displacement . also , since the weight is stationary after the lift , we know that the work that we have done is exactly canceled out by the work done by gravity . the work done by us is $ mgh $ , and the work done by gravity is $ -mgh $ . we will talk more about this when we look into kinetic energy . ok , let 's put some numbers in and find how much of that chocolate bar we would offset by lifting a weight of 50 kg up to a height of 0.5 m. the work done to the weight is $ w= ( 50\mathrm { kg } ) ( 9.81\mathrm { m/s^2 } ) ( 0.5\mathrm { m } ) = 245.25 \mathrm { j } $ ok , so how many 280 calorie—i.e. , $ 1.17 \times 10^6 $ joule—chocolate bars is this ? well , 245.25 j is about $ \dfrac { 1 } { 4770 } $ of a chocolate bar . but remember , our bodies are only about 25 % efficient , so the work done by the person is actually four times larger , about 981.8 j , which is $ \dfrac { 1 } { 1190 } $ chocolate bars . so , if we can lift this weight once every 2 seconds , it will take us about 2380 seconds or 40 minutes of hard work to burn off this chocolate bar ! what about simply holding a weight stationary ? one frequent source of confusion people have with the concept of work comes about when thinking about holding a heavy weight stationary above our heads , against the force of gravity . we are not moving the weight through any distance , so no work is being done to the weight . we could also achieve this by placing the weight on a table ; it is clear that the table is not doing any work to keep the weight in position . yet , we know from our experience that we get tired when doing the same job . so what is going on here ? it turns out that what is actually happening here is that our bodies are doing work on our muscles to maintain the necessary tension to hold the weight up . the body does this by sending a cascade of nerve impulses to each muscle . each impulse causes the muscle to momentarily contract and release . this all happens so fast that we might only notice a slight twitching at first . eventually though , not enough chemical energy is available in the muscle and it can no longer keep up . we then begin to shake and eventually must rest for a while . so work is being done , it is just not being done on the weight .
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in the case of figure 2 , it would be : $ ( 200\text { n } \cdot 30 \text { m } ) + \frac { 1 } { 2 } \left ( ( 500\text { n } -200\text { n } ) \cdot 30 \text { m } \right ) =10500 \mathrm { j } $ for the initial $ 30 \text { m } $ of displacement . similarly , the work done for the final 40 m of displacement would be : $ 500\text { n } \cdot 40 \text { m } =20,000 \mathrm { j } $ what if we are n't pushing straight on ? there is one thing we need to watch out for when doing these problems .
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except that due to low friction the box accelerates during the push and decelerates after i stop pushing thus traveling additional 10 m. i push the object over 10 m but the object is displaced over 20 m. is the work equal to 1000 j or 2000 j ?
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what does energy and work mean ? energy is a word which tends to be used a lot in everyday life . though it is often used quite loosely , it does have a very specific physical meaning . energy is a measurement of the ability of something to do work . it is not a material substance . energy can be stored and measured in many forms . although we often hear people talking about energy consumption , energy is never really destroyed . it is just transferred from one form to another , doing work in the process . some forms of energy are less useful to us than others—for example , low level heat energy . it is better to talk about the consumption or extraction of energy resources , for example coal , oil , or wind , than consumption of energy itself . a speeding bullet has a measurable amount of energy associated with it ; this is known as kinetic energy . the bullet gained this energy because work was done on it by a charge of gunpowder which lost some chemical potential energy in the process . a hot cup of coffee has a measurable amount of thermal energy which it acquired via work done by a microwave oven , which in turn took electrical energy from the electrical grid . in practice , whenever work is done to move energy from one form to another , there is always some loss to other forms of energy such as heat and sound . for example , a traditional light bulb is only about 3 % efficient at converting electrical energy to visible light , while a human being is about 25 % efficient at converting chemical energy from food into work . how do we measure energy and work ? the standard unit used to measure energy and work done in physics is the joule , which has the symbol j . in mechanics , 1 joule is the energy transferred when a force of 1 newton is applied to an object and moves it through a distance of 1 meter . another unit of energy you may have come across is the calorie . the amount of energy in an item of food is often written in calories on the back of the packet . a typical 60 gram chocolate bar for example contains about 280 calories of energy . one calorie is the amount of energy required to raise 1 kg of water by 1 $ ^ { \circ } $ celsius . this is equal to 4184 joules , so one chocolate bar has 1.17 million joules or 1.17 mj of stored energy . that 's a lot of joules ! how long do i have to push a heavy box around to burn off one chocolate bar ? suppose we 're feeling guilty about eating a chocolate bar ; we want to find how much exercise we need to do to offset those extra 280 calories . let 's consider a simple form of exercise : pushing a heavy box around a room , see figure 1 below . using a bathroom scale between ourselves and the box , we find that we can push with a force of 500 n. meanwhile , we use a stopwatch and measuring tape to measure our speed . this comes out to be 0.25 meters per second . so how much work do we need to do to the box to burn off the candy bar ? the definition of work , $ w $ , is below : $ \large w = f\cdot \delta x $ the work we need to do to burn the energy in the candy bar is $ e=280 \mathrm { cal } \cdot 4184 \mathrm { j/cal } =1.17 \mathrm { mj } $ . therefore , the distance , $ δx $ , we need to move the box through is : $ \begin { align } w & amp ; = f\cdot \delta x \ 1.17 \text { mj } & amp ; = ( 500 \text { n } ) \cdot \delta x \ \dfrac { 1.17\times 10^6 \text { j } } { 500 \text { n } } & amp ; = \delta x \ 2,340 \text { m } & amp ; = \delta x \end { align } $ remember , however , that our bodies are about 25 % efficient at transferring stored energy from food into work . the actual energy we will offset is then four times as high as the work done to the box . so , we only need to push the box through a distance of 585 m , which is still over five football fields long . given the known speed of 0.25 m/s that will take us : $ \frac { 585 \mathrm { m } } { 0.25 \mathrm { m/s } } =2340 \mathrm { s } $ exercise : suppose that the force that we apply to the box , see figure 1 above , is initially reduced but increases to a constant value as we warm up . for instance , in the graph below we see that as the box is displaced further—i.e. , $ x $ gets larger—the force , $ f $ , increases for the first 30 m , see figure 2 below . how could we find the work done during the period where the force is changing ? if the force is not constant , one way to determine the work done is to divide the problem up into small sections over which the change is negligible and sum up the work done in each section . just as we have learned when looking at velocity time graphs , this can be done by calculating the area under the curve using geometry . the work done by a force is equal to the area under a force vs. position graph . in the case of figure 2 , it would be : $ ( 200\text { n } \cdot 30 \text { m } ) + \frac { 1 } { 2 } \left ( ( 500\text { n } -200\text { n } ) \cdot 30 \text { m } \right ) =10500 \mathrm { j } $ for the initial $ 30 \text { m } $ of displacement . similarly , the work done for the final 40 m of displacement would be : $ 500\text { n } \cdot 40 \text { m } =20,000 \mathrm { j } $ what if we are n't pushing straight on ? there is one thing we need to watch out for when doing these problems . the previous equation , $ w = f \cdot \delta x $ , does n't take into account situations where the force we are applying is not in the same direction as the motion . for instance , imagine we use a rope to pull on the box . in that case there will be an angle between the rope and the ground . to untangle this situation , we begin by drawing a triangle to separate out the horizontal and vertical components of the applied force . the key point here is that it is only the component of the force , $ f_ { || } $ , that lies parallel to the displacement that does work on an object . in the case of the box shown above , only the horizontal component of the applied force , $ f \text { cos } ( \theta ) $ , is doing work on the box since the box is being displaced horizontally . this means that a more general equation for the work done on the box by a force at an angle θ could be written as : $ w=f_ { || } \cdot\delta x $ $ w= ( f \cos { \theta } ) \cdot\delta x $ which is more often written as , $ \large w=f \delta x\cos { \theta } $ exercise : suppose we use a rope to pull the box , and the angle between the rope and the ground is 30º . this time we pull along the rope with a force of 500 n. how much of a chocolate bar can we eat this time if we pull the box through the same 585 m ? what about lifting weights instead ? in the previous example , we were doing work on a box which we were pushing around a floor . in doing so , we were working against a frictional force . another common form of exercise is lifting weights . in this case we are working against the force of gravity rather than friction . using newton 's laws we can find the force , $ f $ , required to lift a weight with mass $ m $ straight up , placing it on a rack which is at a height $ h $ above us : $ f=mg $ the change in position—previously $ \delta x $ —is simply the height , so the work , $ w $ , that we have done in lifting the weight is then $ w=m g h $ the exercise we have done in lifting the weight has resulted in energy being stored in the form of gravitational potential energy . it is called potential energy because it has the potential to be released at any moment with a crash as the weight falls back to the ground . we did positive work on the weight since we exerted our force in the same direction as the displacement of the weight , i.e. , upward . the work done by gravity on the weight while it was lifted was negative since the force of gravity is directed in the opposite direction to the displacement . also , since the weight is stationary after the lift , we know that the work that we have done is exactly canceled out by the work done by gravity . the work done by us is $ mgh $ , and the work done by gravity is $ -mgh $ . we will talk more about this when we look into kinetic energy . ok , let 's put some numbers in and find how much of that chocolate bar we would offset by lifting a weight of 50 kg up to a height of 0.5 m. the work done to the weight is $ w= ( 50\mathrm { kg } ) ( 9.81\mathrm { m/s^2 } ) ( 0.5\mathrm { m } ) = 245.25 \mathrm { j } $ ok , so how many 280 calorie—i.e. , $ 1.17 \times 10^6 $ joule—chocolate bars is this ? well , 245.25 j is about $ \dfrac { 1 } { 4770 } $ of a chocolate bar . but remember , our bodies are only about 25 % efficient , so the work done by the person is actually four times larger , about 981.8 j , which is $ \dfrac { 1 } { 1190 } $ chocolate bars . so , if we can lift this weight once every 2 seconds , it will take us about 2380 seconds or 40 minutes of hard work to burn off this chocolate bar ! what about simply holding a weight stationary ? one frequent source of confusion people have with the concept of work comes about when thinking about holding a heavy weight stationary above our heads , against the force of gravity . we are not moving the weight through any distance , so no work is being done to the weight . we could also achieve this by placing the weight on a table ; it is clear that the table is not doing any work to keep the weight in position . yet , we know from our experience that we get tired when doing the same job . so what is going on here ? it turns out that what is actually happening here is that our bodies are doing work on our muscles to maintain the necessary tension to hold the weight up . the body does this by sending a cascade of nerve impulses to each muscle . each impulse causes the muscle to momentarily contract and release . this all happens so fast that we might only notice a slight twitching at first . eventually though , not enough chemical energy is available in the muscle and it can no longer keep up . we then begin to shake and eventually must rest for a while . so work is being done , it is just not being done on the weight .
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it is not a material substance . energy can be stored and measured in many forms . although we often hear people talking about energy consumption , energy is never really destroyed . it is just transferred from one form to another , doing work in the process .
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why is energy so important ?
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what does energy and work mean ? energy is a word which tends to be used a lot in everyday life . though it is often used quite loosely , it does have a very specific physical meaning . energy is a measurement of the ability of something to do work . it is not a material substance . energy can be stored and measured in many forms . although we often hear people talking about energy consumption , energy is never really destroyed . it is just transferred from one form to another , doing work in the process . some forms of energy are less useful to us than others—for example , low level heat energy . it is better to talk about the consumption or extraction of energy resources , for example coal , oil , or wind , than consumption of energy itself . a speeding bullet has a measurable amount of energy associated with it ; this is known as kinetic energy . the bullet gained this energy because work was done on it by a charge of gunpowder which lost some chemical potential energy in the process . a hot cup of coffee has a measurable amount of thermal energy which it acquired via work done by a microwave oven , which in turn took electrical energy from the electrical grid . in practice , whenever work is done to move energy from one form to another , there is always some loss to other forms of energy such as heat and sound . for example , a traditional light bulb is only about 3 % efficient at converting electrical energy to visible light , while a human being is about 25 % efficient at converting chemical energy from food into work . how do we measure energy and work ? the standard unit used to measure energy and work done in physics is the joule , which has the symbol j . in mechanics , 1 joule is the energy transferred when a force of 1 newton is applied to an object and moves it through a distance of 1 meter . another unit of energy you may have come across is the calorie . the amount of energy in an item of food is often written in calories on the back of the packet . a typical 60 gram chocolate bar for example contains about 280 calories of energy . one calorie is the amount of energy required to raise 1 kg of water by 1 $ ^ { \circ } $ celsius . this is equal to 4184 joules , so one chocolate bar has 1.17 million joules or 1.17 mj of stored energy . that 's a lot of joules ! how long do i have to push a heavy box around to burn off one chocolate bar ? suppose we 're feeling guilty about eating a chocolate bar ; we want to find how much exercise we need to do to offset those extra 280 calories . let 's consider a simple form of exercise : pushing a heavy box around a room , see figure 1 below . using a bathroom scale between ourselves and the box , we find that we can push with a force of 500 n. meanwhile , we use a stopwatch and measuring tape to measure our speed . this comes out to be 0.25 meters per second . so how much work do we need to do to the box to burn off the candy bar ? the definition of work , $ w $ , is below : $ \large w = f\cdot \delta x $ the work we need to do to burn the energy in the candy bar is $ e=280 \mathrm { cal } \cdot 4184 \mathrm { j/cal } =1.17 \mathrm { mj } $ . therefore , the distance , $ δx $ , we need to move the box through is : $ \begin { align } w & amp ; = f\cdot \delta x \ 1.17 \text { mj } & amp ; = ( 500 \text { n } ) \cdot \delta x \ \dfrac { 1.17\times 10^6 \text { j } } { 500 \text { n } } & amp ; = \delta x \ 2,340 \text { m } & amp ; = \delta x \end { align } $ remember , however , that our bodies are about 25 % efficient at transferring stored energy from food into work . the actual energy we will offset is then four times as high as the work done to the box . so , we only need to push the box through a distance of 585 m , which is still over five football fields long . given the known speed of 0.25 m/s that will take us : $ \frac { 585 \mathrm { m } } { 0.25 \mathrm { m/s } } =2340 \mathrm { s } $ exercise : suppose that the force that we apply to the box , see figure 1 above , is initially reduced but increases to a constant value as we warm up . for instance , in the graph below we see that as the box is displaced further—i.e. , $ x $ gets larger—the force , $ f $ , increases for the first 30 m , see figure 2 below . how could we find the work done during the period where the force is changing ? if the force is not constant , one way to determine the work done is to divide the problem up into small sections over which the change is negligible and sum up the work done in each section . just as we have learned when looking at velocity time graphs , this can be done by calculating the area under the curve using geometry . the work done by a force is equal to the area under a force vs. position graph . in the case of figure 2 , it would be : $ ( 200\text { n } \cdot 30 \text { m } ) + \frac { 1 } { 2 } \left ( ( 500\text { n } -200\text { n } ) \cdot 30 \text { m } \right ) =10500 \mathrm { j } $ for the initial $ 30 \text { m } $ of displacement . similarly , the work done for the final 40 m of displacement would be : $ 500\text { n } \cdot 40 \text { m } =20,000 \mathrm { j } $ what if we are n't pushing straight on ? there is one thing we need to watch out for when doing these problems . the previous equation , $ w = f \cdot \delta x $ , does n't take into account situations where the force we are applying is not in the same direction as the motion . for instance , imagine we use a rope to pull on the box . in that case there will be an angle between the rope and the ground . to untangle this situation , we begin by drawing a triangle to separate out the horizontal and vertical components of the applied force . the key point here is that it is only the component of the force , $ f_ { || } $ , that lies parallel to the displacement that does work on an object . in the case of the box shown above , only the horizontal component of the applied force , $ f \text { cos } ( \theta ) $ , is doing work on the box since the box is being displaced horizontally . this means that a more general equation for the work done on the box by a force at an angle θ could be written as : $ w=f_ { || } \cdot\delta x $ $ w= ( f \cos { \theta } ) \cdot\delta x $ which is more often written as , $ \large w=f \delta x\cos { \theta } $ exercise : suppose we use a rope to pull the box , and the angle between the rope and the ground is 30º . this time we pull along the rope with a force of 500 n. how much of a chocolate bar can we eat this time if we pull the box through the same 585 m ? what about lifting weights instead ? in the previous example , we were doing work on a box which we were pushing around a floor . in doing so , we were working against a frictional force . another common form of exercise is lifting weights . in this case we are working against the force of gravity rather than friction . using newton 's laws we can find the force , $ f $ , required to lift a weight with mass $ m $ straight up , placing it on a rack which is at a height $ h $ above us : $ f=mg $ the change in position—previously $ \delta x $ —is simply the height , so the work , $ w $ , that we have done in lifting the weight is then $ w=m g h $ the exercise we have done in lifting the weight has resulted in energy being stored in the form of gravitational potential energy . it is called potential energy because it has the potential to be released at any moment with a crash as the weight falls back to the ground . we did positive work on the weight since we exerted our force in the same direction as the displacement of the weight , i.e. , upward . the work done by gravity on the weight while it was lifted was negative since the force of gravity is directed in the opposite direction to the displacement . also , since the weight is stationary after the lift , we know that the work that we have done is exactly canceled out by the work done by gravity . the work done by us is $ mgh $ , and the work done by gravity is $ -mgh $ . we will talk more about this when we look into kinetic energy . ok , let 's put some numbers in and find how much of that chocolate bar we would offset by lifting a weight of 50 kg up to a height of 0.5 m. the work done to the weight is $ w= ( 50\mathrm { kg } ) ( 9.81\mathrm { m/s^2 } ) ( 0.5\mathrm { m } ) = 245.25 \mathrm { j } $ ok , so how many 280 calorie—i.e. , $ 1.17 \times 10^6 $ joule—chocolate bars is this ? well , 245.25 j is about $ \dfrac { 1 } { 4770 } $ of a chocolate bar . but remember , our bodies are only about 25 % efficient , so the work done by the person is actually four times larger , about 981.8 j , which is $ \dfrac { 1 } { 1190 } $ chocolate bars . so , if we can lift this weight once every 2 seconds , it will take us about 2380 seconds or 40 minutes of hard work to burn off this chocolate bar ! what about simply holding a weight stationary ? one frequent source of confusion people have with the concept of work comes about when thinking about holding a heavy weight stationary above our heads , against the force of gravity . we are not moving the weight through any distance , so no work is being done to the weight . we could also achieve this by placing the weight on a table ; it is clear that the table is not doing any work to keep the weight in position . yet , we know from our experience that we get tired when doing the same job . so what is going on here ? it turns out that what is actually happening here is that our bodies are doing work on our muscles to maintain the necessary tension to hold the weight up . the body does this by sending a cascade of nerve impulses to each muscle . each impulse causes the muscle to momentarily contract and release . this all happens so fast that we might only notice a slight twitching at first . eventually though , not enough chemical energy is available in the muscle and it can no longer keep up . we then begin to shake and eventually must rest for a while . so work is being done , it is just not being done on the weight .
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one frequent source of confusion people have with the concept of work comes about when thinking about holding a heavy weight stationary above our heads , against the force of gravity . we are not moving the weight through any distance , so no work is being done to the weight . we could also achieve this by placing the weight on a table ; it is clear that the table is not doing any work to keep the weight in position .
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the weight of the object does n't matter ?
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what does energy and work mean ? energy is a word which tends to be used a lot in everyday life . though it is often used quite loosely , it does have a very specific physical meaning . energy is a measurement of the ability of something to do work . it is not a material substance . energy can be stored and measured in many forms . although we often hear people talking about energy consumption , energy is never really destroyed . it is just transferred from one form to another , doing work in the process . some forms of energy are less useful to us than others—for example , low level heat energy . it is better to talk about the consumption or extraction of energy resources , for example coal , oil , or wind , than consumption of energy itself . a speeding bullet has a measurable amount of energy associated with it ; this is known as kinetic energy . the bullet gained this energy because work was done on it by a charge of gunpowder which lost some chemical potential energy in the process . a hot cup of coffee has a measurable amount of thermal energy which it acquired via work done by a microwave oven , which in turn took electrical energy from the electrical grid . in practice , whenever work is done to move energy from one form to another , there is always some loss to other forms of energy such as heat and sound . for example , a traditional light bulb is only about 3 % efficient at converting electrical energy to visible light , while a human being is about 25 % efficient at converting chemical energy from food into work . how do we measure energy and work ? the standard unit used to measure energy and work done in physics is the joule , which has the symbol j . in mechanics , 1 joule is the energy transferred when a force of 1 newton is applied to an object and moves it through a distance of 1 meter . another unit of energy you may have come across is the calorie . the amount of energy in an item of food is often written in calories on the back of the packet . a typical 60 gram chocolate bar for example contains about 280 calories of energy . one calorie is the amount of energy required to raise 1 kg of water by 1 $ ^ { \circ } $ celsius . this is equal to 4184 joules , so one chocolate bar has 1.17 million joules or 1.17 mj of stored energy . that 's a lot of joules ! how long do i have to push a heavy box around to burn off one chocolate bar ? suppose we 're feeling guilty about eating a chocolate bar ; we want to find how much exercise we need to do to offset those extra 280 calories . let 's consider a simple form of exercise : pushing a heavy box around a room , see figure 1 below . using a bathroom scale between ourselves and the box , we find that we can push with a force of 500 n. meanwhile , we use a stopwatch and measuring tape to measure our speed . this comes out to be 0.25 meters per second . so how much work do we need to do to the box to burn off the candy bar ? the definition of work , $ w $ , is below : $ \large w = f\cdot \delta x $ the work we need to do to burn the energy in the candy bar is $ e=280 \mathrm { cal } \cdot 4184 \mathrm { j/cal } =1.17 \mathrm { mj } $ . therefore , the distance , $ δx $ , we need to move the box through is : $ \begin { align } w & amp ; = f\cdot \delta x \ 1.17 \text { mj } & amp ; = ( 500 \text { n } ) \cdot \delta x \ \dfrac { 1.17\times 10^6 \text { j } } { 500 \text { n } } & amp ; = \delta x \ 2,340 \text { m } & amp ; = \delta x \end { align } $ remember , however , that our bodies are about 25 % efficient at transferring stored energy from food into work . the actual energy we will offset is then four times as high as the work done to the box . so , we only need to push the box through a distance of 585 m , which is still over five football fields long . given the known speed of 0.25 m/s that will take us : $ \frac { 585 \mathrm { m } } { 0.25 \mathrm { m/s } } =2340 \mathrm { s } $ exercise : suppose that the force that we apply to the box , see figure 1 above , is initially reduced but increases to a constant value as we warm up . for instance , in the graph below we see that as the box is displaced further—i.e. , $ x $ gets larger—the force , $ f $ , increases for the first 30 m , see figure 2 below . how could we find the work done during the period where the force is changing ? if the force is not constant , one way to determine the work done is to divide the problem up into small sections over which the change is negligible and sum up the work done in each section . just as we have learned when looking at velocity time graphs , this can be done by calculating the area under the curve using geometry . the work done by a force is equal to the area under a force vs. position graph . in the case of figure 2 , it would be : $ ( 200\text { n } \cdot 30 \text { m } ) + \frac { 1 } { 2 } \left ( ( 500\text { n } -200\text { n } ) \cdot 30 \text { m } \right ) =10500 \mathrm { j } $ for the initial $ 30 \text { m } $ of displacement . similarly , the work done for the final 40 m of displacement would be : $ 500\text { n } \cdot 40 \text { m } =20,000 \mathrm { j } $ what if we are n't pushing straight on ? there is one thing we need to watch out for when doing these problems . the previous equation , $ w = f \cdot \delta x $ , does n't take into account situations where the force we are applying is not in the same direction as the motion . for instance , imagine we use a rope to pull on the box . in that case there will be an angle between the rope and the ground . to untangle this situation , we begin by drawing a triangle to separate out the horizontal and vertical components of the applied force . the key point here is that it is only the component of the force , $ f_ { || } $ , that lies parallel to the displacement that does work on an object . in the case of the box shown above , only the horizontal component of the applied force , $ f \text { cos } ( \theta ) $ , is doing work on the box since the box is being displaced horizontally . this means that a more general equation for the work done on the box by a force at an angle θ could be written as : $ w=f_ { || } \cdot\delta x $ $ w= ( f \cos { \theta } ) \cdot\delta x $ which is more often written as , $ \large w=f \delta x\cos { \theta } $ exercise : suppose we use a rope to pull the box , and the angle between the rope and the ground is 30º . this time we pull along the rope with a force of 500 n. how much of a chocolate bar can we eat this time if we pull the box through the same 585 m ? what about lifting weights instead ? in the previous example , we were doing work on a box which we were pushing around a floor . in doing so , we were working against a frictional force . another common form of exercise is lifting weights . in this case we are working against the force of gravity rather than friction . using newton 's laws we can find the force , $ f $ , required to lift a weight with mass $ m $ straight up , placing it on a rack which is at a height $ h $ above us : $ f=mg $ the change in position—previously $ \delta x $ —is simply the height , so the work , $ w $ , that we have done in lifting the weight is then $ w=m g h $ the exercise we have done in lifting the weight has resulted in energy being stored in the form of gravitational potential energy . it is called potential energy because it has the potential to be released at any moment with a crash as the weight falls back to the ground . we did positive work on the weight since we exerted our force in the same direction as the displacement of the weight , i.e. , upward . the work done by gravity on the weight while it was lifted was negative since the force of gravity is directed in the opposite direction to the displacement . also , since the weight is stationary after the lift , we know that the work that we have done is exactly canceled out by the work done by gravity . the work done by us is $ mgh $ , and the work done by gravity is $ -mgh $ . we will talk more about this when we look into kinetic energy . ok , let 's put some numbers in and find how much of that chocolate bar we would offset by lifting a weight of 50 kg up to a height of 0.5 m. the work done to the weight is $ w= ( 50\mathrm { kg } ) ( 9.81\mathrm { m/s^2 } ) ( 0.5\mathrm { m } ) = 245.25 \mathrm { j } $ ok , so how many 280 calorie—i.e. , $ 1.17 \times 10^6 $ joule—chocolate bars is this ? well , 245.25 j is about $ \dfrac { 1 } { 4770 } $ of a chocolate bar . but remember , our bodies are only about 25 % efficient , so the work done by the person is actually four times larger , about 981.8 j , which is $ \dfrac { 1 } { 1190 } $ chocolate bars . so , if we can lift this weight once every 2 seconds , it will take us about 2380 seconds or 40 minutes of hard work to burn off this chocolate bar ! what about simply holding a weight stationary ? one frequent source of confusion people have with the concept of work comes about when thinking about holding a heavy weight stationary above our heads , against the force of gravity . we are not moving the weight through any distance , so no work is being done to the weight . we could also achieve this by placing the weight on a table ; it is clear that the table is not doing any work to keep the weight in position . yet , we know from our experience that we get tired when doing the same job . so what is going on here ? it turns out that what is actually happening here is that our bodies are doing work on our muscles to maintain the necessary tension to hold the weight up . the body does this by sending a cascade of nerve impulses to each muscle . each impulse causes the muscle to momentarily contract and release . this all happens so fast that we might only notice a slight twitching at first . eventually though , not enough chemical energy is available in the muscle and it can no longer keep up . we then begin to shake and eventually must rest for a while . so work is being done , it is just not being done on the weight .
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it is not a material substance . energy can be stored and measured in many forms . although we often hear people talking about energy consumption , energy is never really destroyed . it is just transferred from one form to another , doing work in the process .
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if a person was initially running at a constant velocity , that person is expending energy in order to move his body right ?
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what does energy and work mean ? energy is a word which tends to be used a lot in everyday life . though it is often used quite loosely , it does have a very specific physical meaning . energy is a measurement of the ability of something to do work . it is not a material substance . energy can be stored and measured in many forms . although we often hear people talking about energy consumption , energy is never really destroyed . it is just transferred from one form to another , doing work in the process . some forms of energy are less useful to us than others—for example , low level heat energy . it is better to talk about the consumption or extraction of energy resources , for example coal , oil , or wind , than consumption of energy itself . a speeding bullet has a measurable amount of energy associated with it ; this is known as kinetic energy . the bullet gained this energy because work was done on it by a charge of gunpowder which lost some chemical potential energy in the process . a hot cup of coffee has a measurable amount of thermal energy which it acquired via work done by a microwave oven , which in turn took electrical energy from the electrical grid . in practice , whenever work is done to move energy from one form to another , there is always some loss to other forms of energy such as heat and sound . for example , a traditional light bulb is only about 3 % efficient at converting electrical energy to visible light , while a human being is about 25 % efficient at converting chemical energy from food into work . how do we measure energy and work ? the standard unit used to measure energy and work done in physics is the joule , which has the symbol j . in mechanics , 1 joule is the energy transferred when a force of 1 newton is applied to an object and moves it through a distance of 1 meter . another unit of energy you may have come across is the calorie . the amount of energy in an item of food is often written in calories on the back of the packet . a typical 60 gram chocolate bar for example contains about 280 calories of energy . one calorie is the amount of energy required to raise 1 kg of water by 1 $ ^ { \circ } $ celsius . this is equal to 4184 joules , so one chocolate bar has 1.17 million joules or 1.17 mj of stored energy . that 's a lot of joules ! how long do i have to push a heavy box around to burn off one chocolate bar ? suppose we 're feeling guilty about eating a chocolate bar ; we want to find how much exercise we need to do to offset those extra 280 calories . let 's consider a simple form of exercise : pushing a heavy box around a room , see figure 1 below . using a bathroom scale between ourselves and the box , we find that we can push with a force of 500 n. meanwhile , we use a stopwatch and measuring tape to measure our speed . this comes out to be 0.25 meters per second . so how much work do we need to do to the box to burn off the candy bar ? the definition of work , $ w $ , is below : $ \large w = f\cdot \delta x $ the work we need to do to burn the energy in the candy bar is $ e=280 \mathrm { cal } \cdot 4184 \mathrm { j/cal } =1.17 \mathrm { mj } $ . therefore , the distance , $ δx $ , we need to move the box through is : $ \begin { align } w & amp ; = f\cdot \delta x \ 1.17 \text { mj } & amp ; = ( 500 \text { n } ) \cdot \delta x \ \dfrac { 1.17\times 10^6 \text { j } } { 500 \text { n } } & amp ; = \delta x \ 2,340 \text { m } & amp ; = \delta x \end { align } $ remember , however , that our bodies are about 25 % efficient at transferring stored energy from food into work . the actual energy we will offset is then four times as high as the work done to the box . so , we only need to push the box through a distance of 585 m , which is still over five football fields long . given the known speed of 0.25 m/s that will take us : $ \frac { 585 \mathrm { m } } { 0.25 \mathrm { m/s } } =2340 \mathrm { s } $ exercise : suppose that the force that we apply to the box , see figure 1 above , is initially reduced but increases to a constant value as we warm up . for instance , in the graph below we see that as the box is displaced further—i.e. , $ x $ gets larger—the force , $ f $ , increases for the first 30 m , see figure 2 below . how could we find the work done during the period where the force is changing ? if the force is not constant , one way to determine the work done is to divide the problem up into small sections over which the change is negligible and sum up the work done in each section . just as we have learned when looking at velocity time graphs , this can be done by calculating the area under the curve using geometry . the work done by a force is equal to the area under a force vs. position graph . in the case of figure 2 , it would be : $ ( 200\text { n } \cdot 30 \text { m } ) + \frac { 1 } { 2 } \left ( ( 500\text { n } -200\text { n } ) \cdot 30 \text { m } \right ) =10500 \mathrm { j } $ for the initial $ 30 \text { m } $ of displacement . similarly , the work done for the final 40 m of displacement would be : $ 500\text { n } \cdot 40 \text { m } =20,000 \mathrm { j } $ what if we are n't pushing straight on ? there is one thing we need to watch out for when doing these problems . the previous equation , $ w = f \cdot \delta x $ , does n't take into account situations where the force we are applying is not in the same direction as the motion . for instance , imagine we use a rope to pull on the box . in that case there will be an angle between the rope and the ground . to untangle this situation , we begin by drawing a triangle to separate out the horizontal and vertical components of the applied force . the key point here is that it is only the component of the force , $ f_ { || } $ , that lies parallel to the displacement that does work on an object . in the case of the box shown above , only the horizontal component of the applied force , $ f \text { cos } ( \theta ) $ , is doing work on the box since the box is being displaced horizontally . this means that a more general equation for the work done on the box by a force at an angle θ could be written as : $ w=f_ { || } \cdot\delta x $ $ w= ( f \cos { \theta } ) \cdot\delta x $ which is more often written as , $ \large w=f \delta x\cos { \theta } $ exercise : suppose we use a rope to pull the box , and the angle between the rope and the ground is 30º . this time we pull along the rope with a force of 500 n. how much of a chocolate bar can we eat this time if we pull the box through the same 585 m ? what about lifting weights instead ? in the previous example , we were doing work on a box which we were pushing around a floor . in doing so , we were working against a frictional force . another common form of exercise is lifting weights . in this case we are working against the force of gravity rather than friction . using newton 's laws we can find the force , $ f $ , required to lift a weight with mass $ m $ straight up , placing it on a rack which is at a height $ h $ above us : $ f=mg $ the change in position—previously $ \delta x $ —is simply the height , so the work , $ w $ , that we have done in lifting the weight is then $ w=m g h $ the exercise we have done in lifting the weight has resulted in energy being stored in the form of gravitational potential energy . it is called potential energy because it has the potential to be released at any moment with a crash as the weight falls back to the ground . we did positive work on the weight since we exerted our force in the same direction as the displacement of the weight , i.e. , upward . the work done by gravity on the weight while it was lifted was negative since the force of gravity is directed in the opposite direction to the displacement . also , since the weight is stationary after the lift , we know that the work that we have done is exactly canceled out by the work done by gravity . the work done by us is $ mgh $ , and the work done by gravity is $ -mgh $ . we will talk more about this when we look into kinetic energy . ok , let 's put some numbers in and find how much of that chocolate bar we would offset by lifting a weight of 50 kg up to a height of 0.5 m. the work done to the weight is $ w= ( 50\mathrm { kg } ) ( 9.81\mathrm { m/s^2 } ) ( 0.5\mathrm { m } ) = 245.25 \mathrm { j } $ ok , so how many 280 calorie—i.e. , $ 1.17 \times 10^6 $ joule—chocolate bars is this ? well , 245.25 j is about $ \dfrac { 1 } { 4770 } $ of a chocolate bar . but remember , our bodies are only about 25 % efficient , so the work done by the person is actually four times larger , about 981.8 j , which is $ \dfrac { 1 } { 1190 } $ chocolate bars . so , if we can lift this weight once every 2 seconds , it will take us about 2380 seconds or 40 minutes of hard work to burn off this chocolate bar ! what about simply holding a weight stationary ? one frequent source of confusion people have with the concept of work comes about when thinking about holding a heavy weight stationary above our heads , against the force of gravity . we are not moving the weight through any distance , so no work is being done to the weight . we could also achieve this by placing the weight on a table ; it is clear that the table is not doing any work to keep the weight in position . yet , we know from our experience that we get tired when doing the same job . so what is going on here ? it turns out that what is actually happening here is that our bodies are doing work on our muscles to maintain the necessary tension to hold the weight up . the body does this by sending a cascade of nerve impulses to each muscle . each impulse causes the muscle to momentarily contract and release . this all happens so fast that we might only notice a slight twitching at first . eventually though , not enough chemical energy is available in the muscle and it can no longer keep up . we then begin to shake and eventually must rest for a while . so work is being done , it is just not being done on the weight .
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also , since the weight is stationary after the lift , we know that the work that we have done is exactly canceled out by the work done by gravity . the work done by us is $ mgh $ , and the work done by gravity is $ -mgh $ . we will talk more about this when we look into kinetic energy .
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so , if a person was just running at constant velocity , meaning there is n't acceleration , is there work being done ?
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what does energy and work mean ? energy is a word which tends to be used a lot in everyday life . though it is often used quite loosely , it does have a very specific physical meaning . energy is a measurement of the ability of something to do work . it is not a material substance . energy can be stored and measured in many forms . although we often hear people talking about energy consumption , energy is never really destroyed . it is just transferred from one form to another , doing work in the process . some forms of energy are less useful to us than others—for example , low level heat energy . it is better to talk about the consumption or extraction of energy resources , for example coal , oil , or wind , than consumption of energy itself . a speeding bullet has a measurable amount of energy associated with it ; this is known as kinetic energy . the bullet gained this energy because work was done on it by a charge of gunpowder which lost some chemical potential energy in the process . a hot cup of coffee has a measurable amount of thermal energy which it acquired via work done by a microwave oven , which in turn took electrical energy from the electrical grid . in practice , whenever work is done to move energy from one form to another , there is always some loss to other forms of energy such as heat and sound . for example , a traditional light bulb is only about 3 % efficient at converting electrical energy to visible light , while a human being is about 25 % efficient at converting chemical energy from food into work . how do we measure energy and work ? the standard unit used to measure energy and work done in physics is the joule , which has the symbol j . in mechanics , 1 joule is the energy transferred when a force of 1 newton is applied to an object and moves it through a distance of 1 meter . another unit of energy you may have come across is the calorie . the amount of energy in an item of food is often written in calories on the back of the packet . a typical 60 gram chocolate bar for example contains about 280 calories of energy . one calorie is the amount of energy required to raise 1 kg of water by 1 $ ^ { \circ } $ celsius . this is equal to 4184 joules , so one chocolate bar has 1.17 million joules or 1.17 mj of stored energy . that 's a lot of joules ! how long do i have to push a heavy box around to burn off one chocolate bar ? suppose we 're feeling guilty about eating a chocolate bar ; we want to find how much exercise we need to do to offset those extra 280 calories . let 's consider a simple form of exercise : pushing a heavy box around a room , see figure 1 below . using a bathroom scale between ourselves and the box , we find that we can push with a force of 500 n. meanwhile , we use a stopwatch and measuring tape to measure our speed . this comes out to be 0.25 meters per second . so how much work do we need to do to the box to burn off the candy bar ? the definition of work , $ w $ , is below : $ \large w = f\cdot \delta x $ the work we need to do to burn the energy in the candy bar is $ e=280 \mathrm { cal } \cdot 4184 \mathrm { j/cal } =1.17 \mathrm { mj } $ . therefore , the distance , $ δx $ , we need to move the box through is : $ \begin { align } w & amp ; = f\cdot \delta x \ 1.17 \text { mj } & amp ; = ( 500 \text { n } ) \cdot \delta x \ \dfrac { 1.17\times 10^6 \text { j } } { 500 \text { n } } & amp ; = \delta x \ 2,340 \text { m } & amp ; = \delta x \end { align } $ remember , however , that our bodies are about 25 % efficient at transferring stored energy from food into work . the actual energy we will offset is then four times as high as the work done to the box . so , we only need to push the box through a distance of 585 m , which is still over five football fields long . given the known speed of 0.25 m/s that will take us : $ \frac { 585 \mathrm { m } } { 0.25 \mathrm { m/s } } =2340 \mathrm { s } $ exercise : suppose that the force that we apply to the box , see figure 1 above , is initially reduced but increases to a constant value as we warm up . for instance , in the graph below we see that as the box is displaced further—i.e. , $ x $ gets larger—the force , $ f $ , increases for the first 30 m , see figure 2 below . how could we find the work done during the period where the force is changing ? if the force is not constant , one way to determine the work done is to divide the problem up into small sections over which the change is negligible and sum up the work done in each section . just as we have learned when looking at velocity time graphs , this can be done by calculating the area under the curve using geometry . the work done by a force is equal to the area under a force vs. position graph . in the case of figure 2 , it would be : $ ( 200\text { n } \cdot 30 \text { m } ) + \frac { 1 } { 2 } \left ( ( 500\text { n } -200\text { n } ) \cdot 30 \text { m } \right ) =10500 \mathrm { j } $ for the initial $ 30 \text { m } $ of displacement . similarly , the work done for the final 40 m of displacement would be : $ 500\text { n } \cdot 40 \text { m } =20,000 \mathrm { j } $ what if we are n't pushing straight on ? there is one thing we need to watch out for when doing these problems . the previous equation , $ w = f \cdot \delta x $ , does n't take into account situations where the force we are applying is not in the same direction as the motion . for instance , imagine we use a rope to pull on the box . in that case there will be an angle between the rope and the ground . to untangle this situation , we begin by drawing a triangle to separate out the horizontal and vertical components of the applied force . the key point here is that it is only the component of the force , $ f_ { || } $ , that lies parallel to the displacement that does work on an object . in the case of the box shown above , only the horizontal component of the applied force , $ f \text { cos } ( \theta ) $ , is doing work on the box since the box is being displaced horizontally . this means that a more general equation for the work done on the box by a force at an angle θ could be written as : $ w=f_ { || } \cdot\delta x $ $ w= ( f \cos { \theta } ) \cdot\delta x $ which is more often written as , $ \large w=f \delta x\cos { \theta } $ exercise : suppose we use a rope to pull the box , and the angle between the rope and the ground is 30º . this time we pull along the rope with a force of 500 n. how much of a chocolate bar can we eat this time if we pull the box through the same 585 m ? what about lifting weights instead ? in the previous example , we were doing work on a box which we were pushing around a floor . in doing so , we were working against a frictional force . another common form of exercise is lifting weights . in this case we are working against the force of gravity rather than friction . using newton 's laws we can find the force , $ f $ , required to lift a weight with mass $ m $ straight up , placing it on a rack which is at a height $ h $ above us : $ f=mg $ the change in position—previously $ \delta x $ —is simply the height , so the work , $ w $ , that we have done in lifting the weight is then $ w=m g h $ the exercise we have done in lifting the weight has resulted in energy being stored in the form of gravitational potential energy . it is called potential energy because it has the potential to be released at any moment with a crash as the weight falls back to the ground . we did positive work on the weight since we exerted our force in the same direction as the displacement of the weight , i.e. , upward . the work done by gravity on the weight while it was lifted was negative since the force of gravity is directed in the opposite direction to the displacement . also , since the weight is stationary after the lift , we know that the work that we have done is exactly canceled out by the work done by gravity . the work done by us is $ mgh $ , and the work done by gravity is $ -mgh $ . we will talk more about this when we look into kinetic energy . ok , let 's put some numbers in and find how much of that chocolate bar we would offset by lifting a weight of 50 kg up to a height of 0.5 m. the work done to the weight is $ w= ( 50\mathrm { kg } ) ( 9.81\mathrm { m/s^2 } ) ( 0.5\mathrm { m } ) = 245.25 \mathrm { j } $ ok , so how many 280 calorie—i.e. , $ 1.17 \times 10^6 $ joule—chocolate bars is this ? well , 245.25 j is about $ \dfrac { 1 } { 4770 } $ of a chocolate bar . but remember , our bodies are only about 25 % efficient , so the work done by the person is actually four times larger , about 981.8 j , which is $ \dfrac { 1 } { 1190 } $ chocolate bars . so , if we can lift this weight once every 2 seconds , it will take us about 2380 seconds or 40 minutes of hard work to burn off this chocolate bar ! what about simply holding a weight stationary ? one frequent source of confusion people have with the concept of work comes about when thinking about holding a heavy weight stationary above our heads , against the force of gravity . we are not moving the weight through any distance , so no work is being done to the weight . we could also achieve this by placing the weight on a table ; it is clear that the table is not doing any work to keep the weight in position . yet , we know from our experience that we get tired when doing the same job . so what is going on here ? it turns out that what is actually happening here is that our bodies are doing work on our muscles to maintain the necessary tension to hold the weight up . the body does this by sending a cascade of nerve impulses to each muscle . each impulse causes the muscle to momentarily contract and release . this all happens so fast that we might only notice a slight twitching at first . eventually though , not enough chemical energy is available in the muscle and it can no longer keep up . we then begin to shake and eventually must rest for a while . so work is being done , it is just not being done on the weight .
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it is not a material substance . energy can be stored and measured in many forms . although we often hear people talking about energy consumption , energy is never really destroyed . it is just transferred from one form to another , doing work in the process .
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so can this equation be reversed and does this equation disprove the law of conservation of mass and energy ?
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what does energy and work mean ? energy is a word which tends to be used a lot in everyday life . though it is often used quite loosely , it does have a very specific physical meaning . energy is a measurement of the ability of something to do work . it is not a material substance . energy can be stored and measured in many forms . although we often hear people talking about energy consumption , energy is never really destroyed . it is just transferred from one form to another , doing work in the process . some forms of energy are less useful to us than others—for example , low level heat energy . it is better to talk about the consumption or extraction of energy resources , for example coal , oil , or wind , than consumption of energy itself . a speeding bullet has a measurable amount of energy associated with it ; this is known as kinetic energy . the bullet gained this energy because work was done on it by a charge of gunpowder which lost some chemical potential energy in the process . a hot cup of coffee has a measurable amount of thermal energy which it acquired via work done by a microwave oven , which in turn took electrical energy from the electrical grid . in practice , whenever work is done to move energy from one form to another , there is always some loss to other forms of energy such as heat and sound . for example , a traditional light bulb is only about 3 % efficient at converting electrical energy to visible light , while a human being is about 25 % efficient at converting chemical energy from food into work . how do we measure energy and work ? the standard unit used to measure energy and work done in physics is the joule , which has the symbol j . in mechanics , 1 joule is the energy transferred when a force of 1 newton is applied to an object and moves it through a distance of 1 meter . another unit of energy you may have come across is the calorie . the amount of energy in an item of food is often written in calories on the back of the packet . a typical 60 gram chocolate bar for example contains about 280 calories of energy . one calorie is the amount of energy required to raise 1 kg of water by 1 $ ^ { \circ } $ celsius . this is equal to 4184 joules , so one chocolate bar has 1.17 million joules or 1.17 mj of stored energy . that 's a lot of joules ! how long do i have to push a heavy box around to burn off one chocolate bar ? suppose we 're feeling guilty about eating a chocolate bar ; we want to find how much exercise we need to do to offset those extra 280 calories . let 's consider a simple form of exercise : pushing a heavy box around a room , see figure 1 below . using a bathroom scale between ourselves and the box , we find that we can push with a force of 500 n. meanwhile , we use a stopwatch and measuring tape to measure our speed . this comes out to be 0.25 meters per second . so how much work do we need to do to the box to burn off the candy bar ? the definition of work , $ w $ , is below : $ \large w = f\cdot \delta x $ the work we need to do to burn the energy in the candy bar is $ e=280 \mathrm { cal } \cdot 4184 \mathrm { j/cal } =1.17 \mathrm { mj } $ . therefore , the distance , $ δx $ , we need to move the box through is : $ \begin { align } w & amp ; = f\cdot \delta x \ 1.17 \text { mj } & amp ; = ( 500 \text { n } ) \cdot \delta x \ \dfrac { 1.17\times 10^6 \text { j } } { 500 \text { n } } & amp ; = \delta x \ 2,340 \text { m } & amp ; = \delta x \end { align } $ remember , however , that our bodies are about 25 % efficient at transferring stored energy from food into work . the actual energy we will offset is then four times as high as the work done to the box . so , we only need to push the box through a distance of 585 m , which is still over five football fields long . given the known speed of 0.25 m/s that will take us : $ \frac { 585 \mathrm { m } } { 0.25 \mathrm { m/s } } =2340 \mathrm { s } $ exercise : suppose that the force that we apply to the box , see figure 1 above , is initially reduced but increases to a constant value as we warm up . for instance , in the graph below we see that as the box is displaced further—i.e. , $ x $ gets larger—the force , $ f $ , increases for the first 30 m , see figure 2 below . how could we find the work done during the period where the force is changing ? if the force is not constant , one way to determine the work done is to divide the problem up into small sections over which the change is negligible and sum up the work done in each section . just as we have learned when looking at velocity time graphs , this can be done by calculating the area under the curve using geometry . the work done by a force is equal to the area under a force vs. position graph . in the case of figure 2 , it would be : $ ( 200\text { n } \cdot 30 \text { m } ) + \frac { 1 } { 2 } \left ( ( 500\text { n } -200\text { n } ) \cdot 30 \text { m } \right ) =10500 \mathrm { j } $ for the initial $ 30 \text { m } $ of displacement . similarly , the work done for the final 40 m of displacement would be : $ 500\text { n } \cdot 40 \text { m } =20,000 \mathrm { j } $ what if we are n't pushing straight on ? there is one thing we need to watch out for when doing these problems . the previous equation , $ w = f \cdot \delta x $ , does n't take into account situations where the force we are applying is not in the same direction as the motion . for instance , imagine we use a rope to pull on the box . in that case there will be an angle between the rope and the ground . to untangle this situation , we begin by drawing a triangle to separate out the horizontal and vertical components of the applied force . the key point here is that it is only the component of the force , $ f_ { || } $ , that lies parallel to the displacement that does work on an object . in the case of the box shown above , only the horizontal component of the applied force , $ f \text { cos } ( \theta ) $ , is doing work on the box since the box is being displaced horizontally . this means that a more general equation for the work done on the box by a force at an angle θ could be written as : $ w=f_ { || } \cdot\delta x $ $ w= ( f \cos { \theta } ) \cdot\delta x $ which is more often written as , $ \large w=f \delta x\cos { \theta } $ exercise : suppose we use a rope to pull the box , and the angle between the rope and the ground is 30º . this time we pull along the rope with a force of 500 n. how much of a chocolate bar can we eat this time if we pull the box through the same 585 m ? what about lifting weights instead ? in the previous example , we were doing work on a box which we were pushing around a floor . in doing so , we were working against a frictional force . another common form of exercise is lifting weights . in this case we are working against the force of gravity rather than friction . using newton 's laws we can find the force , $ f $ , required to lift a weight with mass $ m $ straight up , placing it on a rack which is at a height $ h $ above us : $ f=mg $ the change in position—previously $ \delta x $ —is simply the height , so the work , $ w $ , that we have done in lifting the weight is then $ w=m g h $ the exercise we have done in lifting the weight has resulted in energy being stored in the form of gravitational potential energy . it is called potential energy because it has the potential to be released at any moment with a crash as the weight falls back to the ground . we did positive work on the weight since we exerted our force in the same direction as the displacement of the weight , i.e. , upward . the work done by gravity on the weight while it was lifted was negative since the force of gravity is directed in the opposite direction to the displacement . also , since the weight is stationary after the lift , we know that the work that we have done is exactly canceled out by the work done by gravity . the work done by us is $ mgh $ , and the work done by gravity is $ -mgh $ . we will talk more about this when we look into kinetic energy . ok , let 's put some numbers in and find how much of that chocolate bar we would offset by lifting a weight of 50 kg up to a height of 0.5 m. the work done to the weight is $ w= ( 50\mathrm { kg } ) ( 9.81\mathrm { m/s^2 } ) ( 0.5\mathrm { m } ) = 245.25 \mathrm { j } $ ok , so how many 280 calorie—i.e. , $ 1.17 \times 10^6 $ joule—chocolate bars is this ? well , 245.25 j is about $ \dfrac { 1 } { 4770 } $ of a chocolate bar . but remember , our bodies are only about 25 % efficient , so the work done by the person is actually four times larger , about 981.8 j , which is $ \dfrac { 1 } { 1190 } $ chocolate bars . so , if we can lift this weight once every 2 seconds , it will take us about 2380 seconds or 40 minutes of hard work to burn off this chocolate bar ! what about simply holding a weight stationary ? one frequent source of confusion people have with the concept of work comes about when thinking about holding a heavy weight stationary above our heads , against the force of gravity . we are not moving the weight through any distance , so no work is being done to the weight . we could also achieve this by placing the weight on a table ; it is clear that the table is not doing any work to keep the weight in position . yet , we know from our experience that we get tired when doing the same job . so what is going on here ? it turns out that what is actually happening here is that our bodies are doing work on our muscles to maintain the necessary tension to hold the weight up . the body does this by sending a cascade of nerve impulses to each muscle . each impulse causes the muscle to momentarily contract and release . this all happens so fast that we might only notice a slight twitching at first . eventually though , not enough chemical energy is available in the muscle and it can no longer keep up . we then begin to shake and eventually must rest for a while . so work is being done , it is just not being done on the weight .
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in practice , whenever work is done to move energy from one form to another , there is always some loss to other forms of energy such as heat and sound . for example , a traditional light bulb is only about 3 % efficient at converting electrical energy to visible light , while a human being is about 25 % efficient at converting chemical energy from food into work . how do we measure energy and work ? the standard unit used to measure energy and work done in physics is the joule , which has the symbol j .
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humans are about 25 % efficient at transferring stored energy from food into work.could anyone explain the meaning of this and how can i use this type of information while solving exercises ?
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what does energy and work mean ? energy is a word which tends to be used a lot in everyday life . though it is often used quite loosely , it does have a very specific physical meaning . energy is a measurement of the ability of something to do work . it is not a material substance . energy can be stored and measured in many forms . although we often hear people talking about energy consumption , energy is never really destroyed . it is just transferred from one form to another , doing work in the process . some forms of energy are less useful to us than others—for example , low level heat energy . it is better to talk about the consumption or extraction of energy resources , for example coal , oil , or wind , than consumption of energy itself . a speeding bullet has a measurable amount of energy associated with it ; this is known as kinetic energy . the bullet gained this energy because work was done on it by a charge of gunpowder which lost some chemical potential energy in the process . a hot cup of coffee has a measurable amount of thermal energy which it acquired via work done by a microwave oven , which in turn took electrical energy from the electrical grid . in practice , whenever work is done to move energy from one form to another , there is always some loss to other forms of energy such as heat and sound . for example , a traditional light bulb is only about 3 % efficient at converting electrical energy to visible light , while a human being is about 25 % efficient at converting chemical energy from food into work . how do we measure energy and work ? the standard unit used to measure energy and work done in physics is the joule , which has the symbol j . in mechanics , 1 joule is the energy transferred when a force of 1 newton is applied to an object and moves it through a distance of 1 meter . another unit of energy you may have come across is the calorie . the amount of energy in an item of food is often written in calories on the back of the packet . a typical 60 gram chocolate bar for example contains about 280 calories of energy . one calorie is the amount of energy required to raise 1 kg of water by 1 $ ^ { \circ } $ celsius . this is equal to 4184 joules , so one chocolate bar has 1.17 million joules or 1.17 mj of stored energy . that 's a lot of joules ! how long do i have to push a heavy box around to burn off one chocolate bar ? suppose we 're feeling guilty about eating a chocolate bar ; we want to find how much exercise we need to do to offset those extra 280 calories . let 's consider a simple form of exercise : pushing a heavy box around a room , see figure 1 below . using a bathroom scale between ourselves and the box , we find that we can push with a force of 500 n. meanwhile , we use a stopwatch and measuring tape to measure our speed . this comes out to be 0.25 meters per second . so how much work do we need to do to the box to burn off the candy bar ? the definition of work , $ w $ , is below : $ \large w = f\cdot \delta x $ the work we need to do to burn the energy in the candy bar is $ e=280 \mathrm { cal } \cdot 4184 \mathrm { j/cal } =1.17 \mathrm { mj } $ . therefore , the distance , $ δx $ , we need to move the box through is : $ \begin { align } w & amp ; = f\cdot \delta x \ 1.17 \text { mj } & amp ; = ( 500 \text { n } ) \cdot \delta x \ \dfrac { 1.17\times 10^6 \text { j } } { 500 \text { n } } & amp ; = \delta x \ 2,340 \text { m } & amp ; = \delta x \end { align } $ remember , however , that our bodies are about 25 % efficient at transferring stored energy from food into work . the actual energy we will offset is then four times as high as the work done to the box . so , we only need to push the box through a distance of 585 m , which is still over five football fields long . given the known speed of 0.25 m/s that will take us : $ \frac { 585 \mathrm { m } } { 0.25 \mathrm { m/s } } =2340 \mathrm { s } $ exercise : suppose that the force that we apply to the box , see figure 1 above , is initially reduced but increases to a constant value as we warm up . for instance , in the graph below we see that as the box is displaced further—i.e. , $ x $ gets larger—the force , $ f $ , increases for the first 30 m , see figure 2 below . how could we find the work done during the period where the force is changing ? if the force is not constant , one way to determine the work done is to divide the problem up into small sections over which the change is negligible and sum up the work done in each section . just as we have learned when looking at velocity time graphs , this can be done by calculating the area under the curve using geometry . the work done by a force is equal to the area under a force vs. position graph . in the case of figure 2 , it would be : $ ( 200\text { n } \cdot 30 \text { m } ) + \frac { 1 } { 2 } \left ( ( 500\text { n } -200\text { n } ) \cdot 30 \text { m } \right ) =10500 \mathrm { j } $ for the initial $ 30 \text { m } $ of displacement . similarly , the work done for the final 40 m of displacement would be : $ 500\text { n } \cdot 40 \text { m } =20,000 \mathrm { j } $ what if we are n't pushing straight on ? there is one thing we need to watch out for when doing these problems . the previous equation , $ w = f \cdot \delta x $ , does n't take into account situations where the force we are applying is not in the same direction as the motion . for instance , imagine we use a rope to pull on the box . in that case there will be an angle between the rope and the ground . to untangle this situation , we begin by drawing a triangle to separate out the horizontal and vertical components of the applied force . the key point here is that it is only the component of the force , $ f_ { || } $ , that lies parallel to the displacement that does work on an object . in the case of the box shown above , only the horizontal component of the applied force , $ f \text { cos } ( \theta ) $ , is doing work on the box since the box is being displaced horizontally . this means that a more general equation for the work done on the box by a force at an angle θ could be written as : $ w=f_ { || } \cdot\delta x $ $ w= ( f \cos { \theta } ) \cdot\delta x $ which is more often written as , $ \large w=f \delta x\cos { \theta } $ exercise : suppose we use a rope to pull the box , and the angle between the rope and the ground is 30º . this time we pull along the rope with a force of 500 n. how much of a chocolate bar can we eat this time if we pull the box through the same 585 m ? what about lifting weights instead ? in the previous example , we were doing work on a box which we were pushing around a floor . in doing so , we were working against a frictional force . another common form of exercise is lifting weights . in this case we are working against the force of gravity rather than friction . using newton 's laws we can find the force , $ f $ , required to lift a weight with mass $ m $ straight up , placing it on a rack which is at a height $ h $ above us : $ f=mg $ the change in position—previously $ \delta x $ —is simply the height , so the work , $ w $ , that we have done in lifting the weight is then $ w=m g h $ the exercise we have done in lifting the weight has resulted in energy being stored in the form of gravitational potential energy . it is called potential energy because it has the potential to be released at any moment with a crash as the weight falls back to the ground . we did positive work on the weight since we exerted our force in the same direction as the displacement of the weight , i.e. , upward . the work done by gravity on the weight while it was lifted was negative since the force of gravity is directed in the opposite direction to the displacement . also , since the weight is stationary after the lift , we know that the work that we have done is exactly canceled out by the work done by gravity . the work done by us is $ mgh $ , and the work done by gravity is $ -mgh $ . we will talk more about this when we look into kinetic energy . ok , let 's put some numbers in and find how much of that chocolate bar we would offset by lifting a weight of 50 kg up to a height of 0.5 m. the work done to the weight is $ w= ( 50\mathrm { kg } ) ( 9.81\mathrm { m/s^2 } ) ( 0.5\mathrm { m } ) = 245.25 \mathrm { j } $ ok , so how many 280 calorie—i.e. , $ 1.17 \times 10^6 $ joule—chocolate bars is this ? well , 245.25 j is about $ \dfrac { 1 } { 4770 } $ of a chocolate bar . but remember , our bodies are only about 25 % efficient , so the work done by the person is actually four times larger , about 981.8 j , which is $ \dfrac { 1 } { 1190 } $ chocolate bars . so , if we can lift this weight once every 2 seconds , it will take us about 2380 seconds or 40 minutes of hard work to burn off this chocolate bar ! what about simply holding a weight stationary ? one frequent source of confusion people have with the concept of work comes about when thinking about holding a heavy weight stationary above our heads , against the force of gravity . we are not moving the weight through any distance , so no work is being done to the weight . we could also achieve this by placing the weight on a table ; it is clear that the table is not doing any work to keep the weight in position . yet , we know from our experience that we get tired when doing the same job . so what is going on here ? it turns out that what is actually happening here is that our bodies are doing work on our muscles to maintain the necessary tension to hold the weight up . the body does this by sending a cascade of nerve impulses to each muscle . each impulse causes the muscle to momentarily contract and release . this all happens so fast that we might only notice a slight twitching at first . eventually though , not enough chemical energy is available in the muscle and it can no longer keep up . we then begin to shake and eventually must rest for a while . so work is being done , it is just not being done on the weight .
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it is better to talk about the consumption or extraction of energy resources , for example coal , oil , or wind , than consumption of energy itself . a speeding bullet has a measurable amount of energy associated with it ; this is known as kinetic energy . the bullet gained this energy because work was done on it by a charge of gunpowder which lost some chemical potential energy in the process .
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is it intuitive to think of energy as the amount of newtons an object has acquired over a distance ?
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what does energy and work mean ? energy is a word which tends to be used a lot in everyday life . though it is often used quite loosely , it does have a very specific physical meaning . energy is a measurement of the ability of something to do work . it is not a material substance . energy can be stored and measured in many forms . although we often hear people talking about energy consumption , energy is never really destroyed . it is just transferred from one form to another , doing work in the process . some forms of energy are less useful to us than others—for example , low level heat energy . it is better to talk about the consumption or extraction of energy resources , for example coal , oil , or wind , than consumption of energy itself . a speeding bullet has a measurable amount of energy associated with it ; this is known as kinetic energy . the bullet gained this energy because work was done on it by a charge of gunpowder which lost some chemical potential energy in the process . a hot cup of coffee has a measurable amount of thermal energy which it acquired via work done by a microwave oven , which in turn took electrical energy from the electrical grid . in practice , whenever work is done to move energy from one form to another , there is always some loss to other forms of energy such as heat and sound . for example , a traditional light bulb is only about 3 % efficient at converting electrical energy to visible light , while a human being is about 25 % efficient at converting chemical energy from food into work . how do we measure energy and work ? the standard unit used to measure energy and work done in physics is the joule , which has the symbol j . in mechanics , 1 joule is the energy transferred when a force of 1 newton is applied to an object and moves it through a distance of 1 meter . another unit of energy you may have come across is the calorie . the amount of energy in an item of food is often written in calories on the back of the packet . a typical 60 gram chocolate bar for example contains about 280 calories of energy . one calorie is the amount of energy required to raise 1 kg of water by 1 $ ^ { \circ } $ celsius . this is equal to 4184 joules , so one chocolate bar has 1.17 million joules or 1.17 mj of stored energy . that 's a lot of joules ! how long do i have to push a heavy box around to burn off one chocolate bar ? suppose we 're feeling guilty about eating a chocolate bar ; we want to find how much exercise we need to do to offset those extra 280 calories . let 's consider a simple form of exercise : pushing a heavy box around a room , see figure 1 below . using a bathroom scale between ourselves and the box , we find that we can push with a force of 500 n. meanwhile , we use a stopwatch and measuring tape to measure our speed . this comes out to be 0.25 meters per second . so how much work do we need to do to the box to burn off the candy bar ? the definition of work , $ w $ , is below : $ \large w = f\cdot \delta x $ the work we need to do to burn the energy in the candy bar is $ e=280 \mathrm { cal } \cdot 4184 \mathrm { j/cal } =1.17 \mathrm { mj } $ . therefore , the distance , $ δx $ , we need to move the box through is : $ \begin { align } w & amp ; = f\cdot \delta x \ 1.17 \text { mj } & amp ; = ( 500 \text { n } ) \cdot \delta x \ \dfrac { 1.17\times 10^6 \text { j } } { 500 \text { n } } & amp ; = \delta x \ 2,340 \text { m } & amp ; = \delta x \end { align } $ remember , however , that our bodies are about 25 % efficient at transferring stored energy from food into work . the actual energy we will offset is then four times as high as the work done to the box . so , we only need to push the box through a distance of 585 m , which is still over five football fields long . given the known speed of 0.25 m/s that will take us : $ \frac { 585 \mathrm { m } } { 0.25 \mathrm { m/s } } =2340 \mathrm { s } $ exercise : suppose that the force that we apply to the box , see figure 1 above , is initially reduced but increases to a constant value as we warm up . for instance , in the graph below we see that as the box is displaced further—i.e. , $ x $ gets larger—the force , $ f $ , increases for the first 30 m , see figure 2 below . how could we find the work done during the period where the force is changing ? if the force is not constant , one way to determine the work done is to divide the problem up into small sections over which the change is negligible and sum up the work done in each section . just as we have learned when looking at velocity time graphs , this can be done by calculating the area under the curve using geometry . the work done by a force is equal to the area under a force vs. position graph . in the case of figure 2 , it would be : $ ( 200\text { n } \cdot 30 \text { m } ) + \frac { 1 } { 2 } \left ( ( 500\text { n } -200\text { n } ) \cdot 30 \text { m } \right ) =10500 \mathrm { j } $ for the initial $ 30 \text { m } $ of displacement . similarly , the work done for the final 40 m of displacement would be : $ 500\text { n } \cdot 40 \text { m } =20,000 \mathrm { j } $ what if we are n't pushing straight on ? there is one thing we need to watch out for when doing these problems . the previous equation , $ w = f \cdot \delta x $ , does n't take into account situations where the force we are applying is not in the same direction as the motion . for instance , imagine we use a rope to pull on the box . in that case there will be an angle between the rope and the ground . to untangle this situation , we begin by drawing a triangle to separate out the horizontal and vertical components of the applied force . the key point here is that it is only the component of the force , $ f_ { || } $ , that lies parallel to the displacement that does work on an object . in the case of the box shown above , only the horizontal component of the applied force , $ f \text { cos } ( \theta ) $ , is doing work on the box since the box is being displaced horizontally . this means that a more general equation for the work done on the box by a force at an angle θ could be written as : $ w=f_ { || } \cdot\delta x $ $ w= ( f \cos { \theta } ) \cdot\delta x $ which is more often written as , $ \large w=f \delta x\cos { \theta } $ exercise : suppose we use a rope to pull the box , and the angle between the rope and the ground is 30º . this time we pull along the rope with a force of 500 n. how much of a chocolate bar can we eat this time if we pull the box through the same 585 m ? what about lifting weights instead ? in the previous example , we were doing work on a box which we were pushing around a floor . in doing so , we were working against a frictional force . another common form of exercise is lifting weights . in this case we are working against the force of gravity rather than friction . using newton 's laws we can find the force , $ f $ , required to lift a weight with mass $ m $ straight up , placing it on a rack which is at a height $ h $ above us : $ f=mg $ the change in position—previously $ \delta x $ —is simply the height , so the work , $ w $ , that we have done in lifting the weight is then $ w=m g h $ the exercise we have done in lifting the weight has resulted in energy being stored in the form of gravitational potential energy . it is called potential energy because it has the potential to be released at any moment with a crash as the weight falls back to the ground . we did positive work on the weight since we exerted our force in the same direction as the displacement of the weight , i.e. , upward . the work done by gravity on the weight while it was lifted was negative since the force of gravity is directed in the opposite direction to the displacement . also , since the weight is stationary after the lift , we know that the work that we have done is exactly canceled out by the work done by gravity . the work done by us is $ mgh $ , and the work done by gravity is $ -mgh $ . we will talk more about this when we look into kinetic energy . ok , let 's put some numbers in and find how much of that chocolate bar we would offset by lifting a weight of 50 kg up to a height of 0.5 m. the work done to the weight is $ w= ( 50\mathrm { kg } ) ( 9.81\mathrm { m/s^2 } ) ( 0.5\mathrm { m } ) = 245.25 \mathrm { j } $ ok , so how many 280 calorie—i.e. , $ 1.17 \times 10^6 $ joule—chocolate bars is this ? well , 245.25 j is about $ \dfrac { 1 } { 4770 } $ of a chocolate bar . but remember , our bodies are only about 25 % efficient , so the work done by the person is actually four times larger , about 981.8 j , which is $ \dfrac { 1 } { 1190 } $ chocolate bars . so , if we can lift this weight once every 2 seconds , it will take us about 2380 seconds or 40 minutes of hard work to burn off this chocolate bar ! what about simply holding a weight stationary ? one frequent source of confusion people have with the concept of work comes about when thinking about holding a heavy weight stationary above our heads , against the force of gravity . we are not moving the weight through any distance , so no work is being done to the weight . we could also achieve this by placing the weight on a table ; it is clear that the table is not doing any work to keep the weight in position . yet , we know from our experience that we get tired when doing the same job . so what is going on here ? it turns out that what is actually happening here is that our bodies are doing work on our muscles to maintain the necessary tension to hold the weight up . the body does this by sending a cascade of nerve impulses to each muscle . each impulse causes the muscle to momentarily contract and release . this all happens so fast that we might only notice a slight twitching at first . eventually though , not enough chemical energy is available in the muscle and it can no longer keep up . we then begin to shake and eventually must rest for a while . so work is being done , it is just not being done on the weight .
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it is not a material substance . energy can be stored and measured in many forms . although we often hear people talking about energy consumption , energy is never really destroyed . it is just transferred from one form to another , doing work in the process .
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so where does the thermal energy go or in which other form is it transformed ?
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what does energy and work mean ? energy is a word which tends to be used a lot in everyday life . though it is often used quite loosely , it does have a very specific physical meaning . energy is a measurement of the ability of something to do work . it is not a material substance . energy can be stored and measured in many forms . although we often hear people talking about energy consumption , energy is never really destroyed . it is just transferred from one form to another , doing work in the process . some forms of energy are less useful to us than others—for example , low level heat energy . it is better to talk about the consumption or extraction of energy resources , for example coal , oil , or wind , than consumption of energy itself . a speeding bullet has a measurable amount of energy associated with it ; this is known as kinetic energy . the bullet gained this energy because work was done on it by a charge of gunpowder which lost some chemical potential energy in the process . a hot cup of coffee has a measurable amount of thermal energy which it acquired via work done by a microwave oven , which in turn took electrical energy from the electrical grid . in practice , whenever work is done to move energy from one form to another , there is always some loss to other forms of energy such as heat and sound . for example , a traditional light bulb is only about 3 % efficient at converting electrical energy to visible light , while a human being is about 25 % efficient at converting chemical energy from food into work . how do we measure energy and work ? the standard unit used to measure energy and work done in physics is the joule , which has the symbol j . in mechanics , 1 joule is the energy transferred when a force of 1 newton is applied to an object and moves it through a distance of 1 meter . another unit of energy you may have come across is the calorie . the amount of energy in an item of food is often written in calories on the back of the packet . a typical 60 gram chocolate bar for example contains about 280 calories of energy . one calorie is the amount of energy required to raise 1 kg of water by 1 $ ^ { \circ } $ celsius . this is equal to 4184 joules , so one chocolate bar has 1.17 million joules or 1.17 mj of stored energy . that 's a lot of joules ! how long do i have to push a heavy box around to burn off one chocolate bar ? suppose we 're feeling guilty about eating a chocolate bar ; we want to find how much exercise we need to do to offset those extra 280 calories . let 's consider a simple form of exercise : pushing a heavy box around a room , see figure 1 below . using a bathroom scale between ourselves and the box , we find that we can push with a force of 500 n. meanwhile , we use a stopwatch and measuring tape to measure our speed . this comes out to be 0.25 meters per second . so how much work do we need to do to the box to burn off the candy bar ? the definition of work , $ w $ , is below : $ \large w = f\cdot \delta x $ the work we need to do to burn the energy in the candy bar is $ e=280 \mathrm { cal } \cdot 4184 \mathrm { j/cal } =1.17 \mathrm { mj } $ . therefore , the distance , $ δx $ , we need to move the box through is : $ \begin { align } w & amp ; = f\cdot \delta x \ 1.17 \text { mj } & amp ; = ( 500 \text { n } ) \cdot \delta x \ \dfrac { 1.17\times 10^6 \text { j } } { 500 \text { n } } & amp ; = \delta x \ 2,340 \text { m } & amp ; = \delta x \end { align } $ remember , however , that our bodies are about 25 % efficient at transferring stored energy from food into work . the actual energy we will offset is then four times as high as the work done to the box . so , we only need to push the box through a distance of 585 m , which is still over five football fields long . given the known speed of 0.25 m/s that will take us : $ \frac { 585 \mathrm { m } } { 0.25 \mathrm { m/s } } =2340 \mathrm { s } $ exercise : suppose that the force that we apply to the box , see figure 1 above , is initially reduced but increases to a constant value as we warm up . for instance , in the graph below we see that as the box is displaced further—i.e. , $ x $ gets larger—the force , $ f $ , increases for the first 30 m , see figure 2 below . how could we find the work done during the period where the force is changing ? if the force is not constant , one way to determine the work done is to divide the problem up into small sections over which the change is negligible and sum up the work done in each section . just as we have learned when looking at velocity time graphs , this can be done by calculating the area under the curve using geometry . the work done by a force is equal to the area under a force vs. position graph . in the case of figure 2 , it would be : $ ( 200\text { n } \cdot 30 \text { m } ) + \frac { 1 } { 2 } \left ( ( 500\text { n } -200\text { n } ) \cdot 30 \text { m } \right ) =10500 \mathrm { j } $ for the initial $ 30 \text { m } $ of displacement . similarly , the work done for the final 40 m of displacement would be : $ 500\text { n } \cdot 40 \text { m } =20,000 \mathrm { j } $ what if we are n't pushing straight on ? there is one thing we need to watch out for when doing these problems . the previous equation , $ w = f \cdot \delta x $ , does n't take into account situations where the force we are applying is not in the same direction as the motion . for instance , imagine we use a rope to pull on the box . in that case there will be an angle between the rope and the ground . to untangle this situation , we begin by drawing a triangle to separate out the horizontal and vertical components of the applied force . the key point here is that it is only the component of the force , $ f_ { || } $ , that lies parallel to the displacement that does work on an object . in the case of the box shown above , only the horizontal component of the applied force , $ f \text { cos } ( \theta ) $ , is doing work on the box since the box is being displaced horizontally . this means that a more general equation for the work done on the box by a force at an angle θ could be written as : $ w=f_ { || } \cdot\delta x $ $ w= ( f \cos { \theta } ) \cdot\delta x $ which is more often written as , $ \large w=f \delta x\cos { \theta } $ exercise : suppose we use a rope to pull the box , and the angle between the rope and the ground is 30º . this time we pull along the rope with a force of 500 n. how much of a chocolate bar can we eat this time if we pull the box through the same 585 m ? what about lifting weights instead ? in the previous example , we were doing work on a box which we were pushing around a floor . in doing so , we were working against a frictional force . another common form of exercise is lifting weights . in this case we are working against the force of gravity rather than friction . using newton 's laws we can find the force , $ f $ , required to lift a weight with mass $ m $ straight up , placing it on a rack which is at a height $ h $ above us : $ f=mg $ the change in position—previously $ \delta x $ —is simply the height , so the work , $ w $ , that we have done in lifting the weight is then $ w=m g h $ the exercise we have done in lifting the weight has resulted in energy being stored in the form of gravitational potential energy . it is called potential energy because it has the potential to be released at any moment with a crash as the weight falls back to the ground . we did positive work on the weight since we exerted our force in the same direction as the displacement of the weight , i.e. , upward . the work done by gravity on the weight while it was lifted was negative since the force of gravity is directed in the opposite direction to the displacement . also , since the weight is stationary after the lift , we know that the work that we have done is exactly canceled out by the work done by gravity . the work done by us is $ mgh $ , and the work done by gravity is $ -mgh $ . we will talk more about this when we look into kinetic energy . ok , let 's put some numbers in and find how much of that chocolate bar we would offset by lifting a weight of 50 kg up to a height of 0.5 m. the work done to the weight is $ w= ( 50\mathrm { kg } ) ( 9.81\mathrm { m/s^2 } ) ( 0.5\mathrm { m } ) = 245.25 \mathrm { j } $ ok , so how many 280 calorie—i.e. , $ 1.17 \times 10^6 $ joule—chocolate bars is this ? well , 245.25 j is about $ \dfrac { 1 } { 4770 } $ of a chocolate bar . but remember , our bodies are only about 25 % efficient , so the work done by the person is actually four times larger , about 981.8 j , which is $ \dfrac { 1 } { 1190 } $ chocolate bars . so , if we can lift this weight once every 2 seconds , it will take us about 2380 seconds or 40 minutes of hard work to burn off this chocolate bar ! what about simply holding a weight stationary ? one frequent source of confusion people have with the concept of work comes about when thinking about holding a heavy weight stationary above our heads , against the force of gravity . we are not moving the weight through any distance , so no work is being done to the weight . we could also achieve this by placing the weight on a table ; it is clear that the table is not doing any work to keep the weight in position . yet , we know from our experience that we get tired when doing the same job . so what is going on here ? it turns out that what is actually happening here is that our bodies are doing work on our muscles to maintain the necessary tension to hold the weight up . the body does this by sending a cascade of nerve impulses to each muscle . each impulse causes the muscle to momentarily contract and release . this all happens so fast that we might only notice a slight twitching at first . eventually though , not enough chemical energy is available in the muscle and it can no longer keep up . we then begin to shake and eventually must rest for a while . so work is being done , it is just not being done on the weight .
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there is one thing we need to watch out for when doing these problems . the previous equation , $ w = f \cdot \delta x $ , does n't take into account situations where the force we are applying is not in the same direction as the motion . for instance , imagine we use a rope to pull on the box .
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is n't your force already in the horizontal direction without needing to correct for the angle ?
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what does energy and work mean ? energy is a word which tends to be used a lot in everyday life . though it is often used quite loosely , it does have a very specific physical meaning . energy is a measurement of the ability of something to do work . it is not a material substance . energy can be stored and measured in many forms . although we often hear people talking about energy consumption , energy is never really destroyed . it is just transferred from one form to another , doing work in the process . some forms of energy are less useful to us than others—for example , low level heat energy . it is better to talk about the consumption or extraction of energy resources , for example coal , oil , or wind , than consumption of energy itself . a speeding bullet has a measurable amount of energy associated with it ; this is known as kinetic energy . the bullet gained this energy because work was done on it by a charge of gunpowder which lost some chemical potential energy in the process . a hot cup of coffee has a measurable amount of thermal energy which it acquired via work done by a microwave oven , which in turn took electrical energy from the electrical grid . in practice , whenever work is done to move energy from one form to another , there is always some loss to other forms of energy such as heat and sound . for example , a traditional light bulb is only about 3 % efficient at converting electrical energy to visible light , while a human being is about 25 % efficient at converting chemical energy from food into work . how do we measure energy and work ? the standard unit used to measure energy and work done in physics is the joule , which has the symbol j . in mechanics , 1 joule is the energy transferred when a force of 1 newton is applied to an object and moves it through a distance of 1 meter . another unit of energy you may have come across is the calorie . the amount of energy in an item of food is often written in calories on the back of the packet . a typical 60 gram chocolate bar for example contains about 280 calories of energy . one calorie is the amount of energy required to raise 1 kg of water by 1 $ ^ { \circ } $ celsius . this is equal to 4184 joules , so one chocolate bar has 1.17 million joules or 1.17 mj of stored energy . that 's a lot of joules ! how long do i have to push a heavy box around to burn off one chocolate bar ? suppose we 're feeling guilty about eating a chocolate bar ; we want to find how much exercise we need to do to offset those extra 280 calories . let 's consider a simple form of exercise : pushing a heavy box around a room , see figure 1 below . using a bathroom scale between ourselves and the box , we find that we can push with a force of 500 n. meanwhile , we use a stopwatch and measuring tape to measure our speed . this comes out to be 0.25 meters per second . so how much work do we need to do to the box to burn off the candy bar ? the definition of work , $ w $ , is below : $ \large w = f\cdot \delta x $ the work we need to do to burn the energy in the candy bar is $ e=280 \mathrm { cal } \cdot 4184 \mathrm { j/cal } =1.17 \mathrm { mj } $ . therefore , the distance , $ δx $ , we need to move the box through is : $ \begin { align } w & amp ; = f\cdot \delta x \ 1.17 \text { mj } & amp ; = ( 500 \text { n } ) \cdot \delta x \ \dfrac { 1.17\times 10^6 \text { j } } { 500 \text { n } } & amp ; = \delta x \ 2,340 \text { m } & amp ; = \delta x \end { align } $ remember , however , that our bodies are about 25 % efficient at transferring stored energy from food into work . the actual energy we will offset is then four times as high as the work done to the box . so , we only need to push the box through a distance of 585 m , which is still over five football fields long . given the known speed of 0.25 m/s that will take us : $ \frac { 585 \mathrm { m } } { 0.25 \mathrm { m/s } } =2340 \mathrm { s } $ exercise : suppose that the force that we apply to the box , see figure 1 above , is initially reduced but increases to a constant value as we warm up . for instance , in the graph below we see that as the box is displaced further—i.e. , $ x $ gets larger—the force , $ f $ , increases for the first 30 m , see figure 2 below . how could we find the work done during the period where the force is changing ? if the force is not constant , one way to determine the work done is to divide the problem up into small sections over which the change is negligible and sum up the work done in each section . just as we have learned when looking at velocity time graphs , this can be done by calculating the area under the curve using geometry . the work done by a force is equal to the area under a force vs. position graph . in the case of figure 2 , it would be : $ ( 200\text { n } \cdot 30 \text { m } ) + \frac { 1 } { 2 } \left ( ( 500\text { n } -200\text { n } ) \cdot 30 \text { m } \right ) =10500 \mathrm { j } $ for the initial $ 30 \text { m } $ of displacement . similarly , the work done for the final 40 m of displacement would be : $ 500\text { n } \cdot 40 \text { m } =20,000 \mathrm { j } $ what if we are n't pushing straight on ? there is one thing we need to watch out for when doing these problems . the previous equation , $ w = f \cdot \delta x $ , does n't take into account situations where the force we are applying is not in the same direction as the motion . for instance , imagine we use a rope to pull on the box . in that case there will be an angle between the rope and the ground . to untangle this situation , we begin by drawing a triangle to separate out the horizontal and vertical components of the applied force . the key point here is that it is only the component of the force , $ f_ { || } $ , that lies parallel to the displacement that does work on an object . in the case of the box shown above , only the horizontal component of the applied force , $ f \text { cos } ( \theta ) $ , is doing work on the box since the box is being displaced horizontally . this means that a more general equation for the work done on the box by a force at an angle θ could be written as : $ w=f_ { || } \cdot\delta x $ $ w= ( f \cos { \theta } ) \cdot\delta x $ which is more often written as , $ \large w=f \delta x\cos { \theta } $ exercise : suppose we use a rope to pull the box , and the angle between the rope and the ground is 30º . this time we pull along the rope with a force of 500 n. how much of a chocolate bar can we eat this time if we pull the box through the same 585 m ? what about lifting weights instead ? in the previous example , we were doing work on a box which we were pushing around a floor . in doing so , we were working against a frictional force . another common form of exercise is lifting weights . in this case we are working against the force of gravity rather than friction . using newton 's laws we can find the force , $ f $ , required to lift a weight with mass $ m $ straight up , placing it on a rack which is at a height $ h $ above us : $ f=mg $ the change in position—previously $ \delta x $ —is simply the height , so the work , $ w $ , that we have done in lifting the weight is then $ w=m g h $ the exercise we have done in lifting the weight has resulted in energy being stored in the form of gravitational potential energy . it is called potential energy because it has the potential to be released at any moment with a crash as the weight falls back to the ground . we did positive work on the weight since we exerted our force in the same direction as the displacement of the weight , i.e. , upward . the work done by gravity on the weight while it was lifted was negative since the force of gravity is directed in the opposite direction to the displacement . also , since the weight is stationary after the lift , we know that the work that we have done is exactly canceled out by the work done by gravity . the work done by us is $ mgh $ , and the work done by gravity is $ -mgh $ . we will talk more about this when we look into kinetic energy . ok , let 's put some numbers in and find how much of that chocolate bar we would offset by lifting a weight of 50 kg up to a height of 0.5 m. the work done to the weight is $ w= ( 50\mathrm { kg } ) ( 9.81\mathrm { m/s^2 } ) ( 0.5\mathrm { m } ) = 245.25 \mathrm { j } $ ok , so how many 280 calorie—i.e. , $ 1.17 \times 10^6 $ joule—chocolate bars is this ? well , 245.25 j is about $ \dfrac { 1 } { 4770 } $ of a chocolate bar . but remember , our bodies are only about 25 % efficient , so the work done by the person is actually four times larger , about 981.8 j , which is $ \dfrac { 1 } { 1190 } $ chocolate bars . so , if we can lift this weight once every 2 seconds , it will take us about 2380 seconds or 40 minutes of hard work to burn off this chocolate bar ! what about simply holding a weight stationary ? one frequent source of confusion people have with the concept of work comes about when thinking about holding a heavy weight stationary above our heads , against the force of gravity . we are not moving the weight through any distance , so no work is being done to the weight . we could also achieve this by placing the weight on a table ; it is clear that the table is not doing any work to keep the weight in position . yet , we know from our experience that we get tired when doing the same job . so what is going on here ? it turns out that what is actually happening here is that our bodies are doing work on our muscles to maintain the necessary tension to hold the weight up . the body does this by sending a cascade of nerve impulses to each muscle . each impulse causes the muscle to momentarily contract and release . this all happens so fast that we might only notice a slight twitching at first . eventually though , not enough chemical energy is available in the muscle and it can no longer keep up . we then begin to shake and eventually must rest for a while . so work is being done , it is just not being done on the weight .
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also , since the weight is stationary after the lift , we know that the work that we have done is exactly canceled out by the work done by gravity . the work done by us is $ mgh $ , and the work done by gravity is $ -mgh $ . we will talk more about this when we look into kinetic energy .
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in reality , when pulling the box with a rope at an angle , is n't the total work done by the body greater than the work done when pushing the block horizontally over the same distance ; because not all of the force exerted on the box is in the direction of motion ?
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what does energy and work mean ? energy is a word which tends to be used a lot in everyday life . though it is often used quite loosely , it does have a very specific physical meaning . energy is a measurement of the ability of something to do work . it is not a material substance . energy can be stored and measured in many forms . although we often hear people talking about energy consumption , energy is never really destroyed . it is just transferred from one form to another , doing work in the process . some forms of energy are less useful to us than others—for example , low level heat energy . it is better to talk about the consumption or extraction of energy resources , for example coal , oil , or wind , than consumption of energy itself . a speeding bullet has a measurable amount of energy associated with it ; this is known as kinetic energy . the bullet gained this energy because work was done on it by a charge of gunpowder which lost some chemical potential energy in the process . a hot cup of coffee has a measurable amount of thermal energy which it acquired via work done by a microwave oven , which in turn took electrical energy from the electrical grid . in practice , whenever work is done to move energy from one form to another , there is always some loss to other forms of energy such as heat and sound . for example , a traditional light bulb is only about 3 % efficient at converting electrical energy to visible light , while a human being is about 25 % efficient at converting chemical energy from food into work . how do we measure energy and work ? the standard unit used to measure energy and work done in physics is the joule , which has the symbol j . in mechanics , 1 joule is the energy transferred when a force of 1 newton is applied to an object and moves it through a distance of 1 meter . another unit of energy you may have come across is the calorie . the amount of energy in an item of food is often written in calories on the back of the packet . a typical 60 gram chocolate bar for example contains about 280 calories of energy . one calorie is the amount of energy required to raise 1 kg of water by 1 $ ^ { \circ } $ celsius . this is equal to 4184 joules , so one chocolate bar has 1.17 million joules or 1.17 mj of stored energy . that 's a lot of joules ! how long do i have to push a heavy box around to burn off one chocolate bar ? suppose we 're feeling guilty about eating a chocolate bar ; we want to find how much exercise we need to do to offset those extra 280 calories . let 's consider a simple form of exercise : pushing a heavy box around a room , see figure 1 below . using a bathroom scale between ourselves and the box , we find that we can push with a force of 500 n. meanwhile , we use a stopwatch and measuring tape to measure our speed . this comes out to be 0.25 meters per second . so how much work do we need to do to the box to burn off the candy bar ? the definition of work , $ w $ , is below : $ \large w = f\cdot \delta x $ the work we need to do to burn the energy in the candy bar is $ e=280 \mathrm { cal } \cdot 4184 \mathrm { j/cal } =1.17 \mathrm { mj } $ . therefore , the distance , $ δx $ , we need to move the box through is : $ \begin { align } w & amp ; = f\cdot \delta x \ 1.17 \text { mj } & amp ; = ( 500 \text { n } ) \cdot \delta x \ \dfrac { 1.17\times 10^6 \text { j } } { 500 \text { n } } & amp ; = \delta x \ 2,340 \text { m } & amp ; = \delta x \end { align } $ remember , however , that our bodies are about 25 % efficient at transferring stored energy from food into work . the actual energy we will offset is then four times as high as the work done to the box . so , we only need to push the box through a distance of 585 m , which is still over five football fields long . given the known speed of 0.25 m/s that will take us : $ \frac { 585 \mathrm { m } } { 0.25 \mathrm { m/s } } =2340 \mathrm { s } $ exercise : suppose that the force that we apply to the box , see figure 1 above , is initially reduced but increases to a constant value as we warm up . for instance , in the graph below we see that as the box is displaced further—i.e. , $ x $ gets larger—the force , $ f $ , increases for the first 30 m , see figure 2 below . how could we find the work done during the period where the force is changing ? if the force is not constant , one way to determine the work done is to divide the problem up into small sections over which the change is negligible and sum up the work done in each section . just as we have learned when looking at velocity time graphs , this can be done by calculating the area under the curve using geometry . the work done by a force is equal to the area under a force vs. position graph . in the case of figure 2 , it would be : $ ( 200\text { n } \cdot 30 \text { m } ) + \frac { 1 } { 2 } \left ( ( 500\text { n } -200\text { n } ) \cdot 30 \text { m } \right ) =10500 \mathrm { j } $ for the initial $ 30 \text { m } $ of displacement . similarly , the work done for the final 40 m of displacement would be : $ 500\text { n } \cdot 40 \text { m } =20,000 \mathrm { j } $ what if we are n't pushing straight on ? there is one thing we need to watch out for when doing these problems . the previous equation , $ w = f \cdot \delta x $ , does n't take into account situations where the force we are applying is not in the same direction as the motion . for instance , imagine we use a rope to pull on the box . in that case there will be an angle between the rope and the ground . to untangle this situation , we begin by drawing a triangle to separate out the horizontal and vertical components of the applied force . the key point here is that it is only the component of the force , $ f_ { || } $ , that lies parallel to the displacement that does work on an object . in the case of the box shown above , only the horizontal component of the applied force , $ f \text { cos } ( \theta ) $ , is doing work on the box since the box is being displaced horizontally . this means that a more general equation for the work done on the box by a force at an angle θ could be written as : $ w=f_ { || } \cdot\delta x $ $ w= ( f \cos { \theta } ) \cdot\delta x $ which is more often written as , $ \large w=f \delta x\cos { \theta } $ exercise : suppose we use a rope to pull the box , and the angle between the rope and the ground is 30º . this time we pull along the rope with a force of 500 n. how much of a chocolate bar can we eat this time if we pull the box through the same 585 m ? what about lifting weights instead ? in the previous example , we were doing work on a box which we were pushing around a floor . in doing so , we were working against a frictional force . another common form of exercise is lifting weights . in this case we are working against the force of gravity rather than friction . using newton 's laws we can find the force , $ f $ , required to lift a weight with mass $ m $ straight up , placing it on a rack which is at a height $ h $ above us : $ f=mg $ the change in position—previously $ \delta x $ —is simply the height , so the work , $ w $ , that we have done in lifting the weight is then $ w=m g h $ the exercise we have done in lifting the weight has resulted in energy being stored in the form of gravitational potential energy . it is called potential energy because it has the potential to be released at any moment with a crash as the weight falls back to the ground . we did positive work on the weight since we exerted our force in the same direction as the displacement of the weight , i.e. , upward . the work done by gravity on the weight while it was lifted was negative since the force of gravity is directed in the opposite direction to the displacement . also , since the weight is stationary after the lift , we know that the work that we have done is exactly canceled out by the work done by gravity . the work done by us is $ mgh $ , and the work done by gravity is $ -mgh $ . we will talk more about this when we look into kinetic energy . ok , let 's put some numbers in and find how much of that chocolate bar we would offset by lifting a weight of 50 kg up to a height of 0.5 m. the work done to the weight is $ w= ( 50\mathrm { kg } ) ( 9.81\mathrm { m/s^2 } ) ( 0.5\mathrm { m } ) = 245.25 \mathrm { j } $ ok , so how many 280 calorie—i.e. , $ 1.17 \times 10^6 $ joule—chocolate bars is this ? well , 245.25 j is about $ \dfrac { 1 } { 4770 } $ of a chocolate bar . but remember , our bodies are only about 25 % efficient , so the work done by the person is actually four times larger , about 981.8 j , which is $ \dfrac { 1 } { 1190 } $ chocolate bars . so , if we can lift this weight once every 2 seconds , it will take us about 2380 seconds or 40 minutes of hard work to burn off this chocolate bar ! what about simply holding a weight stationary ? one frequent source of confusion people have with the concept of work comes about when thinking about holding a heavy weight stationary above our heads , against the force of gravity . we are not moving the weight through any distance , so no work is being done to the weight . we could also achieve this by placing the weight on a table ; it is clear that the table is not doing any work to keep the weight in position . yet , we know from our experience that we get tired when doing the same job . so what is going on here ? it turns out that what is actually happening here is that our bodies are doing work on our muscles to maintain the necessary tension to hold the weight up . the body does this by sending a cascade of nerve impulses to each muscle . each impulse causes the muscle to momentarily contract and release . this all happens so fast that we might only notice a slight twitching at first . eventually though , not enough chemical energy is available in the muscle and it can no longer keep up . we then begin to shake and eventually must rest for a while . so work is being done , it is just not being done on the weight .
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the previous equation , $ w = f \cdot \delta x $ , does n't take into account situations where the force we are applying is not in the same direction as the motion . for instance , imagine we use a rope to pull on the box . in that case there will be an angle between the rope and the ground .
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so would n't you actually need to eat more calories to pull the box with the rope at an angle ?
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what does energy and work mean ? energy is a word which tends to be used a lot in everyday life . though it is often used quite loosely , it does have a very specific physical meaning . energy is a measurement of the ability of something to do work . it is not a material substance . energy can be stored and measured in many forms . although we often hear people talking about energy consumption , energy is never really destroyed . it is just transferred from one form to another , doing work in the process . some forms of energy are less useful to us than others—for example , low level heat energy . it is better to talk about the consumption or extraction of energy resources , for example coal , oil , or wind , than consumption of energy itself . a speeding bullet has a measurable amount of energy associated with it ; this is known as kinetic energy . the bullet gained this energy because work was done on it by a charge of gunpowder which lost some chemical potential energy in the process . a hot cup of coffee has a measurable amount of thermal energy which it acquired via work done by a microwave oven , which in turn took electrical energy from the electrical grid . in practice , whenever work is done to move energy from one form to another , there is always some loss to other forms of energy such as heat and sound . for example , a traditional light bulb is only about 3 % efficient at converting electrical energy to visible light , while a human being is about 25 % efficient at converting chemical energy from food into work . how do we measure energy and work ? the standard unit used to measure energy and work done in physics is the joule , which has the symbol j . in mechanics , 1 joule is the energy transferred when a force of 1 newton is applied to an object and moves it through a distance of 1 meter . another unit of energy you may have come across is the calorie . the amount of energy in an item of food is often written in calories on the back of the packet . a typical 60 gram chocolate bar for example contains about 280 calories of energy . one calorie is the amount of energy required to raise 1 kg of water by 1 $ ^ { \circ } $ celsius . this is equal to 4184 joules , so one chocolate bar has 1.17 million joules or 1.17 mj of stored energy . that 's a lot of joules ! how long do i have to push a heavy box around to burn off one chocolate bar ? suppose we 're feeling guilty about eating a chocolate bar ; we want to find how much exercise we need to do to offset those extra 280 calories . let 's consider a simple form of exercise : pushing a heavy box around a room , see figure 1 below . using a bathroom scale between ourselves and the box , we find that we can push with a force of 500 n. meanwhile , we use a stopwatch and measuring tape to measure our speed . this comes out to be 0.25 meters per second . so how much work do we need to do to the box to burn off the candy bar ? the definition of work , $ w $ , is below : $ \large w = f\cdot \delta x $ the work we need to do to burn the energy in the candy bar is $ e=280 \mathrm { cal } \cdot 4184 \mathrm { j/cal } =1.17 \mathrm { mj } $ . therefore , the distance , $ δx $ , we need to move the box through is : $ \begin { align } w & amp ; = f\cdot \delta x \ 1.17 \text { mj } & amp ; = ( 500 \text { n } ) \cdot \delta x \ \dfrac { 1.17\times 10^6 \text { j } } { 500 \text { n } } & amp ; = \delta x \ 2,340 \text { m } & amp ; = \delta x \end { align } $ remember , however , that our bodies are about 25 % efficient at transferring stored energy from food into work . the actual energy we will offset is then four times as high as the work done to the box . so , we only need to push the box through a distance of 585 m , which is still over five football fields long . given the known speed of 0.25 m/s that will take us : $ \frac { 585 \mathrm { m } } { 0.25 \mathrm { m/s } } =2340 \mathrm { s } $ exercise : suppose that the force that we apply to the box , see figure 1 above , is initially reduced but increases to a constant value as we warm up . for instance , in the graph below we see that as the box is displaced further—i.e. , $ x $ gets larger—the force , $ f $ , increases for the first 30 m , see figure 2 below . how could we find the work done during the period where the force is changing ? if the force is not constant , one way to determine the work done is to divide the problem up into small sections over which the change is negligible and sum up the work done in each section . just as we have learned when looking at velocity time graphs , this can be done by calculating the area under the curve using geometry . the work done by a force is equal to the area under a force vs. position graph . in the case of figure 2 , it would be : $ ( 200\text { n } \cdot 30 \text { m } ) + \frac { 1 } { 2 } \left ( ( 500\text { n } -200\text { n } ) \cdot 30 \text { m } \right ) =10500 \mathrm { j } $ for the initial $ 30 \text { m } $ of displacement . similarly , the work done for the final 40 m of displacement would be : $ 500\text { n } \cdot 40 \text { m } =20,000 \mathrm { j } $ what if we are n't pushing straight on ? there is one thing we need to watch out for when doing these problems . the previous equation , $ w = f \cdot \delta x $ , does n't take into account situations where the force we are applying is not in the same direction as the motion . for instance , imagine we use a rope to pull on the box . in that case there will be an angle between the rope and the ground . to untangle this situation , we begin by drawing a triangle to separate out the horizontal and vertical components of the applied force . the key point here is that it is only the component of the force , $ f_ { || } $ , that lies parallel to the displacement that does work on an object . in the case of the box shown above , only the horizontal component of the applied force , $ f \text { cos } ( \theta ) $ , is doing work on the box since the box is being displaced horizontally . this means that a more general equation for the work done on the box by a force at an angle θ could be written as : $ w=f_ { || } \cdot\delta x $ $ w= ( f \cos { \theta } ) \cdot\delta x $ which is more often written as , $ \large w=f \delta x\cos { \theta } $ exercise : suppose we use a rope to pull the box , and the angle between the rope and the ground is 30º . this time we pull along the rope with a force of 500 n. how much of a chocolate bar can we eat this time if we pull the box through the same 585 m ? what about lifting weights instead ? in the previous example , we were doing work on a box which we were pushing around a floor . in doing so , we were working against a frictional force . another common form of exercise is lifting weights . in this case we are working against the force of gravity rather than friction . using newton 's laws we can find the force , $ f $ , required to lift a weight with mass $ m $ straight up , placing it on a rack which is at a height $ h $ above us : $ f=mg $ the change in position—previously $ \delta x $ —is simply the height , so the work , $ w $ , that we have done in lifting the weight is then $ w=m g h $ the exercise we have done in lifting the weight has resulted in energy being stored in the form of gravitational potential energy . it is called potential energy because it has the potential to be released at any moment with a crash as the weight falls back to the ground . we did positive work on the weight since we exerted our force in the same direction as the displacement of the weight , i.e. , upward . the work done by gravity on the weight while it was lifted was negative since the force of gravity is directed in the opposite direction to the displacement . also , since the weight is stationary after the lift , we know that the work that we have done is exactly canceled out by the work done by gravity . the work done by us is $ mgh $ , and the work done by gravity is $ -mgh $ . we will talk more about this when we look into kinetic energy . ok , let 's put some numbers in and find how much of that chocolate bar we would offset by lifting a weight of 50 kg up to a height of 0.5 m. the work done to the weight is $ w= ( 50\mathrm { kg } ) ( 9.81\mathrm { m/s^2 } ) ( 0.5\mathrm { m } ) = 245.25 \mathrm { j } $ ok , so how many 280 calorie—i.e. , $ 1.17 \times 10^6 $ joule—chocolate bars is this ? well , 245.25 j is about $ \dfrac { 1 } { 4770 } $ of a chocolate bar . but remember , our bodies are only about 25 % efficient , so the work done by the person is actually four times larger , about 981.8 j , which is $ \dfrac { 1 } { 1190 } $ chocolate bars . so , if we can lift this weight once every 2 seconds , it will take us about 2380 seconds or 40 minutes of hard work to burn off this chocolate bar ! what about simply holding a weight stationary ? one frequent source of confusion people have with the concept of work comes about when thinking about holding a heavy weight stationary above our heads , against the force of gravity . we are not moving the weight through any distance , so no work is being done to the weight . we could also achieve this by placing the weight on a table ; it is clear that the table is not doing any work to keep the weight in position . yet , we know from our experience that we get tired when doing the same job . so what is going on here ? it turns out that what is actually happening here is that our bodies are doing work on our muscles to maintain the necessary tension to hold the weight up . the body does this by sending a cascade of nerve impulses to each muscle . each impulse causes the muscle to momentarily contract and release . this all happens so fast that we might only notice a slight twitching at first . eventually though , not enough chemical energy is available in the muscle and it can no longer keep up . we then begin to shake and eventually must rest for a while . so work is being done , it is just not being done on the weight .
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using a bathroom scale between ourselves and the box , we find that we can push with a force of 500 n. meanwhile , we use a stopwatch and measuring tape to measure our speed . this comes out to be 0.25 meters per second . so how much work do we need to do to the box to burn off the candy bar ?
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if a man having 25kg on his head moves a distance of 25 meters horizontally , does he really do any work , provided no force in the horizontal direction by him ?
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what does energy and work mean ? energy is a word which tends to be used a lot in everyday life . though it is often used quite loosely , it does have a very specific physical meaning . energy is a measurement of the ability of something to do work . it is not a material substance . energy can be stored and measured in many forms . although we often hear people talking about energy consumption , energy is never really destroyed . it is just transferred from one form to another , doing work in the process . some forms of energy are less useful to us than others—for example , low level heat energy . it is better to talk about the consumption or extraction of energy resources , for example coal , oil , or wind , than consumption of energy itself . a speeding bullet has a measurable amount of energy associated with it ; this is known as kinetic energy . the bullet gained this energy because work was done on it by a charge of gunpowder which lost some chemical potential energy in the process . a hot cup of coffee has a measurable amount of thermal energy which it acquired via work done by a microwave oven , which in turn took electrical energy from the electrical grid . in practice , whenever work is done to move energy from one form to another , there is always some loss to other forms of energy such as heat and sound . for example , a traditional light bulb is only about 3 % efficient at converting electrical energy to visible light , while a human being is about 25 % efficient at converting chemical energy from food into work . how do we measure energy and work ? the standard unit used to measure energy and work done in physics is the joule , which has the symbol j . in mechanics , 1 joule is the energy transferred when a force of 1 newton is applied to an object and moves it through a distance of 1 meter . another unit of energy you may have come across is the calorie . the amount of energy in an item of food is often written in calories on the back of the packet . a typical 60 gram chocolate bar for example contains about 280 calories of energy . one calorie is the amount of energy required to raise 1 kg of water by 1 $ ^ { \circ } $ celsius . this is equal to 4184 joules , so one chocolate bar has 1.17 million joules or 1.17 mj of stored energy . that 's a lot of joules ! how long do i have to push a heavy box around to burn off one chocolate bar ? suppose we 're feeling guilty about eating a chocolate bar ; we want to find how much exercise we need to do to offset those extra 280 calories . let 's consider a simple form of exercise : pushing a heavy box around a room , see figure 1 below . using a bathroom scale between ourselves and the box , we find that we can push with a force of 500 n. meanwhile , we use a stopwatch and measuring tape to measure our speed . this comes out to be 0.25 meters per second . so how much work do we need to do to the box to burn off the candy bar ? the definition of work , $ w $ , is below : $ \large w = f\cdot \delta x $ the work we need to do to burn the energy in the candy bar is $ e=280 \mathrm { cal } \cdot 4184 \mathrm { j/cal } =1.17 \mathrm { mj } $ . therefore , the distance , $ δx $ , we need to move the box through is : $ \begin { align } w & amp ; = f\cdot \delta x \ 1.17 \text { mj } & amp ; = ( 500 \text { n } ) \cdot \delta x \ \dfrac { 1.17\times 10^6 \text { j } } { 500 \text { n } } & amp ; = \delta x \ 2,340 \text { m } & amp ; = \delta x \end { align } $ remember , however , that our bodies are about 25 % efficient at transferring stored energy from food into work . the actual energy we will offset is then four times as high as the work done to the box . so , we only need to push the box through a distance of 585 m , which is still over five football fields long . given the known speed of 0.25 m/s that will take us : $ \frac { 585 \mathrm { m } } { 0.25 \mathrm { m/s } } =2340 \mathrm { s } $ exercise : suppose that the force that we apply to the box , see figure 1 above , is initially reduced but increases to a constant value as we warm up . for instance , in the graph below we see that as the box is displaced further—i.e. , $ x $ gets larger—the force , $ f $ , increases for the first 30 m , see figure 2 below . how could we find the work done during the period where the force is changing ? if the force is not constant , one way to determine the work done is to divide the problem up into small sections over which the change is negligible and sum up the work done in each section . just as we have learned when looking at velocity time graphs , this can be done by calculating the area under the curve using geometry . the work done by a force is equal to the area under a force vs. position graph . in the case of figure 2 , it would be : $ ( 200\text { n } \cdot 30 \text { m } ) + \frac { 1 } { 2 } \left ( ( 500\text { n } -200\text { n } ) \cdot 30 \text { m } \right ) =10500 \mathrm { j } $ for the initial $ 30 \text { m } $ of displacement . similarly , the work done for the final 40 m of displacement would be : $ 500\text { n } \cdot 40 \text { m } =20,000 \mathrm { j } $ what if we are n't pushing straight on ? there is one thing we need to watch out for when doing these problems . the previous equation , $ w = f \cdot \delta x $ , does n't take into account situations where the force we are applying is not in the same direction as the motion . for instance , imagine we use a rope to pull on the box . in that case there will be an angle between the rope and the ground . to untangle this situation , we begin by drawing a triangle to separate out the horizontal and vertical components of the applied force . the key point here is that it is only the component of the force , $ f_ { || } $ , that lies parallel to the displacement that does work on an object . in the case of the box shown above , only the horizontal component of the applied force , $ f \text { cos } ( \theta ) $ , is doing work on the box since the box is being displaced horizontally . this means that a more general equation for the work done on the box by a force at an angle θ could be written as : $ w=f_ { || } \cdot\delta x $ $ w= ( f \cos { \theta } ) \cdot\delta x $ which is more often written as , $ \large w=f \delta x\cos { \theta } $ exercise : suppose we use a rope to pull the box , and the angle between the rope and the ground is 30º . this time we pull along the rope with a force of 500 n. how much of a chocolate bar can we eat this time if we pull the box through the same 585 m ? what about lifting weights instead ? in the previous example , we were doing work on a box which we were pushing around a floor . in doing so , we were working against a frictional force . another common form of exercise is lifting weights . in this case we are working against the force of gravity rather than friction . using newton 's laws we can find the force , $ f $ , required to lift a weight with mass $ m $ straight up , placing it on a rack which is at a height $ h $ above us : $ f=mg $ the change in position—previously $ \delta x $ —is simply the height , so the work , $ w $ , that we have done in lifting the weight is then $ w=m g h $ the exercise we have done in lifting the weight has resulted in energy being stored in the form of gravitational potential energy . it is called potential energy because it has the potential to be released at any moment with a crash as the weight falls back to the ground . we did positive work on the weight since we exerted our force in the same direction as the displacement of the weight , i.e. , upward . the work done by gravity on the weight while it was lifted was negative since the force of gravity is directed in the opposite direction to the displacement . also , since the weight is stationary after the lift , we know that the work that we have done is exactly canceled out by the work done by gravity . the work done by us is $ mgh $ , and the work done by gravity is $ -mgh $ . we will talk more about this when we look into kinetic energy . ok , let 's put some numbers in and find how much of that chocolate bar we would offset by lifting a weight of 50 kg up to a height of 0.5 m. the work done to the weight is $ w= ( 50\mathrm { kg } ) ( 9.81\mathrm { m/s^2 } ) ( 0.5\mathrm { m } ) = 245.25 \mathrm { j } $ ok , so how many 280 calorie—i.e. , $ 1.17 \times 10^6 $ joule—chocolate bars is this ? well , 245.25 j is about $ \dfrac { 1 } { 4770 } $ of a chocolate bar . but remember , our bodies are only about 25 % efficient , so the work done by the person is actually four times larger , about 981.8 j , which is $ \dfrac { 1 } { 1190 } $ chocolate bars . so , if we can lift this weight once every 2 seconds , it will take us about 2380 seconds or 40 minutes of hard work to burn off this chocolate bar ! what about simply holding a weight stationary ? one frequent source of confusion people have with the concept of work comes about when thinking about holding a heavy weight stationary above our heads , against the force of gravity . we are not moving the weight through any distance , so no work is being done to the weight . we could also achieve this by placing the weight on a table ; it is clear that the table is not doing any work to keep the weight in position . yet , we know from our experience that we get tired when doing the same job . so what is going on here ? it turns out that what is actually happening here is that our bodies are doing work on our muscles to maintain the necessary tension to hold the weight up . the body does this by sending a cascade of nerve impulses to each muscle . each impulse causes the muscle to momentarily contract and release . this all happens so fast that we might only notice a slight twitching at first . eventually though , not enough chemical energy is available in the muscle and it can no longer keep up . we then begin to shake and eventually must rest for a while . so work is being done , it is just not being done on the weight .
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what does energy and work mean ? energy is a word which tends to be used a lot in everyday life .
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does work and energy have the same measurement as joules ?
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what does energy and work mean ? energy is a word which tends to be used a lot in everyday life . though it is often used quite loosely , it does have a very specific physical meaning . energy is a measurement of the ability of something to do work . it is not a material substance . energy can be stored and measured in many forms . although we often hear people talking about energy consumption , energy is never really destroyed . it is just transferred from one form to another , doing work in the process . some forms of energy are less useful to us than others—for example , low level heat energy . it is better to talk about the consumption or extraction of energy resources , for example coal , oil , or wind , than consumption of energy itself . a speeding bullet has a measurable amount of energy associated with it ; this is known as kinetic energy . the bullet gained this energy because work was done on it by a charge of gunpowder which lost some chemical potential energy in the process . a hot cup of coffee has a measurable amount of thermal energy which it acquired via work done by a microwave oven , which in turn took electrical energy from the electrical grid . in practice , whenever work is done to move energy from one form to another , there is always some loss to other forms of energy such as heat and sound . for example , a traditional light bulb is only about 3 % efficient at converting electrical energy to visible light , while a human being is about 25 % efficient at converting chemical energy from food into work . how do we measure energy and work ? the standard unit used to measure energy and work done in physics is the joule , which has the symbol j . in mechanics , 1 joule is the energy transferred when a force of 1 newton is applied to an object and moves it through a distance of 1 meter . another unit of energy you may have come across is the calorie . the amount of energy in an item of food is often written in calories on the back of the packet . a typical 60 gram chocolate bar for example contains about 280 calories of energy . one calorie is the amount of energy required to raise 1 kg of water by 1 $ ^ { \circ } $ celsius . this is equal to 4184 joules , so one chocolate bar has 1.17 million joules or 1.17 mj of stored energy . that 's a lot of joules ! how long do i have to push a heavy box around to burn off one chocolate bar ? suppose we 're feeling guilty about eating a chocolate bar ; we want to find how much exercise we need to do to offset those extra 280 calories . let 's consider a simple form of exercise : pushing a heavy box around a room , see figure 1 below . using a bathroom scale between ourselves and the box , we find that we can push with a force of 500 n. meanwhile , we use a stopwatch and measuring tape to measure our speed . this comes out to be 0.25 meters per second . so how much work do we need to do to the box to burn off the candy bar ? the definition of work , $ w $ , is below : $ \large w = f\cdot \delta x $ the work we need to do to burn the energy in the candy bar is $ e=280 \mathrm { cal } \cdot 4184 \mathrm { j/cal } =1.17 \mathrm { mj } $ . therefore , the distance , $ δx $ , we need to move the box through is : $ \begin { align } w & amp ; = f\cdot \delta x \ 1.17 \text { mj } & amp ; = ( 500 \text { n } ) \cdot \delta x \ \dfrac { 1.17\times 10^6 \text { j } } { 500 \text { n } } & amp ; = \delta x \ 2,340 \text { m } & amp ; = \delta x \end { align } $ remember , however , that our bodies are about 25 % efficient at transferring stored energy from food into work . the actual energy we will offset is then four times as high as the work done to the box . so , we only need to push the box through a distance of 585 m , which is still over five football fields long . given the known speed of 0.25 m/s that will take us : $ \frac { 585 \mathrm { m } } { 0.25 \mathrm { m/s } } =2340 \mathrm { s } $ exercise : suppose that the force that we apply to the box , see figure 1 above , is initially reduced but increases to a constant value as we warm up . for instance , in the graph below we see that as the box is displaced further—i.e. , $ x $ gets larger—the force , $ f $ , increases for the first 30 m , see figure 2 below . how could we find the work done during the period where the force is changing ? if the force is not constant , one way to determine the work done is to divide the problem up into small sections over which the change is negligible and sum up the work done in each section . just as we have learned when looking at velocity time graphs , this can be done by calculating the area under the curve using geometry . the work done by a force is equal to the area under a force vs. position graph . in the case of figure 2 , it would be : $ ( 200\text { n } \cdot 30 \text { m } ) + \frac { 1 } { 2 } \left ( ( 500\text { n } -200\text { n } ) \cdot 30 \text { m } \right ) =10500 \mathrm { j } $ for the initial $ 30 \text { m } $ of displacement . similarly , the work done for the final 40 m of displacement would be : $ 500\text { n } \cdot 40 \text { m } =20,000 \mathrm { j } $ what if we are n't pushing straight on ? there is one thing we need to watch out for when doing these problems . the previous equation , $ w = f \cdot \delta x $ , does n't take into account situations where the force we are applying is not in the same direction as the motion . for instance , imagine we use a rope to pull on the box . in that case there will be an angle between the rope and the ground . to untangle this situation , we begin by drawing a triangle to separate out the horizontal and vertical components of the applied force . the key point here is that it is only the component of the force , $ f_ { || } $ , that lies parallel to the displacement that does work on an object . in the case of the box shown above , only the horizontal component of the applied force , $ f \text { cos } ( \theta ) $ , is doing work on the box since the box is being displaced horizontally . this means that a more general equation for the work done on the box by a force at an angle θ could be written as : $ w=f_ { || } \cdot\delta x $ $ w= ( f \cos { \theta } ) \cdot\delta x $ which is more often written as , $ \large w=f \delta x\cos { \theta } $ exercise : suppose we use a rope to pull the box , and the angle between the rope and the ground is 30º . this time we pull along the rope with a force of 500 n. how much of a chocolate bar can we eat this time if we pull the box through the same 585 m ? what about lifting weights instead ? in the previous example , we were doing work on a box which we were pushing around a floor . in doing so , we were working against a frictional force . another common form of exercise is lifting weights . in this case we are working against the force of gravity rather than friction . using newton 's laws we can find the force , $ f $ , required to lift a weight with mass $ m $ straight up , placing it on a rack which is at a height $ h $ above us : $ f=mg $ the change in position—previously $ \delta x $ —is simply the height , so the work , $ w $ , that we have done in lifting the weight is then $ w=m g h $ the exercise we have done in lifting the weight has resulted in energy being stored in the form of gravitational potential energy . it is called potential energy because it has the potential to be released at any moment with a crash as the weight falls back to the ground . we did positive work on the weight since we exerted our force in the same direction as the displacement of the weight , i.e. , upward . the work done by gravity on the weight while it was lifted was negative since the force of gravity is directed in the opposite direction to the displacement . also , since the weight is stationary after the lift , we know that the work that we have done is exactly canceled out by the work done by gravity . the work done by us is $ mgh $ , and the work done by gravity is $ -mgh $ . we will talk more about this when we look into kinetic energy . ok , let 's put some numbers in and find how much of that chocolate bar we would offset by lifting a weight of 50 kg up to a height of 0.5 m. the work done to the weight is $ w= ( 50\mathrm { kg } ) ( 9.81\mathrm { m/s^2 } ) ( 0.5\mathrm { m } ) = 245.25 \mathrm { j } $ ok , so how many 280 calorie—i.e. , $ 1.17 \times 10^6 $ joule—chocolate bars is this ? well , 245.25 j is about $ \dfrac { 1 } { 4770 } $ of a chocolate bar . but remember , our bodies are only about 25 % efficient , so the work done by the person is actually four times larger , about 981.8 j , which is $ \dfrac { 1 } { 1190 } $ chocolate bars . so , if we can lift this weight once every 2 seconds , it will take us about 2380 seconds or 40 minutes of hard work to burn off this chocolate bar ! what about simply holding a weight stationary ? one frequent source of confusion people have with the concept of work comes about when thinking about holding a heavy weight stationary above our heads , against the force of gravity . we are not moving the weight through any distance , so no work is being done to the weight . we could also achieve this by placing the weight on a table ; it is clear that the table is not doing any work to keep the weight in position . yet , we know from our experience that we get tired when doing the same job . so what is going on here ? it turns out that what is actually happening here is that our bodies are doing work on our muscles to maintain the necessary tension to hold the weight up . the body does this by sending a cascade of nerve impulses to each muscle . each impulse causes the muscle to momentarily contract and release . this all happens so fast that we might only notice a slight twitching at first . eventually though , not enough chemical energy is available in the muscle and it can no longer keep up . we then begin to shake and eventually must rest for a while . so work is being done , it is just not being done on the weight .
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what does energy and work mean ? energy is a word which tends to be used a lot in everyday life .
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so , when lifting a weight : - chemical energy gets transferred into kinetic energy by me ( positive work ) - the same kinetic energy gets transferred into gravitational energy , by gravity ( negative work ) - the decrease in my chemical energy is equal to the increase in gravitational energy when dropping the weight , slowly : - gravitational energy gets transferred into kinetic energy , by gravity ( positive work ) - chemical energy gets transferred into ?
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what does energy and work mean ? energy is a word which tends to be used a lot in everyday life . though it is often used quite loosely , it does have a very specific physical meaning . energy is a measurement of the ability of something to do work . it is not a material substance . energy can be stored and measured in many forms . although we often hear people talking about energy consumption , energy is never really destroyed . it is just transferred from one form to another , doing work in the process . some forms of energy are less useful to us than others—for example , low level heat energy . it is better to talk about the consumption or extraction of energy resources , for example coal , oil , or wind , than consumption of energy itself . a speeding bullet has a measurable amount of energy associated with it ; this is known as kinetic energy . the bullet gained this energy because work was done on it by a charge of gunpowder which lost some chemical potential energy in the process . a hot cup of coffee has a measurable amount of thermal energy which it acquired via work done by a microwave oven , which in turn took electrical energy from the electrical grid . in practice , whenever work is done to move energy from one form to another , there is always some loss to other forms of energy such as heat and sound . for example , a traditional light bulb is only about 3 % efficient at converting electrical energy to visible light , while a human being is about 25 % efficient at converting chemical energy from food into work . how do we measure energy and work ? the standard unit used to measure energy and work done in physics is the joule , which has the symbol j . in mechanics , 1 joule is the energy transferred when a force of 1 newton is applied to an object and moves it through a distance of 1 meter . another unit of energy you may have come across is the calorie . the amount of energy in an item of food is often written in calories on the back of the packet . a typical 60 gram chocolate bar for example contains about 280 calories of energy . one calorie is the amount of energy required to raise 1 kg of water by 1 $ ^ { \circ } $ celsius . this is equal to 4184 joules , so one chocolate bar has 1.17 million joules or 1.17 mj of stored energy . that 's a lot of joules ! how long do i have to push a heavy box around to burn off one chocolate bar ? suppose we 're feeling guilty about eating a chocolate bar ; we want to find how much exercise we need to do to offset those extra 280 calories . let 's consider a simple form of exercise : pushing a heavy box around a room , see figure 1 below . using a bathroom scale between ourselves and the box , we find that we can push with a force of 500 n. meanwhile , we use a stopwatch and measuring tape to measure our speed . this comes out to be 0.25 meters per second . so how much work do we need to do to the box to burn off the candy bar ? the definition of work , $ w $ , is below : $ \large w = f\cdot \delta x $ the work we need to do to burn the energy in the candy bar is $ e=280 \mathrm { cal } \cdot 4184 \mathrm { j/cal } =1.17 \mathrm { mj } $ . therefore , the distance , $ δx $ , we need to move the box through is : $ \begin { align } w & amp ; = f\cdot \delta x \ 1.17 \text { mj } & amp ; = ( 500 \text { n } ) \cdot \delta x \ \dfrac { 1.17\times 10^6 \text { j } } { 500 \text { n } } & amp ; = \delta x \ 2,340 \text { m } & amp ; = \delta x \end { align } $ remember , however , that our bodies are about 25 % efficient at transferring stored energy from food into work . the actual energy we will offset is then four times as high as the work done to the box . so , we only need to push the box through a distance of 585 m , which is still over five football fields long . given the known speed of 0.25 m/s that will take us : $ \frac { 585 \mathrm { m } } { 0.25 \mathrm { m/s } } =2340 \mathrm { s } $ exercise : suppose that the force that we apply to the box , see figure 1 above , is initially reduced but increases to a constant value as we warm up . for instance , in the graph below we see that as the box is displaced further—i.e. , $ x $ gets larger—the force , $ f $ , increases for the first 30 m , see figure 2 below . how could we find the work done during the period where the force is changing ? if the force is not constant , one way to determine the work done is to divide the problem up into small sections over which the change is negligible and sum up the work done in each section . just as we have learned when looking at velocity time graphs , this can be done by calculating the area under the curve using geometry . the work done by a force is equal to the area under a force vs. position graph . in the case of figure 2 , it would be : $ ( 200\text { n } \cdot 30 \text { m } ) + \frac { 1 } { 2 } \left ( ( 500\text { n } -200\text { n } ) \cdot 30 \text { m } \right ) =10500 \mathrm { j } $ for the initial $ 30 \text { m } $ of displacement . similarly , the work done for the final 40 m of displacement would be : $ 500\text { n } \cdot 40 \text { m } =20,000 \mathrm { j } $ what if we are n't pushing straight on ? there is one thing we need to watch out for when doing these problems . the previous equation , $ w = f \cdot \delta x $ , does n't take into account situations where the force we are applying is not in the same direction as the motion . for instance , imagine we use a rope to pull on the box . in that case there will be an angle between the rope and the ground . to untangle this situation , we begin by drawing a triangle to separate out the horizontal and vertical components of the applied force . the key point here is that it is only the component of the force , $ f_ { || } $ , that lies parallel to the displacement that does work on an object . in the case of the box shown above , only the horizontal component of the applied force , $ f \text { cos } ( \theta ) $ , is doing work on the box since the box is being displaced horizontally . this means that a more general equation for the work done on the box by a force at an angle θ could be written as : $ w=f_ { || } \cdot\delta x $ $ w= ( f \cos { \theta } ) \cdot\delta x $ which is more often written as , $ \large w=f \delta x\cos { \theta } $ exercise : suppose we use a rope to pull the box , and the angle between the rope and the ground is 30º . this time we pull along the rope with a force of 500 n. how much of a chocolate bar can we eat this time if we pull the box through the same 585 m ? what about lifting weights instead ? in the previous example , we were doing work on a box which we were pushing around a floor . in doing so , we were working against a frictional force . another common form of exercise is lifting weights . in this case we are working against the force of gravity rather than friction . using newton 's laws we can find the force , $ f $ , required to lift a weight with mass $ m $ straight up , placing it on a rack which is at a height $ h $ above us : $ f=mg $ the change in position—previously $ \delta x $ —is simply the height , so the work , $ w $ , that we have done in lifting the weight is then $ w=m g h $ the exercise we have done in lifting the weight has resulted in energy being stored in the form of gravitational potential energy . it is called potential energy because it has the potential to be released at any moment with a crash as the weight falls back to the ground . we did positive work on the weight since we exerted our force in the same direction as the displacement of the weight , i.e. , upward . the work done by gravity on the weight while it was lifted was negative since the force of gravity is directed in the opposite direction to the displacement . also , since the weight is stationary after the lift , we know that the work that we have done is exactly canceled out by the work done by gravity . the work done by us is $ mgh $ , and the work done by gravity is $ -mgh $ . we will talk more about this when we look into kinetic energy . ok , let 's put some numbers in and find how much of that chocolate bar we would offset by lifting a weight of 50 kg up to a height of 0.5 m. the work done to the weight is $ w= ( 50\mathrm { kg } ) ( 9.81\mathrm { m/s^2 } ) ( 0.5\mathrm { m } ) = 245.25 \mathrm { j } $ ok , so how many 280 calorie—i.e. , $ 1.17 \times 10^6 $ joule—chocolate bars is this ? well , 245.25 j is about $ \dfrac { 1 } { 4770 } $ of a chocolate bar . but remember , our bodies are only about 25 % efficient , so the work done by the person is actually four times larger , about 981.8 j , which is $ \dfrac { 1 } { 1190 } $ chocolate bars . so , if we can lift this weight once every 2 seconds , it will take us about 2380 seconds or 40 minutes of hard work to burn off this chocolate bar ! what about simply holding a weight stationary ? one frequent source of confusion people have with the concept of work comes about when thinking about holding a heavy weight stationary above our heads , against the force of gravity . we are not moving the weight through any distance , so no work is being done to the weight . we could also achieve this by placing the weight on a table ; it is clear that the table is not doing any work to keep the weight in position . yet , we know from our experience that we get tired when doing the same job . so what is going on here ? it turns out that what is actually happening here is that our bodies are doing work on our muscles to maintain the necessary tension to hold the weight up . the body does this by sending a cascade of nerve impulses to each muscle . each impulse causes the muscle to momentarily contract and release . this all happens so fast that we might only notice a slight twitching at first . eventually though , not enough chemical energy is available in the muscle and it can no longer keep up . we then begin to shake and eventually must rest for a while . so work is being done , it is just not being done on the weight .
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in practice , whenever work is done to move energy from one form to another , there is always some loss to other forms of energy such as heat and sound . for example , a traditional light bulb is only about 3 % efficient at converting electrical energy to visible light , while a human being is about 25 % efficient at converting chemical energy from food into work . how do we measure energy and work ?
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why are we dividing the distance travelled by 4 and what does `` our body is 25 % efficient at transferring chemical energy from food into work ' ?
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what does energy and work mean ? energy is a word which tends to be used a lot in everyday life . though it is often used quite loosely , it does have a very specific physical meaning . energy is a measurement of the ability of something to do work . it is not a material substance . energy can be stored and measured in many forms . although we often hear people talking about energy consumption , energy is never really destroyed . it is just transferred from one form to another , doing work in the process . some forms of energy are less useful to us than others—for example , low level heat energy . it is better to talk about the consumption or extraction of energy resources , for example coal , oil , or wind , than consumption of energy itself . a speeding bullet has a measurable amount of energy associated with it ; this is known as kinetic energy . the bullet gained this energy because work was done on it by a charge of gunpowder which lost some chemical potential energy in the process . a hot cup of coffee has a measurable amount of thermal energy which it acquired via work done by a microwave oven , which in turn took electrical energy from the electrical grid . in practice , whenever work is done to move energy from one form to another , there is always some loss to other forms of energy such as heat and sound . for example , a traditional light bulb is only about 3 % efficient at converting electrical energy to visible light , while a human being is about 25 % efficient at converting chemical energy from food into work . how do we measure energy and work ? the standard unit used to measure energy and work done in physics is the joule , which has the symbol j . in mechanics , 1 joule is the energy transferred when a force of 1 newton is applied to an object and moves it through a distance of 1 meter . another unit of energy you may have come across is the calorie . the amount of energy in an item of food is often written in calories on the back of the packet . a typical 60 gram chocolate bar for example contains about 280 calories of energy . one calorie is the amount of energy required to raise 1 kg of water by 1 $ ^ { \circ } $ celsius . this is equal to 4184 joules , so one chocolate bar has 1.17 million joules or 1.17 mj of stored energy . that 's a lot of joules ! how long do i have to push a heavy box around to burn off one chocolate bar ? suppose we 're feeling guilty about eating a chocolate bar ; we want to find how much exercise we need to do to offset those extra 280 calories . let 's consider a simple form of exercise : pushing a heavy box around a room , see figure 1 below . using a bathroom scale between ourselves and the box , we find that we can push with a force of 500 n. meanwhile , we use a stopwatch and measuring tape to measure our speed . this comes out to be 0.25 meters per second . so how much work do we need to do to the box to burn off the candy bar ? the definition of work , $ w $ , is below : $ \large w = f\cdot \delta x $ the work we need to do to burn the energy in the candy bar is $ e=280 \mathrm { cal } \cdot 4184 \mathrm { j/cal } =1.17 \mathrm { mj } $ . therefore , the distance , $ δx $ , we need to move the box through is : $ \begin { align } w & amp ; = f\cdot \delta x \ 1.17 \text { mj } & amp ; = ( 500 \text { n } ) \cdot \delta x \ \dfrac { 1.17\times 10^6 \text { j } } { 500 \text { n } } & amp ; = \delta x \ 2,340 \text { m } & amp ; = \delta x \end { align } $ remember , however , that our bodies are about 25 % efficient at transferring stored energy from food into work . the actual energy we will offset is then four times as high as the work done to the box . so , we only need to push the box through a distance of 585 m , which is still over five football fields long . given the known speed of 0.25 m/s that will take us : $ \frac { 585 \mathrm { m } } { 0.25 \mathrm { m/s } } =2340 \mathrm { s } $ exercise : suppose that the force that we apply to the box , see figure 1 above , is initially reduced but increases to a constant value as we warm up . for instance , in the graph below we see that as the box is displaced further—i.e. , $ x $ gets larger—the force , $ f $ , increases for the first 30 m , see figure 2 below . how could we find the work done during the period where the force is changing ? if the force is not constant , one way to determine the work done is to divide the problem up into small sections over which the change is negligible and sum up the work done in each section . just as we have learned when looking at velocity time graphs , this can be done by calculating the area under the curve using geometry . the work done by a force is equal to the area under a force vs. position graph . in the case of figure 2 , it would be : $ ( 200\text { n } \cdot 30 \text { m } ) + \frac { 1 } { 2 } \left ( ( 500\text { n } -200\text { n } ) \cdot 30 \text { m } \right ) =10500 \mathrm { j } $ for the initial $ 30 \text { m } $ of displacement . similarly , the work done for the final 40 m of displacement would be : $ 500\text { n } \cdot 40 \text { m } =20,000 \mathrm { j } $ what if we are n't pushing straight on ? there is one thing we need to watch out for when doing these problems . the previous equation , $ w = f \cdot \delta x $ , does n't take into account situations where the force we are applying is not in the same direction as the motion . for instance , imagine we use a rope to pull on the box . in that case there will be an angle between the rope and the ground . to untangle this situation , we begin by drawing a triangle to separate out the horizontal and vertical components of the applied force . the key point here is that it is only the component of the force , $ f_ { || } $ , that lies parallel to the displacement that does work on an object . in the case of the box shown above , only the horizontal component of the applied force , $ f \text { cos } ( \theta ) $ , is doing work on the box since the box is being displaced horizontally . this means that a more general equation for the work done on the box by a force at an angle θ could be written as : $ w=f_ { || } \cdot\delta x $ $ w= ( f \cos { \theta } ) \cdot\delta x $ which is more often written as , $ \large w=f \delta x\cos { \theta } $ exercise : suppose we use a rope to pull the box , and the angle between the rope and the ground is 30º . this time we pull along the rope with a force of 500 n. how much of a chocolate bar can we eat this time if we pull the box through the same 585 m ? what about lifting weights instead ? in the previous example , we were doing work on a box which we were pushing around a floor . in doing so , we were working against a frictional force . another common form of exercise is lifting weights . in this case we are working against the force of gravity rather than friction . using newton 's laws we can find the force , $ f $ , required to lift a weight with mass $ m $ straight up , placing it on a rack which is at a height $ h $ above us : $ f=mg $ the change in position—previously $ \delta x $ —is simply the height , so the work , $ w $ , that we have done in lifting the weight is then $ w=m g h $ the exercise we have done in lifting the weight has resulted in energy being stored in the form of gravitational potential energy . it is called potential energy because it has the potential to be released at any moment with a crash as the weight falls back to the ground . we did positive work on the weight since we exerted our force in the same direction as the displacement of the weight , i.e. , upward . the work done by gravity on the weight while it was lifted was negative since the force of gravity is directed in the opposite direction to the displacement . also , since the weight is stationary after the lift , we know that the work that we have done is exactly canceled out by the work done by gravity . the work done by us is $ mgh $ , and the work done by gravity is $ -mgh $ . we will talk more about this when we look into kinetic energy . ok , let 's put some numbers in and find how much of that chocolate bar we would offset by lifting a weight of 50 kg up to a height of 0.5 m. the work done to the weight is $ w= ( 50\mathrm { kg } ) ( 9.81\mathrm { m/s^2 } ) ( 0.5\mathrm { m } ) = 245.25 \mathrm { j } $ ok , so how many 280 calorie—i.e. , $ 1.17 \times 10^6 $ joule—chocolate bars is this ? well , 245.25 j is about $ \dfrac { 1 } { 4770 } $ of a chocolate bar . but remember , our bodies are only about 25 % efficient , so the work done by the person is actually four times larger , about 981.8 j , which is $ \dfrac { 1 } { 1190 } $ chocolate bars . so , if we can lift this weight once every 2 seconds , it will take us about 2380 seconds or 40 minutes of hard work to burn off this chocolate bar ! what about simply holding a weight stationary ? one frequent source of confusion people have with the concept of work comes about when thinking about holding a heavy weight stationary above our heads , against the force of gravity . we are not moving the weight through any distance , so no work is being done to the weight . we could also achieve this by placing the weight on a table ; it is clear that the table is not doing any work to keep the weight in position . yet , we know from our experience that we get tired when doing the same job . so what is going on here ? it turns out that what is actually happening here is that our bodies are doing work on our muscles to maintain the necessary tension to hold the weight up . the body does this by sending a cascade of nerve impulses to each muscle . each impulse causes the muscle to momentarily contract and release . this all happens so fast that we might only notice a slight twitching at first . eventually though , not enough chemical energy is available in the muscle and it can no longer keep up . we then begin to shake and eventually must rest for a while . so work is being done , it is just not being done on the weight .
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in this case we are working against the force of gravity rather than friction . using newton 's laws we can find the force , $ f $ , required to lift a weight with mass $ m $ straight up , placing it on a rack which is at a height $ h $ above us : $ f=mg $ the change in position—previously $ \delta x $ —is simply the height , so the work , $ w $ , that we have done in lifting the weight is then $ w=m g h $ the exercise we have done in lifting the weight has resulted in energy being stored in the form of gravitational potential energy . it is called potential energy because it has the potential to be released at any moment with a crash as the weight falls back to the ground .
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what will be the maximum height attained by the ball ?
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what does energy and work mean ? energy is a word which tends to be used a lot in everyday life . though it is often used quite loosely , it does have a very specific physical meaning . energy is a measurement of the ability of something to do work . it is not a material substance . energy can be stored and measured in many forms . although we often hear people talking about energy consumption , energy is never really destroyed . it is just transferred from one form to another , doing work in the process . some forms of energy are less useful to us than others—for example , low level heat energy . it is better to talk about the consumption or extraction of energy resources , for example coal , oil , or wind , than consumption of energy itself . a speeding bullet has a measurable amount of energy associated with it ; this is known as kinetic energy . the bullet gained this energy because work was done on it by a charge of gunpowder which lost some chemical potential energy in the process . a hot cup of coffee has a measurable amount of thermal energy which it acquired via work done by a microwave oven , which in turn took electrical energy from the electrical grid . in practice , whenever work is done to move energy from one form to another , there is always some loss to other forms of energy such as heat and sound . for example , a traditional light bulb is only about 3 % efficient at converting electrical energy to visible light , while a human being is about 25 % efficient at converting chemical energy from food into work . how do we measure energy and work ? the standard unit used to measure energy and work done in physics is the joule , which has the symbol j . in mechanics , 1 joule is the energy transferred when a force of 1 newton is applied to an object and moves it through a distance of 1 meter . another unit of energy you may have come across is the calorie . the amount of energy in an item of food is often written in calories on the back of the packet . a typical 60 gram chocolate bar for example contains about 280 calories of energy . one calorie is the amount of energy required to raise 1 kg of water by 1 $ ^ { \circ } $ celsius . this is equal to 4184 joules , so one chocolate bar has 1.17 million joules or 1.17 mj of stored energy . that 's a lot of joules ! how long do i have to push a heavy box around to burn off one chocolate bar ? suppose we 're feeling guilty about eating a chocolate bar ; we want to find how much exercise we need to do to offset those extra 280 calories . let 's consider a simple form of exercise : pushing a heavy box around a room , see figure 1 below . using a bathroom scale between ourselves and the box , we find that we can push with a force of 500 n. meanwhile , we use a stopwatch and measuring tape to measure our speed . this comes out to be 0.25 meters per second . so how much work do we need to do to the box to burn off the candy bar ? the definition of work , $ w $ , is below : $ \large w = f\cdot \delta x $ the work we need to do to burn the energy in the candy bar is $ e=280 \mathrm { cal } \cdot 4184 \mathrm { j/cal } =1.17 \mathrm { mj } $ . therefore , the distance , $ δx $ , we need to move the box through is : $ \begin { align } w & amp ; = f\cdot \delta x \ 1.17 \text { mj } & amp ; = ( 500 \text { n } ) \cdot \delta x \ \dfrac { 1.17\times 10^6 \text { j } } { 500 \text { n } } & amp ; = \delta x \ 2,340 \text { m } & amp ; = \delta x \end { align } $ remember , however , that our bodies are about 25 % efficient at transferring stored energy from food into work . the actual energy we will offset is then four times as high as the work done to the box . so , we only need to push the box through a distance of 585 m , which is still over five football fields long . given the known speed of 0.25 m/s that will take us : $ \frac { 585 \mathrm { m } } { 0.25 \mathrm { m/s } } =2340 \mathrm { s } $ exercise : suppose that the force that we apply to the box , see figure 1 above , is initially reduced but increases to a constant value as we warm up . for instance , in the graph below we see that as the box is displaced further—i.e. , $ x $ gets larger—the force , $ f $ , increases for the first 30 m , see figure 2 below . how could we find the work done during the period where the force is changing ? if the force is not constant , one way to determine the work done is to divide the problem up into small sections over which the change is negligible and sum up the work done in each section . just as we have learned when looking at velocity time graphs , this can be done by calculating the area under the curve using geometry . the work done by a force is equal to the area under a force vs. position graph . in the case of figure 2 , it would be : $ ( 200\text { n } \cdot 30 \text { m } ) + \frac { 1 } { 2 } \left ( ( 500\text { n } -200\text { n } ) \cdot 30 \text { m } \right ) =10500 \mathrm { j } $ for the initial $ 30 \text { m } $ of displacement . similarly , the work done for the final 40 m of displacement would be : $ 500\text { n } \cdot 40 \text { m } =20,000 \mathrm { j } $ what if we are n't pushing straight on ? there is one thing we need to watch out for when doing these problems . the previous equation , $ w = f \cdot \delta x $ , does n't take into account situations where the force we are applying is not in the same direction as the motion . for instance , imagine we use a rope to pull on the box . in that case there will be an angle between the rope and the ground . to untangle this situation , we begin by drawing a triangle to separate out the horizontal and vertical components of the applied force . the key point here is that it is only the component of the force , $ f_ { || } $ , that lies parallel to the displacement that does work on an object . in the case of the box shown above , only the horizontal component of the applied force , $ f \text { cos } ( \theta ) $ , is doing work on the box since the box is being displaced horizontally . this means that a more general equation for the work done on the box by a force at an angle θ could be written as : $ w=f_ { || } \cdot\delta x $ $ w= ( f \cos { \theta } ) \cdot\delta x $ which is more often written as , $ \large w=f \delta x\cos { \theta } $ exercise : suppose we use a rope to pull the box , and the angle between the rope and the ground is 30º . this time we pull along the rope with a force of 500 n. how much of a chocolate bar can we eat this time if we pull the box through the same 585 m ? what about lifting weights instead ? in the previous example , we were doing work on a box which we were pushing around a floor . in doing so , we were working against a frictional force . another common form of exercise is lifting weights . in this case we are working against the force of gravity rather than friction . using newton 's laws we can find the force , $ f $ , required to lift a weight with mass $ m $ straight up , placing it on a rack which is at a height $ h $ above us : $ f=mg $ the change in position—previously $ \delta x $ —is simply the height , so the work , $ w $ , that we have done in lifting the weight is then $ w=m g h $ the exercise we have done in lifting the weight has resulted in energy being stored in the form of gravitational potential energy . it is called potential energy because it has the potential to be released at any moment with a crash as the weight falls back to the ground . we did positive work on the weight since we exerted our force in the same direction as the displacement of the weight , i.e. , upward . the work done by gravity on the weight while it was lifted was negative since the force of gravity is directed in the opposite direction to the displacement . also , since the weight is stationary after the lift , we know that the work that we have done is exactly canceled out by the work done by gravity . the work done by us is $ mgh $ , and the work done by gravity is $ -mgh $ . we will talk more about this when we look into kinetic energy . ok , let 's put some numbers in and find how much of that chocolate bar we would offset by lifting a weight of 50 kg up to a height of 0.5 m. the work done to the weight is $ w= ( 50\mathrm { kg } ) ( 9.81\mathrm { m/s^2 } ) ( 0.5\mathrm { m } ) = 245.25 \mathrm { j } $ ok , so how many 280 calorie—i.e. , $ 1.17 \times 10^6 $ joule—chocolate bars is this ? well , 245.25 j is about $ \dfrac { 1 } { 4770 } $ of a chocolate bar . but remember , our bodies are only about 25 % efficient , so the work done by the person is actually four times larger , about 981.8 j , which is $ \dfrac { 1 } { 1190 } $ chocolate bars . so , if we can lift this weight once every 2 seconds , it will take us about 2380 seconds or 40 minutes of hard work to burn off this chocolate bar ! what about simply holding a weight stationary ? one frequent source of confusion people have with the concept of work comes about when thinking about holding a heavy weight stationary above our heads , against the force of gravity . we are not moving the weight through any distance , so no work is being done to the weight . we could also achieve this by placing the weight on a table ; it is clear that the table is not doing any work to keep the weight in position . yet , we know from our experience that we get tired when doing the same job . so what is going on here ? it turns out that what is actually happening here is that our bodies are doing work on our muscles to maintain the necessary tension to hold the weight up . the body does this by sending a cascade of nerve impulses to each muscle . each impulse causes the muscle to momentarily contract and release . this all happens so fast that we might only notice a slight twitching at first . eventually though , not enough chemical energy is available in the muscle and it can no longer keep up . we then begin to shake and eventually must rest for a while . so work is being done , it is just not being done on the weight .
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so how much work do we need to do to the box to burn off the candy bar ? the definition of work , $ w $ , is below : $ \large w = f\cdot \delta x $ the work we need to do to burn the energy in the candy bar is $ e=280 \mathrm { cal } \cdot 4184 \mathrm { j/cal } =1.17 \mathrm { mj } $ . therefore , the distance , $ δx $ , we need to move the box through is : $ \begin { align } w & amp ; = f\cdot \delta x \ 1.17 \text { mj } & amp ; = ( 500 \text { n } ) \cdot \delta x \ \dfrac { 1.17\times 10^6 \text { j } } { 500 \text { n } } & amp ; = \delta x \ 2,340 \text { m } & amp ; = \delta x \end { align } $ remember , however , that our bodies are about 25 % efficient at transferring stored energy from food into work .
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how is j/cal a distance ?
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what does energy and work mean ? energy is a word which tends to be used a lot in everyday life . though it is often used quite loosely , it does have a very specific physical meaning . energy is a measurement of the ability of something to do work . it is not a material substance . energy can be stored and measured in many forms . although we often hear people talking about energy consumption , energy is never really destroyed . it is just transferred from one form to another , doing work in the process . some forms of energy are less useful to us than others—for example , low level heat energy . it is better to talk about the consumption or extraction of energy resources , for example coal , oil , or wind , than consumption of energy itself . a speeding bullet has a measurable amount of energy associated with it ; this is known as kinetic energy . the bullet gained this energy because work was done on it by a charge of gunpowder which lost some chemical potential energy in the process . a hot cup of coffee has a measurable amount of thermal energy which it acquired via work done by a microwave oven , which in turn took electrical energy from the electrical grid . in practice , whenever work is done to move energy from one form to another , there is always some loss to other forms of energy such as heat and sound . for example , a traditional light bulb is only about 3 % efficient at converting electrical energy to visible light , while a human being is about 25 % efficient at converting chemical energy from food into work . how do we measure energy and work ? the standard unit used to measure energy and work done in physics is the joule , which has the symbol j . in mechanics , 1 joule is the energy transferred when a force of 1 newton is applied to an object and moves it through a distance of 1 meter . another unit of energy you may have come across is the calorie . the amount of energy in an item of food is often written in calories on the back of the packet . a typical 60 gram chocolate bar for example contains about 280 calories of energy . one calorie is the amount of energy required to raise 1 kg of water by 1 $ ^ { \circ } $ celsius . this is equal to 4184 joules , so one chocolate bar has 1.17 million joules or 1.17 mj of stored energy . that 's a lot of joules ! how long do i have to push a heavy box around to burn off one chocolate bar ? suppose we 're feeling guilty about eating a chocolate bar ; we want to find how much exercise we need to do to offset those extra 280 calories . let 's consider a simple form of exercise : pushing a heavy box around a room , see figure 1 below . using a bathroom scale between ourselves and the box , we find that we can push with a force of 500 n. meanwhile , we use a stopwatch and measuring tape to measure our speed . this comes out to be 0.25 meters per second . so how much work do we need to do to the box to burn off the candy bar ? the definition of work , $ w $ , is below : $ \large w = f\cdot \delta x $ the work we need to do to burn the energy in the candy bar is $ e=280 \mathrm { cal } \cdot 4184 \mathrm { j/cal } =1.17 \mathrm { mj } $ . therefore , the distance , $ δx $ , we need to move the box through is : $ \begin { align } w & amp ; = f\cdot \delta x \ 1.17 \text { mj } & amp ; = ( 500 \text { n } ) \cdot \delta x \ \dfrac { 1.17\times 10^6 \text { j } } { 500 \text { n } } & amp ; = \delta x \ 2,340 \text { m } & amp ; = \delta x \end { align } $ remember , however , that our bodies are about 25 % efficient at transferring stored energy from food into work . the actual energy we will offset is then four times as high as the work done to the box . so , we only need to push the box through a distance of 585 m , which is still over five football fields long . given the known speed of 0.25 m/s that will take us : $ \frac { 585 \mathrm { m } } { 0.25 \mathrm { m/s } } =2340 \mathrm { s } $ exercise : suppose that the force that we apply to the box , see figure 1 above , is initially reduced but increases to a constant value as we warm up . for instance , in the graph below we see that as the box is displaced further—i.e. , $ x $ gets larger—the force , $ f $ , increases for the first 30 m , see figure 2 below . how could we find the work done during the period where the force is changing ? if the force is not constant , one way to determine the work done is to divide the problem up into small sections over which the change is negligible and sum up the work done in each section . just as we have learned when looking at velocity time graphs , this can be done by calculating the area under the curve using geometry . the work done by a force is equal to the area under a force vs. position graph . in the case of figure 2 , it would be : $ ( 200\text { n } \cdot 30 \text { m } ) + \frac { 1 } { 2 } \left ( ( 500\text { n } -200\text { n } ) \cdot 30 \text { m } \right ) =10500 \mathrm { j } $ for the initial $ 30 \text { m } $ of displacement . similarly , the work done for the final 40 m of displacement would be : $ 500\text { n } \cdot 40 \text { m } =20,000 \mathrm { j } $ what if we are n't pushing straight on ? there is one thing we need to watch out for when doing these problems . the previous equation , $ w = f \cdot \delta x $ , does n't take into account situations where the force we are applying is not in the same direction as the motion . for instance , imagine we use a rope to pull on the box . in that case there will be an angle between the rope and the ground . to untangle this situation , we begin by drawing a triangle to separate out the horizontal and vertical components of the applied force . the key point here is that it is only the component of the force , $ f_ { || } $ , that lies parallel to the displacement that does work on an object . in the case of the box shown above , only the horizontal component of the applied force , $ f \text { cos } ( \theta ) $ , is doing work on the box since the box is being displaced horizontally . this means that a more general equation for the work done on the box by a force at an angle θ could be written as : $ w=f_ { || } \cdot\delta x $ $ w= ( f \cos { \theta } ) \cdot\delta x $ which is more often written as , $ \large w=f \delta x\cos { \theta } $ exercise : suppose we use a rope to pull the box , and the angle between the rope and the ground is 30º . this time we pull along the rope with a force of 500 n. how much of a chocolate bar can we eat this time if we pull the box through the same 585 m ? what about lifting weights instead ? in the previous example , we were doing work on a box which we were pushing around a floor . in doing so , we were working against a frictional force . another common form of exercise is lifting weights . in this case we are working against the force of gravity rather than friction . using newton 's laws we can find the force , $ f $ , required to lift a weight with mass $ m $ straight up , placing it on a rack which is at a height $ h $ above us : $ f=mg $ the change in position—previously $ \delta x $ —is simply the height , so the work , $ w $ , that we have done in lifting the weight is then $ w=m g h $ the exercise we have done in lifting the weight has resulted in energy being stored in the form of gravitational potential energy . it is called potential energy because it has the potential to be released at any moment with a crash as the weight falls back to the ground . we did positive work on the weight since we exerted our force in the same direction as the displacement of the weight , i.e. , upward . the work done by gravity on the weight while it was lifted was negative since the force of gravity is directed in the opposite direction to the displacement . also , since the weight is stationary after the lift , we know that the work that we have done is exactly canceled out by the work done by gravity . the work done by us is $ mgh $ , and the work done by gravity is $ -mgh $ . we will talk more about this when we look into kinetic energy . ok , let 's put some numbers in and find how much of that chocolate bar we would offset by lifting a weight of 50 kg up to a height of 0.5 m. the work done to the weight is $ w= ( 50\mathrm { kg } ) ( 9.81\mathrm { m/s^2 } ) ( 0.5\mathrm { m } ) = 245.25 \mathrm { j } $ ok , so how many 280 calorie—i.e. , $ 1.17 \times 10^6 $ joule—chocolate bars is this ? well , 245.25 j is about $ \dfrac { 1 } { 4770 } $ of a chocolate bar . but remember , our bodies are only about 25 % efficient , so the work done by the person is actually four times larger , about 981.8 j , which is $ \dfrac { 1 } { 1190 } $ chocolate bars . so , if we can lift this weight once every 2 seconds , it will take us about 2380 seconds or 40 minutes of hard work to burn off this chocolate bar ! what about simply holding a weight stationary ? one frequent source of confusion people have with the concept of work comes about when thinking about holding a heavy weight stationary above our heads , against the force of gravity . we are not moving the weight through any distance , so no work is being done to the weight . we could also achieve this by placing the weight on a table ; it is clear that the table is not doing any work to keep the weight in position . yet , we know from our experience that we get tired when doing the same job . so what is going on here ? it turns out that what is actually happening here is that our bodies are doing work on our muscles to maintain the necessary tension to hold the weight up . the body does this by sending a cascade of nerve impulses to each muscle . each impulse causes the muscle to momentarily contract and release . this all happens so fast that we might only notice a slight twitching at first . eventually though , not enough chemical energy is available in the muscle and it can no longer keep up . we then begin to shake and eventually must rest for a while . so work is being done , it is just not being done on the weight .
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another common form of exercise is lifting weights . in this case we are working against the force of gravity rather than friction . using newton 's laws we can find the force , $ f $ , required to lift a weight with mass $ m $ straight up , placing it on a rack which is at a height $ h $ above us : $ f=mg $ the change in position—previously $ \delta x $ —is simply the height , so the work , $ w $ , that we have done in lifting the weight is then $ w=m g h $ the exercise we have done in lifting the weight has resulted in energy being stored in the form of gravitational potential energy .
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what if there is a force of friction when we push an object ?
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what does energy and work mean ? energy is a word which tends to be used a lot in everyday life . though it is often used quite loosely , it does have a very specific physical meaning . energy is a measurement of the ability of something to do work . it is not a material substance . energy can be stored and measured in many forms . although we often hear people talking about energy consumption , energy is never really destroyed . it is just transferred from one form to another , doing work in the process . some forms of energy are less useful to us than others—for example , low level heat energy . it is better to talk about the consumption or extraction of energy resources , for example coal , oil , or wind , than consumption of energy itself . a speeding bullet has a measurable amount of energy associated with it ; this is known as kinetic energy . the bullet gained this energy because work was done on it by a charge of gunpowder which lost some chemical potential energy in the process . a hot cup of coffee has a measurable amount of thermal energy which it acquired via work done by a microwave oven , which in turn took electrical energy from the electrical grid . in practice , whenever work is done to move energy from one form to another , there is always some loss to other forms of energy such as heat and sound . for example , a traditional light bulb is only about 3 % efficient at converting electrical energy to visible light , while a human being is about 25 % efficient at converting chemical energy from food into work . how do we measure energy and work ? the standard unit used to measure energy and work done in physics is the joule , which has the symbol j . in mechanics , 1 joule is the energy transferred when a force of 1 newton is applied to an object and moves it through a distance of 1 meter . another unit of energy you may have come across is the calorie . the amount of energy in an item of food is often written in calories on the back of the packet . a typical 60 gram chocolate bar for example contains about 280 calories of energy . one calorie is the amount of energy required to raise 1 kg of water by 1 $ ^ { \circ } $ celsius . this is equal to 4184 joules , so one chocolate bar has 1.17 million joules or 1.17 mj of stored energy . that 's a lot of joules ! how long do i have to push a heavy box around to burn off one chocolate bar ? suppose we 're feeling guilty about eating a chocolate bar ; we want to find how much exercise we need to do to offset those extra 280 calories . let 's consider a simple form of exercise : pushing a heavy box around a room , see figure 1 below . using a bathroom scale between ourselves and the box , we find that we can push with a force of 500 n. meanwhile , we use a stopwatch and measuring tape to measure our speed . this comes out to be 0.25 meters per second . so how much work do we need to do to the box to burn off the candy bar ? the definition of work , $ w $ , is below : $ \large w = f\cdot \delta x $ the work we need to do to burn the energy in the candy bar is $ e=280 \mathrm { cal } \cdot 4184 \mathrm { j/cal } =1.17 \mathrm { mj } $ . therefore , the distance , $ δx $ , we need to move the box through is : $ \begin { align } w & amp ; = f\cdot \delta x \ 1.17 \text { mj } & amp ; = ( 500 \text { n } ) \cdot \delta x \ \dfrac { 1.17\times 10^6 \text { j } } { 500 \text { n } } & amp ; = \delta x \ 2,340 \text { m } & amp ; = \delta x \end { align } $ remember , however , that our bodies are about 25 % efficient at transferring stored energy from food into work . the actual energy we will offset is then four times as high as the work done to the box . so , we only need to push the box through a distance of 585 m , which is still over five football fields long . given the known speed of 0.25 m/s that will take us : $ \frac { 585 \mathrm { m } } { 0.25 \mathrm { m/s } } =2340 \mathrm { s } $ exercise : suppose that the force that we apply to the box , see figure 1 above , is initially reduced but increases to a constant value as we warm up . for instance , in the graph below we see that as the box is displaced further—i.e. , $ x $ gets larger—the force , $ f $ , increases for the first 30 m , see figure 2 below . how could we find the work done during the period where the force is changing ? if the force is not constant , one way to determine the work done is to divide the problem up into small sections over which the change is negligible and sum up the work done in each section . just as we have learned when looking at velocity time graphs , this can be done by calculating the area under the curve using geometry . the work done by a force is equal to the area under a force vs. position graph . in the case of figure 2 , it would be : $ ( 200\text { n } \cdot 30 \text { m } ) + \frac { 1 } { 2 } \left ( ( 500\text { n } -200\text { n } ) \cdot 30 \text { m } \right ) =10500 \mathrm { j } $ for the initial $ 30 \text { m } $ of displacement . similarly , the work done for the final 40 m of displacement would be : $ 500\text { n } \cdot 40 \text { m } =20,000 \mathrm { j } $ what if we are n't pushing straight on ? there is one thing we need to watch out for when doing these problems . the previous equation , $ w = f \cdot \delta x $ , does n't take into account situations where the force we are applying is not in the same direction as the motion . for instance , imagine we use a rope to pull on the box . in that case there will be an angle between the rope and the ground . to untangle this situation , we begin by drawing a triangle to separate out the horizontal and vertical components of the applied force . the key point here is that it is only the component of the force , $ f_ { || } $ , that lies parallel to the displacement that does work on an object . in the case of the box shown above , only the horizontal component of the applied force , $ f \text { cos } ( \theta ) $ , is doing work on the box since the box is being displaced horizontally . this means that a more general equation for the work done on the box by a force at an angle θ could be written as : $ w=f_ { || } \cdot\delta x $ $ w= ( f \cos { \theta } ) \cdot\delta x $ which is more often written as , $ \large w=f \delta x\cos { \theta } $ exercise : suppose we use a rope to pull the box , and the angle between the rope and the ground is 30º . this time we pull along the rope with a force of 500 n. how much of a chocolate bar can we eat this time if we pull the box through the same 585 m ? what about lifting weights instead ? in the previous example , we were doing work on a box which we were pushing around a floor . in doing so , we were working against a frictional force . another common form of exercise is lifting weights . in this case we are working against the force of gravity rather than friction . using newton 's laws we can find the force , $ f $ , required to lift a weight with mass $ m $ straight up , placing it on a rack which is at a height $ h $ above us : $ f=mg $ the change in position—previously $ \delta x $ —is simply the height , so the work , $ w $ , that we have done in lifting the weight is then $ w=m g h $ the exercise we have done in lifting the weight has resulted in energy being stored in the form of gravitational potential energy . it is called potential energy because it has the potential to be released at any moment with a crash as the weight falls back to the ground . we did positive work on the weight since we exerted our force in the same direction as the displacement of the weight , i.e. , upward . the work done by gravity on the weight while it was lifted was negative since the force of gravity is directed in the opposite direction to the displacement . also , since the weight is stationary after the lift , we know that the work that we have done is exactly canceled out by the work done by gravity . the work done by us is $ mgh $ , and the work done by gravity is $ -mgh $ . we will talk more about this when we look into kinetic energy . ok , let 's put some numbers in and find how much of that chocolate bar we would offset by lifting a weight of 50 kg up to a height of 0.5 m. the work done to the weight is $ w= ( 50\mathrm { kg } ) ( 9.81\mathrm { m/s^2 } ) ( 0.5\mathrm { m } ) = 245.25 \mathrm { j } $ ok , so how many 280 calorie—i.e. , $ 1.17 \times 10^6 $ joule—chocolate bars is this ? well , 245.25 j is about $ \dfrac { 1 } { 4770 } $ of a chocolate bar . but remember , our bodies are only about 25 % efficient , so the work done by the person is actually four times larger , about 981.8 j , which is $ \dfrac { 1 } { 1190 } $ chocolate bars . so , if we can lift this weight once every 2 seconds , it will take us about 2380 seconds or 40 minutes of hard work to burn off this chocolate bar ! what about simply holding a weight stationary ? one frequent source of confusion people have with the concept of work comes about when thinking about holding a heavy weight stationary above our heads , against the force of gravity . we are not moving the weight through any distance , so no work is being done to the weight . we could also achieve this by placing the weight on a table ; it is clear that the table is not doing any work to keep the weight in position . yet , we know from our experience that we get tired when doing the same job . so what is going on here ? it turns out that what is actually happening here is that our bodies are doing work on our muscles to maintain the necessary tension to hold the weight up . the body does this by sending a cascade of nerve impulses to each muscle . each impulse causes the muscle to momentarily contract and release . this all happens so fast that we might only notice a slight twitching at first . eventually though , not enough chemical energy is available in the muscle and it can no longer keep up . we then begin to shake and eventually must rest for a while . so work is being done , it is just not being done on the weight .
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it is not a material substance . energy can be stored and measured in many forms . although we often hear people talking about energy consumption , energy is never really destroyed . it is just transferred from one form to another , doing work in the process .
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what type of energy would i be producing ?
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what does energy and work mean ? energy is a word which tends to be used a lot in everyday life . though it is often used quite loosely , it does have a very specific physical meaning . energy is a measurement of the ability of something to do work . it is not a material substance . energy can be stored and measured in many forms . although we often hear people talking about energy consumption , energy is never really destroyed . it is just transferred from one form to another , doing work in the process . some forms of energy are less useful to us than others—for example , low level heat energy . it is better to talk about the consumption or extraction of energy resources , for example coal , oil , or wind , than consumption of energy itself . a speeding bullet has a measurable amount of energy associated with it ; this is known as kinetic energy . the bullet gained this energy because work was done on it by a charge of gunpowder which lost some chemical potential energy in the process . a hot cup of coffee has a measurable amount of thermal energy which it acquired via work done by a microwave oven , which in turn took electrical energy from the electrical grid . in practice , whenever work is done to move energy from one form to another , there is always some loss to other forms of energy such as heat and sound . for example , a traditional light bulb is only about 3 % efficient at converting electrical energy to visible light , while a human being is about 25 % efficient at converting chemical energy from food into work . how do we measure energy and work ? the standard unit used to measure energy and work done in physics is the joule , which has the symbol j . in mechanics , 1 joule is the energy transferred when a force of 1 newton is applied to an object and moves it through a distance of 1 meter . another unit of energy you may have come across is the calorie . the amount of energy in an item of food is often written in calories on the back of the packet . a typical 60 gram chocolate bar for example contains about 280 calories of energy . one calorie is the amount of energy required to raise 1 kg of water by 1 $ ^ { \circ } $ celsius . this is equal to 4184 joules , so one chocolate bar has 1.17 million joules or 1.17 mj of stored energy . that 's a lot of joules ! how long do i have to push a heavy box around to burn off one chocolate bar ? suppose we 're feeling guilty about eating a chocolate bar ; we want to find how much exercise we need to do to offset those extra 280 calories . let 's consider a simple form of exercise : pushing a heavy box around a room , see figure 1 below . using a bathroom scale between ourselves and the box , we find that we can push with a force of 500 n. meanwhile , we use a stopwatch and measuring tape to measure our speed . this comes out to be 0.25 meters per second . so how much work do we need to do to the box to burn off the candy bar ? the definition of work , $ w $ , is below : $ \large w = f\cdot \delta x $ the work we need to do to burn the energy in the candy bar is $ e=280 \mathrm { cal } \cdot 4184 \mathrm { j/cal } =1.17 \mathrm { mj } $ . therefore , the distance , $ δx $ , we need to move the box through is : $ \begin { align } w & amp ; = f\cdot \delta x \ 1.17 \text { mj } & amp ; = ( 500 \text { n } ) \cdot \delta x \ \dfrac { 1.17\times 10^6 \text { j } } { 500 \text { n } } & amp ; = \delta x \ 2,340 \text { m } & amp ; = \delta x \end { align } $ remember , however , that our bodies are about 25 % efficient at transferring stored energy from food into work . the actual energy we will offset is then four times as high as the work done to the box . so , we only need to push the box through a distance of 585 m , which is still over five football fields long . given the known speed of 0.25 m/s that will take us : $ \frac { 585 \mathrm { m } } { 0.25 \mathrm { m/s } } =2340 \mathrm { s } $ exercise : suppose that the force that we apply to the box , see figure 1 above , is initially reduced but increases to a constant value as we warm up . for instance , in the graph below we see that as the box is displaced further—i.e. , $ x $ gets larger—the force , $ f $ , increases for the first 30 m , see figure 2 below . how could we find the work done during the period where the force is changing ? if the force is not constant , one way to determine the work done is to divide the problem up into small sections over which the change is negligible and sum up the work done in each section . just as we have learned when looking at velocity time graphs , this can be done by calculating the area under the curve using geometry . the work done by a force is equal to the area under a force vs. position graph . in the case of figure 2 , it would be : $ ( 200\text { n } \cdot 30 \text { m } ) + \frac { 1 } { 2 } \left ( ( 500\text { n } -200\text { n } ) \cdot 30 \text { m } \right ) =10500 \mathrm { j } $ for the initial $ 30 \text { m } $ of displacement . similarly , the work done for the final 40 m of displacement would be : $ 500\text { n } \cdot 40 \text { m } =20,000 \mathrm { j } $ what if we are n't pushing straight on ? there is one thing we need to watch out for when doing these problems . the previous equation , $ w = f \cdot \delta x $ , does n't take into account situations where the force we are applying is not in the same direction as the motion . for instance , imagine we use a rope to pull on the box . in that case there will be an angle between the rope and the ground . to untangle this situation , we begin by drawing a triangle to separate out the horizontal and vertical components of the applied force . the key point here is that it is only the component of the force , $ f_ { || } $ , that lies parallel to the displacement that does work on an object . in the case of the box shown above , only the horizontal component of the applied force , $ f \text { cos } ( \theta ) $ , is doing work on the box since the box is being displaced horizontally . this means that a more general equation for the work done on the box by a force at an angle θ could be written as : $ w=f_ { || } \cdot\delta x $ $ w= ( f \cos { \theta } ) \cdot\delta x $ which is more often written as , $ \large w=f \delta x\cos { \theta } $ exercise : suppose we use a rope to pull the box , and the angle between the rope and the ground is 30º . this time we pull along the rope with a force of 500 n. how much of a chocolate bar can we eat this time if we pull the box through the same 585 m ? what about lifting weights instead ? in the previous example , we were doing work on a box which we were pushing around a floor . in doing so , we were working against a frictional force . another common form of exercise is lifting weights . in this case we are working against the force of gravity rather than friction . using newton 's laws we can find the force , $ f $ , required to lift a weight with mass $ m $ straight up , placing it on a rack which is at a height $ h $ above us : $ f=mg $ the change in position—previously $ \delta x $ —is simply the height , so the work , $ w $ , that we have done in lifting the weight is then $ w=m g h $ the exercise we have done in lifting the weight has resulted in energy being stored in the form of gravitational potential energy . it is called potential energy because it has the potential to be released at any moment with a crash as the weight falls back to the ground . we did positive work on the weight since we exerted our force in the same direction as the displacement of the weight , i.e. , upward . the work done by gravity on the weight while it was lifted was negative since the force of gravity is directed in the opposite direction to the displacement . also , since the weight is stationary after the lift , we know that the work that we have done is exactly canceled out by the work done by gravity . the work done by us is $ mgh $ , and the work done by gravity is $ -mgh $ . we will talk more about this when we look into kinetic energy . ok , let 's put some numbers in and find how much of that chocolate bar we would offset by lifting a weight of 50 kg up to a height of 0.5 m. the work done to the weight is $ w= ( 50\mathrm { kg } ) ( 9.81\mathrm { m/s^2 } ) ( 0.5\mathrm { m } ) = 245.25 \mathrm { j } $ ok , so how many 280 calorie—i.e. , $ 1.17 \times 10^6 $ joule—chocolate bars is this ? well , 245.25 j is about $ \dfrac { 1 } { 4770 } $ of a chocolate bar . but remember , our bodies are only about 25 % efficient , so the work done by the person is actually four times larger , about 981.8 j , which is $ \dfrac { 1 } { 1190 } $ chocolate bars . so , if we can lift this weight once every 2 seconds , it will take us about 2380 seconds or 40 minutes of hard work to burn off this chocolate bar ! what about simply holding a weight stationary ? one frequent source of confusion people have with the concept of work comes about when thinking about holding a heavy weight stationary above our heads , against the force of gravity . we are not moving the weight through any distance , so no work is being done to the weight . we could also achieve this by placing the weight on a table ; it is clear that the table is not doing any work to keep the weight in position . yet , we know from our experience that we get tired when doing the same job . so what is going on here ? it turns out that what is actually happening here is that our bodies are doing work on our muscles to maintain the necessary tension to hold the weight up . the body does this by sending a cascade of nerve impulses to each muscle . each impulse causes the muscle to momentarily contract and release . this all happens so fast that we might only notice a slight twitching at first . eventually though , not enough chemical energy is available in the muscle and it can no longer keep up . we then begin to shake and eventually must rest for a while . so work is being done , it is just not being done on the weight .
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it is just transferred from one form to another , doing work in the process . some forms of energy are less useful to us than others—for example , low level heat energy . it is better to talk about the consumption or extraction of energy resources , for example coal , oil , or wind , than consumption of energy itself . a speeding bullet has a measurable amount of energy associated with it ; this is known as kinetic energy . the bullet gained this energy because work was done on it by a charge of gunpowder which lost some chemical potential energy in the process . a hot cup of coffee has a measurable amount of thermal energy which it acquired via work done by a microwave oven , which in turn took electrical energy from the electrical grid .
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is n't the energy we have in our bodies after eating the chocolate at type of chemical potential energy ?
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what does energy and work mean ? energy is a word which tends to be used a lot in everyday life . though it is often used quite loosely , it does have a very specific physical meaning . energy is a measurement of the ability of something to do work . it is not a material substance . energy can be stored and measured in many forms . although we often hear people talking about energy consumption , energy is never really destroyed . it is just transferred from one form to another , doing work in the process . some forms of energy are less useful to us than others—for example , low level heat energy . it is better to talk about the consumption or extraction of energy resources , for example coal , oil , or wind , than consumption of energy itself . a speeding bullet has a measurable amount of energy associated with it ; this is known as kinetic energy . the bullet gained this energy because work was done on it by a charge of gunpowder which lost some chemical potential energy in the process . a hot cup of coffee has a measurable amount of thermal energy which it acquired via work done by a microwave oven , which in turn took electrical energy from the electrical grid . in practice , whenever work is done to move energy from one form to another , there is always some loss to other forms of energy such as heat and sound . for example , a traditional light bulb is only about 3 % efficient at converting electrical energy to visible light , while a human being is about 25 % efficient at converting chemical energy from food into work . how do we measure energy and work ? the standard unit used to measure energy and work done in physics is the joule , which has the symbol j . in mechanics , 1 joule is the energy transferred when a force of 1 newton is applied to an object and moves it through a distance of 1 meter . another unit of energy you may have come across is the calorie . the amount of energy in an item of food is often written in calories on the back of the packet . a typical 60 gram chocolate bar for example contains about 280 calories of energy . one calorie is the amount of energy required to raise 1 kg of water by 1 $ ^ { \circ } $ celsius . this is equal to 4184 joules , so one chocolate bar has 1.17 million joules or 1.17 mj of stored energy . that 's a lot of joules ! how long do i have to push a heavy box around to burn off one chocolate bar ? suppose we 're feeling guilty about eating a chocolate bar ; we want to find how much exercise we need to do to offset those extra 280 calories . let 's consider a simple form of exercise : pushing a heavy box around a room , see figure 1 below . using a bathroom scale between ourselves and the box , we find that we can push with a force of 500 n. meanwhile , we use a stopwatch and measuring tape to measure our speed . this comes out to be 0.25 meters per second . so how much work do we need to do to the box to burn off the candy bar ? the definition of work , $ w $ , is below : $ \large w = f\cdot \delta x $ the work we need to do to burn the energy in the candy bar is $ e=280 \mathrm { cal } \cdot 4184 \mathrm { j/cal } =1.17 \mathrm { mj } $ . therefore , the distance , $ δx $ , we need to move the box through is : $ \begin { align } w & amp ; = f\cdot \delta x \ 1.17 \text { mj } & amp ; = ( 500 \text { n } ) \cdot \delta x \ \dfrac { 1.17\times 10^6 \text { j } } { 500 \text { n } } & amp ; = \delta x \ 2,340 \text { m } & amp ; = \delta x \end { align } $ remember , however , that our bodies are about 25 % efficient at transferring stored energy from food into work . the actual energy we will offset is then four times as high as the work done to the box . so , we only need to push the box through a distance of 585 m , which is still over five football fields long . given the known speed of 0.25 m/s that will take us : $ \frac { 585 \mathrm { m } } { 0.25 \mathrm { m/s } } =2340 \mathrm { s } $ exercise : suppose that the force that we apply to the box , see figure 1 above , is initially reduced but increases to a constant value as we warm up . for instance , in the graph below we see that as the box is displaced further—i.e. , $ x $ gets larger—the force , $ f $ , increases for the first 30 m , see figure 2 below . how could we find the work done during the period where the force is changing ? if the force is not constant , one way to determine the work done is to divide the problem up into small sections over which the change is negligible and sum up the work done in each section . just as we have learned when looking at velocity time graphs , this can be done by calculating the area under the curve using geometry . the work done by a force is equal to the area under a force vs. position graph . in the case of figure 2 , it would be : $ ( 200\text { n } \cdot 30 \text { m } ) + \frac { 1 } { 2 } \left ( ( 500\text { n } -200\text { n } ) \cdot 30 \text { m } \right ) =10500 \mathrm { j } $ for the initial $ 30 \text { m } $ of displacement . similarly , the work done for the final 40 m of displacement would be : $ 500\text { n } \cdot 40 \text { m } =20,000 \mathrm { j } $ what if we are n't pushing straight on ? there is one thing we need to watch out for when doing these problems . the previous equation , $ w = f \cdot \delta x $ , does n't take into account situations where the force we are applying is not in the same direction as the motion . for instance , imagine we use a rope to pull on the box . in that case there will be an angle between the rope and the ground . to untangle this situation , we begin by drawing a triangle to separate out the horizontal and vertical components of the applied force . the key point here is that it is only the component of the force , $ f_ { || } $ , that lies parallel to the displacement that does work on an object . in the case of the box shown above , only the horizontal component of the applied force , $ f \text { cos } ( \theta ) $ , is doing work on the box since the box is being displaced horizontally . this means that a more general equation for the work done on the box by a force at an angle θ could be written as : $ w=f_ { || } \cdot\delta x $ $ w= ( f \cos { \theta } ) \cdot\delta x $ which is more often written as , $ \large w=f \delta x\cos { \theta } $ exercise : suppose we use a rope to pull the box , and the angle between the rope and the ground is 30º . this time we pull along the rope with a force of 500 n. how much of a chocolate bar can we eat this time if we pull the box through the same 585 m ? what about lifting weights instead ? in the previous example , we were doing work on a box which we were pushing around a floor . in doing so , we were working against a frictional force . another common form of exercise is lifting weights . in this case we are working against the force of gravity rather than friction . using newton 's laws we can find the force , $ f $ , required to lift a weight with mass $ m $ straight up , placing it on a rack which is at a height $ h $ above us : $ f=mg $ the change in position—previously $ \delta x $ —is simply the height , so the work , $ w $ , that we have done in lifting the weight is then $ w=m g h $ the exercise we have done in lifting the weight has resulted in energy being stored in the form of gravitational potential energy . it is called potential energy because it has the potential to be released at any moment with a crash as the weight falls back to the ground . we did positive work on the weight since we exerted our force in the same direction as the displacement of the weight , i.e. , upward . the work done by gravity on the weight while it was lifted was negative since the force of gravity is directed in the opposite direction to the displacement . also , since the weight is stationary after the lift , we know that the work that we have done is exactly canceled out by the work done by gravity . the work done by us is $ mgh $ , and the work done by gravity is $ -mgh $ . we will talk more about this when we look into kinetic energy . ok , let 's put some numbers in and find how much of that chocolate bar we would offset by lifting a weight of 50 kg up to a height of 0.5 m. the work done to the weight is $ w= ( 50\mathrm { kg } ) ( 9.81\mathrm { m/s^2 } ) ( 0.5\mathrm { m } ) = 245.25 \mathrm { j } $ ok , so how many 280 calorie—i.e. , $ 1.17 \times 10^6 $ joule—chocolate bars is this ? well , 245.25 j is about $ \dfrac { 1 } { 4770 } $ of a chocolate bar . but remember , our bodies are only about 25 % efficient , so the work done by the person is actually four times larger , about 981.8 j , which is $ \dfrac { 1 } { 1190 } $ chocolate bars . so , if we can lift this weight once every 2 seconds , it will take us about 2380 seconds or 40 minutes of hard work to burn off this chocolate bar ! what about simply holding a weight stationary ? one frequent source of confusion people have with the concept of work comes about when thinking about holding a heavy weight stationary above our heads , against the force of gravity . we are not moving the weight through any distance , so no work is being done to the weight . we could also achieve this by placing the weight on a table ; it is clear that the table is not doing any work to keep the weight in position . yet , we know from our experience that we get tired when doing the same job . so what is going on here ? it turns out that what is actually happening here is that our bodies are doing work on our muscles to maintain the necessary tension to hold the weight up . the body does this by sending a cascade of nerve impulses to each muscle . each impulse causes the muscle to momentarily contract and release . this all happens so fast that we might only notice a slight twitching at first . eventually though , not enough chemical energy is available in the muscle and it can no longer keep up . we then begin to shake and eventually must rest for a while . so work is being done , it is just not being done on the weight .
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also , since the weight is stationary after the lift , we know that the work that we have done is exactly canceled out by the work done by gravity . the work done by us is $ mgh $ , and the work done by gravity is $ -mgh $ . we will talk more about this when we look into kinetic energy .
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hence ; p.e is directly proportional to work is directly proportional to k.e p.e is given to start a work and it is finished by k.e work is done by between p.e & k.e ... .am i right ?
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introduction in multicellular organisms ( such as yourself ) , cell-cell signaling allows cells to coordinate their activities , ensuring that tissues , organs , and organ systems function correctly . does that mean that unicellular organisms , like yeast and bacteria , don ’ t use cell-cell signaling pathways ? as a matter of fact , these organisms do still need to “ talk ” to one another . the cells may not be part of the same organism , but they belong to the same population , and – just like people in a human population – need ways to communicate about matters of interpersonal or community importance . bacteria , for example , use chemical signals to detect population density ( how many other bacteria are in the area ) and change their behavior accordingly , while yeast produce chemical signals that allow them to find mates . here , we ’ ll take a closer look at how unicellular organisms `` chat '' with one another using chemical signals . quorum sensing in bacteria for many years , it was thought that bacteria were mostly loners , making decisions at the individual rather than the community level . more recently , it ’ s become clear that many types of bacteria engage in a mode of cell-cell signaling called quorum sensing . in quorum sensing , bacteria monitor the density of the population ( the number of other bacteria in the area ) based on chemical signals . when the signaling reaches a threshold level , all the bacteria in the population will change their behavior or gene expression at the same time . quorum sensing in symbiosis quorum sensing was discovered first discovered in aliivibrio fischeri , a bacterium that has a symbiotic ( mutually beneficial ) relationship with the hawaiian bobtail squid $ ^1 $ . a. fischeri form colonies inside the squid ’ s “ light organ . '' the squid gives the bacteria food , and in return , the bacteria bioluminesce ( emit light ) . the glow of the bacteria prevents the squid from casting a shadow , hiding it from predators swimming beneath . when a. fischeri bacteria are inside of a squid ’ s light organ , they glow , but when they ’ re free-living in the ocean , they don ’ t . through decades of work , scientists discovered that the bacteria use quorum sensing to decide when to produce bioluminescence . it would be a metabolic waste for a lone bacterium in the open ocean to carry out chemical reactions that emit light , since they provide no benefit without a squid host . when many bacteria are tightly packed in a light organ , however , glowing in unison provides an advantage : it allows the bacteria to fulfill their end of the symbiotic bargain , keeping their squid host ( their food source ) from being eaten by predators . mechanisms of quorum sensing quorum sensing is based on the production and detection of autoinducers , signaling molecules continually secreted by bacteria to announce their presence to their neighbors ( typically , neighbors of the same species ) . autoinducers let bacteria sense population density and change their behavior in a synchronized fashion when the density reaches a certain threshold . in some types of bacteria , the secreted autoinducers are small , hydrophobic molecules such as acyl-homoserine lactone ( ahl ) . ahl is the autoinducer made by a. fischeri , the bacteria that occupy a squid ’ s light organ . in other types of bacteria , the autoinducers may instead be peptides ( short proteins ) or other types of small molecules $ ^3 $ . because ahl is small and hydrophobic , it can diffuse freely across the membranes of the bacterial cells . when there are few cells in the area , the little ahl that 's made will diffuse into the environment , and the levels of ahl inside the cells will remain low . when more bacteria are present , a larger amount of ahl will be produced ( thanks to the greater number of contributors ) . if ahl levels get high enough , indicating a critical density of bacteria , the ahl will bind to and activate a receptor protein inside the cells . the active receptor acts as a transcription factor , attaching to specific sites on the bacterium ’ s dna and changing the activity of nearby target genes . in a. fischeri , the transcription factor turns on genes that encode enzymes and substrates required for bioluminescence , as well as the gene for the enzyme that makes ahl itself ( amplifying the response in a positive feedback loop ) $ ^4 $ . in general , each species of bacteria has its own autoinducer , with a matching receptor that ’ s highly specific ( won ’ t be activated by the autoinducer of a different bacterium ) . however , some types of autoinducers can be produced and detected by multiple species of bacteria . scientists are investigating how these molecules may allow for between-species communication $ ^6 $ . quorum sensing and biofilms some species of quorum-sensing bacteria form biofilms , surface-attached communities of bacterial cells that stick to one another and to their substrate ( underlying surface ) . biofilms can be quite complex , with bacterial cells organizing to form ordered structures , and some biofilms contain multiple species of coexisting bacteria . although much remains unknown about biofilms , it ’ s increasingly clear that they play crucial roles in human health and disease . for instance , the s. aureus colonizing the surface of a catheter above are organized in a biofilm . quorum sensing can play an important role in the formation , maintenance , and breakdown of biofilms . signaling in yeasts the yeasts that ferment grapes into wine , or cause bread to rise , are single-celled eukaryotes . they 're neither animals nor plants , but actually a type of fungus . ( yum ! ) some baker 's yeasts are shown in the microscope image below . one of the best-studied signaling pathways in yeast is the mating factor pathway . budding yeasts can mate in a process similar to sexual reproduction , in which two haploid cells ( cells with a single set of chromosomes , like human sperm and eggs ) combine to form a diploid cell ( a cell with two sets of chromosomes , like human body cells ) . the diploid cell can then go through meiosis to make haploid cells with new combinations of genetic material . to find another haploid yeast cell that is prepared to mate , budding yeasts secrete a signaling molecule called mating factor . mating factor comes in two different versions , as does its receptor , and this system may help yeasts mate with other yeasts that are not close relatives . binding of mating factor to a compatible receptor triggers a signaling cascade that causes the yeast to `` shmoo , '' or produce an outgrowth so it can fuse with its mate . you can see the details of this pathway in the video on cell signaling in yeast reproduction $ ^ { 7,8 } $ . if you look closely at the mating factor signaling pathway , you ’ ll find that it includes types of molecules familiar from humans . for instance , the mating factor receptor is a g protein-coupled receptor , and it acts through a map kinase signaling pathway like the one used in human growth factor signaling $ ^9 $ .
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when the signaling reaches a threshold level , all the bacteria in the population will change their behavior or gene expression at the same time . quorum sensing in symbiosis quorum sensing was discovered first discovered in aliivibrio fischeri , a bacterium that has a symbiotic ( mutually beneficial ) relationship with the hawaiian bobtail squid $ ^1 $ . a. fischeri form colonies inside the squid ’ s “ light organ . ''
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are there any medical applications of quorum sensing yet ?
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introduction in multicellular organisms ( such as yourself ) , cell-cell signaling allows cells to coordinate their activities , ensuring that tissues , organs , and organ systems function correctly . does that mean that unicellular organisms , like yeast and bacteria , don ’ t use cell-cell signaling pathways ? as a matter of fact , these organisms do still need to “ talk ” to one another . the cells may not be part of the same organism , but they belong to the same population , and – just like people in a human population – need ways to communicate about matters of interpersonal or community importance . bacteria , for example , use chemical signals to detect population density ( how many other bacteria are in the area ) and change their behavior accordingly , while yeast produce chemical signals that allow them to find mates . here , we ’ ll take a closer look at how unicellular organisms `` chat '' with one another using chemical signals . quorum sensing in bacteria for many years , it was thought that bacteria were mostly loners , making decisions at the individual rather than the community level . more recently , it ’ s become clear that many types of bacteria engage in a mode of cell-cell signaling called quorum sensing . in quorum sensing , bacteria monitor the density of the population ( the number of other bacteria in the area ) based on chemical signals . when the signaling reaches a threshold level , all the bacteria in the population will change their behavior or gene expression at the same time . quorum sensing in symbiosis quorum sensing was discovered first discovered in aliivibrio fischeri , a bacterium that has a symbiotic ( mutually beneficial ) relationship with the hawaiian bobtail squid $ ^1 $ . a. fischeri form colonies inside the squid ’ s “ light organ . '' the squid gives the bacteria food , and in return , the bacteria bioluminesce ( emit light ) . the glow of the bacteria prevents the squid from casting a shadow , hiding it from predators swimming beneath . when a. fischeri bacteria are inside of a squid ’ s light organ , they glow , but when they ’ re free-living in the ocean , they don ’ t . through decades of work , scientists discovered that the bacteria use quorum sensing to decide when to produce bioluminescence . it would be a metabolic waste for a lone bacterium in the open ocean to carry out chemical reactions that emit light , since they provide no benefit without a squid host . when many bacteria are tightly packed in a light organ , however , glowing in unison provides an advantage : it allows the bacteria to fulfill their end of the symbiotic bargain , keeping their squid host ( their food source ) from being eaten by predators . mechanisms of quorum sensing quorum sensing is based on the production and detection of autoinducers , signaling molecules continually secreted by bacteria to announce their presence to their neighbors ( typically , neighbors of the same species ) . autoinducers let bacteria sense population density and change their behavior in a synchronized fashion when the density reaches a certain threshold . in some types of bacteria , the secreted autoinducers are small , hydrophobic molecules such as acyl-homoserine lactone ( ahl ) . ahl is the autoinducer made by a. fischeri , the bacteria that occupy a squid ’ s light organ . in other types of bacteria , the autoinducers may instead be peptides ( short proteins ) or other types of small molecules $ ^3 $ . because ahl is small and hydrophobic , it can diffuse freely across the membranes of the bacterial cells . when there are few cells in the area , the little ahl that 's made will diffuse into the environment , and the levels of ahl inside the cells will remain low . when more bacteria are present , a larger amount of ahl will be produced ( thanks to the greater number of contributors ) . if ahl levels get high enough , indicating a critical density of bacteria , the ahl will bind to and activate a receptor protein inside the cells . the active receptor acts as a transcription factor , attaching to specific sites on the bacterium ’ s dna and changing the activity of nearby target genes . in a. fischeri , the transcription factor turns on genes that encode enzymes and substrates required for bioluminescence , as well as the gene for the enzyme that makes ahl itself ( amplifying the response in a positive feedback loop ) $ ^4 $ . in general , each species of bacteria has its own autoinducer , with a matching receptor that ’ s highly specific ( won ’ t be activated by the autoinducer of a different bacterium ) . however , some types of autoinducers can be produced and detected by multiple species of bacteria . scientists are investigating how these molecules may allow for between-species communication $ ^6 $ . quorum sensing and biofilms some species of quorum-sensing bacteria form biofilms , surface-attached communities of bacterial cells that stick to one another and to their substrate ( underlying surface ) . biofilms can be quite complex , with bacterial cells organizing to form ordered structures , and some biofilms contain multiple species of coexisting bacteria . although much remains unknown about biofilms , it ’ s increasingly clear that they play crucial roles in human health and disease . for instance , the s. aureus colonizing the surface of a catheter above are organized in a biofilm . quorum sensing can play an important role in the formation , maintenance , and breakdown of biofilms . signaling in yeasts the yeasts that ferment grapes into wine , or cause bread to rise , are single-celled eukaryotes . they 're neither animals nor plants , but actually a type of fungus . ( yum ! ) some baker 's yeasts are shown in the microscope image below . one of the best-studied signaling pathways in yeast is the mating factor pathway . budding yeasts can mate in a process similar to sexual reproduction , in which two haploid cells ( cells with a single set of chromosomes , like human sperm and eggs ) combine to form a diploid cell ( a cell with two sets of chromosomes , like human body cells ) . the diploid cell can then go through meiosis to make haploid cells with new combinations of genetic material . to find another haploid yeast cell that is prepared to mate , budding yeasts secrete a signaling molecule called mating factor . mating factor comes in two different versions , as does its receptor , and this system may help yeasts mate with other yeasts that are not close relatives . binding of mating factor to a compatible receptor triggers a signaling cascade that causes the yeast to `` shmoo , '' or produce an outgrowth so it can fuse with its mate . you can see the details of this pathway in the video on cell signaling in yeast reproduction $ ^ { 7,8 } $ . if you look closely at the mating factor signaling pathway , you ’ ll find that it includes types of molecules familiar from humans . for instance , the mating factor receptor is a g protein-coupled receptor , and it acts through a map kinase signaling pathway like the one used in human growth factor signaling $ ^9 $ .
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the active receptor acts as a transcription factor , attaching to specific sites on the bacterium ’ s dna and changing the activity of nearby target genes . in a. fischeri , the transcription factor turns on genes that encode enzymes and substrates required for bioluminescence , as well as the gene for the enzyme that makes ahl itself ( amplifying the response in a positive feedback loop ) $ ^4 $ . in general , each species of bacteria has its own autoinducer , with a matching receptor that ’ s highly specific ( won ’ t be activated by the autoinducer of a different bacterium ) .
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what does that have to do with the transcription factor of the gene for the enzyme that makes ahl ?
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navigation between the islands the marshall islands in eastern micronesia consist of thirty-four coral atolls consisting of more than one thousand islands and islets spread out across an area of several hundred miles . in order to maintain links between the islands , the marshall islanders built seafaring canoes . these vessels were both quick and manoeuvrable . the islanders developed a reputation for navigation between the islands—not a simple matter , since they are all so low that none can be seen from more than a few miles away . in order to determine a system of piloting and navigation the islanders devised charts that marked not only the locations of the islands , but their knowledge of the swell and wave patterns as well . the charts were composed of wooden sticks ; the horizontal and vertical sticks act as supports , while diagonal and curved ones represent wave swells . cowrie or other small shells represent the position of the islands . the information was memorized and the charts would not be carried on voyages . rebbelib this chart ( above ) is of a type known as a rebbelib , which cover either a large section or all of the marshall islands . other types of chart more commonly show a smaller area . this example represents the two chains of islands which form the marshall islands . it was collected by admiral e.h.m . davis during the cruise of hms royalist from 1890 to 1893 . mattang this chart ( below ) is of the type known as a mattang , specifically made for the purpose of training people selected to be navigators . such charts depict general information about swell movements around one or more small islands . trainees were taught by experienced navigators . navigation charts continue to be made , often simpler in form , to be sold as souvenirs . suggested readings : tw . davenport. , `` marshall islands cartography , '' the bulletin of the university museum of pennsylvania , 6 : 4 ( summer 1964 ) , pp . 10 -13 . j. feldman and d.h. rubinstein , the art of micronesia ( honolulu , the university of hawaii art gallery , 1988 ) . a.c. haddon and j. hornell , canoes of oceania ( honolulu , bernice p. bishop museum special publications 27-29 , reprinted as one volume , 1975 ) . t.a joyce , `` note on a native chart from the marshall islands in the british museum , '' man-1 , 8 ( 1908 ) , no . 81 , pp . 146-49 .
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trainees were taught by experienced navigators . navigation charts continue to be made , often simpler in form , to be sold as souvenirs . suggested readings : tw .
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i know that the charts are still being made for sale to tourists , but do any marshall islanders use these charts nowadays ?
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navigation between the islands the marshall islands in eastern micronesia consist of thirty-four coral atolls consisting of more than one thousand islands and islets spread out across an area of several hundred miles . in order to maintain links between the islands , the marshall islanders built seafaring canoes . these vessels were both quick and manoeuvrable . the islanders developed a reputation for navigation between the islands—not a simple matter , since they are all so low that none can be seen from more than a few miles away . in order to determine a system of piloting and navigation the islanders devised charts that marked not only the locations of the islands , but their knowledge of the swell and wave patterns as well . the charts were composed of wooden sticks ; the horizontal and vertical sticks act as supports , while diagonal and curved ones represent wave swells . cowrie or other small shells represent the position of the islands . the information was memorized and the charts would not be carried on voyages . rebbelib this chart ( above ) is of a type known as a rebbelib , which cover either a large section or all of the marshall islands . other types of chart more commonly show a smaller area . this example represents the two chains of islands which form the marshall islands . it was collected by admiral e.h.m . davis during the cruise of hms royalist from 1890 to 1893 . mattang this chart ( below ) is of the type known as a mattang , specifically made for the purpose of training people selected to be navigators . such charts depict general information about swell movements around one or more small islands . trainees were taught by experienced navigators . navigation charts continue to be made , often simpler in form , to be sold as souvenirs . suggested readings : tw . davenport. , `` marshall islands cartography , '' the bulletin of the university museum of pennsylvania , 6 : 4 ( summer 1964 ) , pp . 10 -13 . j. feldman and d.h. rubinstein , the art of micronesia ( honolulu , the university of hawaii art gallery , 1988 ) . a.c. haddon and j. hornell , canoes of oceania ( honolulu , bernice p. bishop museum special publications 27-29 , reprinted as one volume , 1975 ) . t.a joyce , `` note on a native chart from the marshall islands in the british museum , '' man-1 , 8 ( 1908 ) , no . 81 , pp . 146-49 .
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trainees were taught by experienced navigators . navigation charts continue to be made , often simpler in form , to be sold as souvenirs . suggested readings : tw .
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do the navigation maps still lead you to those places ?
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navigation between the islands the marshall islands in eastern micronesia consist of thirty-four coral atolls consisting of more than one thousand islands and islets spread out across an area of several hundred miles . in order to maintain links between the islands , the marshall islanders built seafaring canoes . these vessels were both quick and manoeuvrable . the islanders developed a reputation for navigation between the islands—not a simple matter , since they are all so low that none can be seen from more than a few miles away . in order to determine a system of piloting and navigation the islanders devised charts that marked not only the locations of the islands , but their knowledge of the swell and wave patterns as well . the charts were composed of wooden sticks ; the horizontal and vertical sticks act as supports , while diagonal and curved ones represent wave swells . cowrie or other small shells represent the position of the islands . the information was memorized and the charts would not be carried on voyages . rebbelib this chart ( above ) is of a type known as a rebbelib , which cover either a large section or all of the marshall islands . other types of chart more commonly show a smaller area . this example represents the two chains of islands which form the marshall islands . it was collected by admiral e.h.m . davis during the cruise of hms royalist from 1890 to 1893 . mattang this chart ( below ) is of the type known as a mattang , specifically made for the purpose of training people selected to be navigators . such charts depict general information about swell movements around one or more small islands . trainees were taught by experienced navigators . navigation charts continue to be made , often simpler in form , to be sold as souvenirs . suggested readings : tw . davenport. , `` marshall islands cartography , '' the bulletin of the university museum of pennsylvania , 6 : 4 ( summer 1964 ) , pp . 10 -13 . j. feldman and d.h. rubinstein , the art of micronesia ( honolulu , the university of hawaii art gallery , 1988 ) . a.c. haddon and j. hornell , canoes of oceania ( honolulu , bernice p. bishop museum special publications 27-29 , reprinted as one volume , 1975 ) . t.a joyce , `` note on a native chart from the marshall islands in the british museum , '' man-1 , 8 ( 1908 ) , no . 81 , pp . 146-49 .
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rebbelib this chart ( above ) is of a type known as a rebbelib , which cover either a large section or all of the marshall islands . other types of chart more commonly show a smaller area . this example represents the two chains of islands which form the marshall islands .
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how difficult would it be to memorize such a chart ?
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key points heat , $ \text q $ , is thermal energy transferred from a hotter system to a cooler system that are in contact . temperature is a measure of the average kinetic energy of the atoms or molecules in the system . the zeroth law of thermodynamics says that no heat is transferred between two objects in thermal equilibrium ; therefore , they are the same temperature . we can calculate the heat released or absorbed using the specific heat capacity $ \text c $ , the mass of the substance $ \text m $ , and the change in temperature $ \delta \text t $ in the equation : $ \text q = \text { m } \times \text c \times \delta \text t $ heat in thermodynamics what contains more heat , a cup of coffee or a glass of iced tea ? in chemistry class , that would be a trick question ( sorry ! ) . in thermodynamics , heat has a very specific meaning that is different from how we might use the word in everyday speech . scientists define heat as thermal energy transferred between two systems at different temperatures that come in contact . heat is written with the symbol q or q , and it has units of joules ( $ \text j $ ) . heat is sometimes called a process quantity , because it is defined in the context of a process by which energy can be transferred . we do n't talk about a cup of coffee containing heat , but we can talk about the heat transferred from the cup of hot coffee to your hand . heat is also an extensive property , so the change in temperature resulting from heat transferred to a system depends on how many molecules are in the system . relationship between heat and temperature heat and temperature are two different but closely related concepts . note that they have different units : temperature typically has units of degrees celsius ( $ ^\circ\text c $ ) or kelvin ( $ \text k $ ) , and heat has units of energy , joules ( $ \text j $ ) . temperature is a measure of the average kinetic energy of the atoms or molecules in the system . the water molecules in a cup of hot coffee have a higher average kinetic energy than the water molecules in a cup of iced tea , which also means they are moving at a higher velocity . temperature is also an intensive property , which means that the temperature does n't change no matter how much of a substance you have ( as long as it is all at the same temperature ! ) . this is why chemists can use the melting point to help identify a pure substance $ - $ the temperature at which it melts is a property of the substance with no dependence on the mass of a sample . on an atomic level , the molecules in each object are constantly in motion and colliding with each other . every time molecules collide , kinetic energy can be transferred . when the two systems are in contact , heat will be transferred through molecular collisions from the hotter system to the cooler system . the thermal energy will flow in that direction until the two objects are at the same temperature . when the two systems in contact are at the same temperature , we say they are in thermal equilibrium . zeroth law of thermodynamics : defining thermal equilibrium the zeroth law of thermodynamics defines thermal equilibrium within an isolated system . the zeroth law says when two objects at thermal equilibrium are in contact , there is no net heat transfer between the objects ; therefore , they are the same temperature . another way to state the zeroth law is to say that if two objects are both separately in thermal equilibrium with a third object , then they are in thermal equilibrium with each other . the zeroth law allows us to measure the temperature of objects . any time we use a thermometer , we are using the zeroth law of thermodynamics . let 's say we are measuring the temperature of a water bath . in order to make sure the reading is accurate , we usually want to wait for the temperature reading to stay constant . we are waiting for the thermometer and the water to reach thermal equilibrium ! at thermal equilibrium , the temperature of the thermometer bulb and the water bath will be the same , and there should be no net heat transfer from one object to the other ( assuming no other loss of heat to the surroundings ) . heat capacity : converting between heat and change in temperature how can we measure heat ? here are some things we know about heat so far : when a system absorbs or loses heat , the average kinetic energy of the molecules will change . thus , heat transfer results in a change in the system 's temperature as long as the system is not undergoing a phase change . the change in temperature resulting from heat transferred to or from a system depends on how many molecules are in the system . we can use a thermometer to measure the change in a system 's temperature . how can we use the change in temperature to calculate the heat transferred ? in order to figure out how the heat transferred to a system will change the temperature of the system , we need to know at least $ 2 $ things : the number of molecules in the system the heat capacity of the system the heat capacity tells us how much energy is needed to change the temperature of a given substance assuming that no phase changes are occurring . there are two main ways that heat capacity is reported . the specific heat capacity ( also called specific heat ) , represented by the symbol $ \text c $ or $ \text c $ , is how much energy is needed to increase the temperature of one gram of a substance by $ 1~^ { \circ } \text c $ or $ 1\ , \text k $ . specific heat capacity usually has units of $ \dfrac { \text j } { \text { grams } \cdot\text k } $ . the molar heat capacity , $ \text c_\text m $ or $ \text c_ { \text { mol } } $ , measures the amount of thermal energy it takes to raise the temperature of one mole of a substance by $ 1~^ { \circ } \text c $ or $ 1\ , \text k $ , and it usually has units of $ \dfrac { \text j } { \text { mol } \cdot\text k } $ . for example , the heat capacity of lead might be given as the specific heat capacity , $ 0.129\ , \dfrac { \text j } { \text { g } \cdot\text k } $ , or the molar heat capacity , $ 26.65\ , \dfrac { \text j } { \text { mol } \cdot\text k } $ . calculating $ \text q $ using the heat capacity we can use the heat capacity to determine the heat released or absorbed by a material using the following formula : $ \text q = \text { m } \times \text c \times \delta \text t $ where $ \text { m } $ is the mass of the substance ( in grams ) , $ \text { c } $ is the specific heat capacity , and $ \delta \text t $ is the change in temperature during the heat transfer . note that both mass and specific heat capacity can only have positive values , so the sign of $ \text q $ will depend on the sign of $ \delta \text t $ . we can calculate $ \delta \text t $ using the following equation : $ \delta \text t=\text t_ { \text { final } } -\text t_ { \text { initial } } $ where $ \text t_ { \text { final } } $ and $ \text t_ { \text { initial } } $ can have units of either $ ~^ { \circ } \text c $ or $ \text k $ . based on this equation , if $ \text q $ is positive ( energy of the system increases ) , then our system increases in temperature and $ \text t_ { \text { final } } & gt ; \text t_ { \text { initial } } $ . if $ \text q $ is negative ( energy of the system decreases ) , then our system 's temperature decreases and $ \text t_ { \text { final } } & lt ; \text t_ { \text { initial } } $ . example problem : cooling a cup of tea let 's say that we have $ 250\ , \text { ml } $ of hot tea which we would like to cool down before we try to drink it . the tea is currently at $ 370\ , \text k $ , and we 'd like to cool it down to $ 350\ , \text k $ . how much thermal energy has to be transferred from the tea to the surroundings to cool the tea ? we are going to assume that the tea is mostly water , so we can use the density and heat capacity of water in our calculations . the specific heat capacity of water is $ 4.18\ , \dfrac { \text j } { \text g \cdot \text k } $ , and the density of water is $ 1.00\ , \dfrac { \text g } { \text { ml } } $ . we can calculate the energy transferred in the process of cooling the tea using the following steps : 1 . calculate the mass of the substance we can calculate the mass of the tea/water using the volume and density of water : $ \text m=250\ , \cancel { \text { ml } } \times 1.00\ , \dfrac { \text g } { \cancel { \text { ml } } } =250\ , \text g $ 2 . calculate the change in temperature , $ \delta \text t $ we can calculate the change in temperature , $ \delta \text t $ , from the initial and final temperatures : $ \begin { align } \delta \text t & amp ; =\text t_ { \text { final } } -\text t_ { \text { initial } } \ \ & amp ; =350\ , \text k-370\ , \text k\ \ & amp ; =-20\ , \text k\end { align } $ since the temperature of the tea is decreasing and $ \delta \text t $ is negative , we would expect $ \text q $ to also be negative since our system is losing thermal energy . 3 . solve for $ \text q $ now we can solve for the heat transferred from the hot tea using the equation for heat : $ \begin { align } \text q & amp ; = \text { m } \times \text c \times \delta \text t\ & amp ; =250\ , \cancel { \text g } \times4.18\ , \dfrac { \text j } { \cancel { \text g } \cdot \cancel { \text k } } \times -20\ , \cancel { \text k } \ & amp ; =-21000\ , \text j\end { align } $ thus , we calculated that the tea will transfer $ 21000\ , \text j $ of energy to the surroundings when it cools down from $ 370\ , \text k $ to $ 350\ , \text k $ . conclusions in thermodynamics , heat and temperature are closely related concepts with precise definitions . heat , $ \text q $ , is thermal energy transferred from a hotter system to a cooler system that are in contact . temperature is a measure of the average kinetic energy of the atoms or molecules in the system . the zeroth law of thermodynamics says that no heat is transferred between two objects in thermal equilibrium ; therefore , they are the same temperature . we can calculate the heat released or absorbed using the specific heat capacity $ \text c $ , the mass of the substance $ \text m $ , and the change in temperature $ \delta \text t $ in the following equation : $ \text q = \text { m } \times \text c \times \delta \text t $
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we are going to assume that the tea is mostly water , so we can use the density and heat capacity of water in our calculations . the specific heat capacity of water is $ 4.18\ , \dfrac { \text j } { \text g \cdot \text k } $ , and the density of water is $ 1.00\ , \dfrac { \text g } { \text { ml } } $ . we can calculate the energy transferred in the process of cooling the tea using the following steps : 1 .
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250 * 4.18 * -20 = -20900 how is the above answer -21000j ?
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key points heat , $ \text q $ , is thermal energy transferred from a hotter system to a cooler system that are in contact . temperature is a measure of the average kinetic energy of the atoms or molecules in the system . the zeroth law of thermodynamics says that no heat is transferred between two objects in thermal equilibrium ; therefore , they are the same temperature . we can calculate the heat released or absorbed using the specific heat capacity $ \text c $ , the mass of the substance $ \text m $ , and the change in temperature $ \delta \text t $ in the equation : $ \text q = \text { m } \times \text c \times \delta \text t $ heat in thermodynamics what contains more heat , a cup of coffee or a glass of iced tea ? in chemistry class , that would be a trick question ( sorry ! ) . in thermodynamics , heat has a very specific meaning that is different from how we might use the word in everyday speech . scientists define heat as thermal energy transferred between two systems at different temperatures that come in contact . heat is written with the symbol q or q , and it has units of joules ( $ \text j $ ) . heat is sometimes called a process quantity , because it is defined in the context of a process by which energy can be transferred . we do n't talk about a cup of coffee containing heat , but we can talk about the heat transferred from the cup of hot coffee to your hand . heat is also an extensive property , so the change in temperature resulting from heat transferred to a system depends on how many molecules are in the system . relationship between heat and temperature heat and temperature are two different but closely related concepts . note that they have different units : temperature typically has units of degrees celsius ( $ ^\circ\text c $ ) or kelvin ( $ \text k $ ) , and heat has units of energy , joules ( $ \text j $ ) . temperature is a measure of the average kinetic energy of the atoms or molecules in the system . the water molecules in a cup of hot coffee have a higher average kinetic energy than the water molecules in a cup of iced tea , which also means they are moving at a higher velocity . temperature is also an intensive property , which means that the temperature does n't change no matter how much of a substance you have ( as long as it is all at the same temperature ! ) . this is why chemists can use the melting point to help identify a pure substance $ - $ the temperature at which it melts is a property of the substance with no dependence on the mass of a sample . on an atomic level , the molecules in each object are constantly in motion and colliding with each other . every time molecules collide , kinetic energy can be transferred . when the two systems are in contact , heat will be transferred through molecular collisions from the hotter system to the cooler system . the thermal energy will flow in that direction until the two objects are at the same temperature . when the two systems in contact are at the same temperature , we say they are in thermal equilibrium . zeroth law of thermodynamics : defining thermal equilibrium the zeroth law of thermodynamics defines thermal equilibrium within an isolated system . the zeroth law says when two objects at thermal equilibrium are in contact , there is no net heat transfer between the objects ; therefore , they are the same temperature . another way to state the zeroth law is to say that if two objects are both separately in thermal equilibrium with a third object , then they are in thermal equilibrium with each other . the zeroth law allows us to measure the temperature of objects . any time we use a thermometer , we are using the zeroth law of thermodynamics . let 's say we are measuring the temperature of a water bath . in order to make sure the reading is accurate , we usually want to wait for the temperature reading to stay constant . we are waiting for the thermometer and the water to reach thermal equilibrium ! at thermal equilibrium , the temperature of the thermometer bulb and the water bath will be the same , and there should be no net heat transfer from one object to the other ( assuming no other loss of heat to the surroundings ) . heat capacity : converting between heat and change in temperature how can we measure heat ? here are some things we know about heat so far : when a system absorbs or loses heat , the average kinetic energy of the molecules will change . thus , heat transfer results in a change in the system 's temperature as long as the system is not undergoing a phase change . the change in temperature resulting from heat transferred to or from a system depends on how many molecules are in the system . we can use a thermometer to measure the change in a system 's temperature . how can we use the change in temperature to calculate the heat transferred ? in order to figure out how the heat transferred to a system will change the temperature of the system , we need to know at least $ 2 $ things : the number of molecules in the system the heat capacity of the system the heat capacity tells us how much energy is needed to change the temperature of a given substance assuming that no phase changes are occurring . there are two main ways that heat capacity is reported . the specific heat capacity ( also called specific heat ) , represented by the symbol $ \text c $ or $ \text c $ , is how much energy is needed to increase the temperature of one gram of a substance by $ 1~^ { \circ } \text c $ or $ 1\ , \text k $ . specific heat capacity usually has units of $ \dfrac { \text j } { \text { grams } \cdot\text k } $ . the molar heat capacity , $ \text c_\text m $ or $ \text c_ { \text { mol } } $ , measures the amount of thermal energy it takes to raise the temperature of one mole of a substance by $ 1~^ { \circ } \text c $ or $ 1\ , \text k $ , and it usually has units of $ \dfrac { \text j } { \text { mol } \cdot\text k } $ . for example , the heat capacity of lead might be given as the specific heat capacity , $ 0.129\ , \dfrac { \text j } { \text { g } \cdot\text k } $ , or the molar heat capacity , $ 26.65\ , \dfrac { \text j } { \text { mol } \cdot\text k } $ . calculating $ \text q $ using the heat capacity we can use the heat capacity to determine the heat released or absorbed by a material using the following formula : $ \text q = \text { m } \times \text c \times \delta \text t $ where $ \text { m } $ is the mass of the substance ( in grams ) , $ \text { c } $ is the specific heat capacity , and $ \delta \text t $ is the change in temperature during the heat transfer . note that both mass and specific heat capacity can only have positive values , so the sign of $ \text q $ will depend on the sign of $ \delta \text t $ . we can calculate $ \delta \text t $ using the following equation : $ \delta \text t=\text t_ { \text { final } } -\text t_ { \text { initial } } $ where $ \text t_ { \text { final } } $ and $ \text t_ { \text { initial } } $ can have units of either $ ~^ { \circ } \text c $ or $ \text k $ . based on this equation , if $ \text q $ is positive ( energy of the system increases ) , then our system increases in temperature and $ \text t_ { \text { final } } & gt ; \text t_ { \text { initial } } $ . if $ \text q $ is negative ( energy of the system decreases ) , then our system 's temperature decreases and $ \text t_ { \text { final } } & lt ; \text t_ { \text { initial } } $ . example problem : cooling a cup of tea let 's say that we have $ 250\ , \text { ml } $ of hot tea which we would like to cool down before we try to drink it . the tea is currently at $ 370\ , \text k $ , and we 'd like to cool it down to $ 350\ , \text k $ . how much thermal energy has to be transferred from the tea to the surroundings to cool the tea ? we are going to assume that the tea is mostly water , so we can use the density and heat capacity of water in our calculations . the specific heat capacity of water is $ 4.18\ , \dfrac { \text j } { \text g \cdot \text k } $ , and the density of water is $ 1.00\ , \dfrac { \text g } { \text { ml } } $ . we can calculate the energy transferred in the process of cooling the tea using the following steps : 1 . calculate the mass of the substance we can calculate the mass of the tea/water using the volume and density of water : $ \text m=250\ , \cancel { \text { ml } } \times 1.00\ , \dfrac { \text g } { \cancel { \text { ml } } } =250\ , \text g $ 2 . calculate the change in temperature , $ \delta \text t $ we can calculate the change in temperature , $ \delta \text t $ , from the initial and final temperatures : $ \begin { align } \delta \text t & amp ; =\text t_ { \text { final } } -\text t_ { \text { initial } } \ \ & amp ; =350\ , \text k-370\ , \text k\ \ & amp ; =-20\ , \text k\end { align } $ since the temperature of the tea is decreasing and $ \delta \text t $ is negative , we would expect $ \text q $ to also be negative since our system is losing thermal energy . 3 . solve for $ \text q $ now we can solve for the heat transferred from the hot tea using the equation for heat : $ \begin { align } \text q & amp ; = \text { m } \times \text c \times \delta \text t\ & amp ; =250\ , \cancel { \text g } \times4.18\ , \dfrac { \text j } { \cancel { \text g } \cdot \cancel { \text k } } \times -20\ , \cancel { \text k } \ & amp ; =-21000\ , \text j\end { align } $ thus , we calculated that the tea will transfer $ 21000\ , \text j $ of energy to the surroundings when it cools down from $ 370\ , \text k $ to $ 350\ , \text k $ . conclusions in thermodynamics , heat and temperature are closely related concepts with precise definitions . heat , $ \text q $ , is thermal energy transferred from a hotter system to a cooler system that are in contact . temperature is a measure of the average kinetic energy of the atoms or molecules in the system . the zeroth law of thermodynamics says that no heat is transferred between two objects in thermal equilibrium ; therefore , they are the same temperature . we can calculate the heat released or absorbed using the specific heat capacity $ \text c $ , the mass of the substance $ \text m $ , and the change in temperature $ \delta \text t $ in the following equation : $ \text q = \text { m } \times \text c \times \delta \text t $
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heat capacity : converting between heat and change in temperature how can we measure heat ? here are some things we know about heat so far : when a system absorbs or loses heat , the average kinetic energy of the molecules will change . thus , heat transfer results in a change in the system 's temperature as long as the system is not undergoing a phase change .
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for i know that kelvin is always positive , but why in the example , why kevin degree is negative ?
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key points heat , $ \text q $ , is thermal energy transferred from a hotter system to a cooler system that are in contact . temperature is a measure of the average kinetic energy of the atoms or molecules in the system . the zeroth law of thermodynamics says that no heat is transferred between two objects in thermal equilibrium ; therefore , they are the same temperature . we can calculate the heat released or absorbed using the specific heat capacity $ \text c $ , the mass of the substance $ \text m $ , and the change in temperature $ \delta \text t $ in the equation : $ \text q = \text { m } \times \text c \times \delta \text t $ heat in thermodynamics what contains more heat , a cup of coffee or a glass of iced tea ? in chemistry class , that would be a trick question ( sorry ! ) . in thermodynamics , heat has a very specific meaning that is different from how we might use the word in everyday speech . scientists define heat as thermal energy transferred between two systems at different temperatures that come in contact . heat is written with the symbol q or q , and it has units of joules ( $ \text j $ ) . heat is sometimes called a process quantity , because it is defined in the context of a process by which energy can be transferred . we do n't talk about a cup of coffee containing heat , but we can talk about the heat transferred from the cup of hot coffee to your hand . heat is also an extensive property , so the change in temperature resulting from heat transferred to a system depends on how many molecules are in the system . relationship between heat and temperature heat and temperature are two different but closely related concepts . note that they have different units : temperature typically has units of degrees celsius ( $ ^\circ\text c $ ) or kelvin ( $ \text k $ ) , and heat has units of energy , joules ( $ \text j $ ) . temperature is a measure of the average kinetic energy of the atoms or molecules in the system . the water molecules in a cup of hot coffee have a higher average kinetic energy than the water molecules in a cup of iced tea , which also means they are moving at a higher velocity . temperature is also an intensive property , which means that the temperature does n't change no matter how much of a substance you have ( as long as it is all at the same temperature ! ) . this is why chemists can use the melting point to help identify a pure substance $ - $ the temperature at which it melts is a property of the substance with no dependence on the mass of a sample . on an atomic level , the molecules in each object are constantly in motion and colliding with each other . every time molecules collide , kinetic energy can be transferred . when the two systems are in contact , heat will be transferred through molecular collisions from the hotter system to the cooler system . the thermal energy will flow in that direction until the two objects are at the same temperature . when the two systems in contact are at the same temperature , we say they are in thermal equilibrium . zeroth law of thermodynamics : defining thermal equilibrium the zeroth law of thermodynamics defines thermal equilibrium within an isolated system . the zeroth law says when two objects at thermal equilibrium are in contact , there is no net heat transfer between the objects ; therefore , they are the same temperature . another way to state the zeroth law is to say that if two objects are both separately in thermal equilibrium with a third object , then they are in thermal equilibrium with each other . the zeroth law allows us to measure the temperature of objects . any time we use a thermometer , we are using the zeroth law of thermodynamics . let 's say we are measuring the temperature of a water bath . in order to make sure the reading is accurate , we usually want to wait for the temperature reading to stay constant . we are waiting for the thermometer and the water to reach thermal equilibrium ! at thermal equilibrium , the temperature of the thermometer bulb and the water bath will be the same , and there should be no net heat transfer from one object to the other ( assuming no other loss of heat to the surroundings ) . heat capacity : converting between heat and change in temperature how can we measure heat ? here are some things we know about heat so far : when a system absorbs or loses heat , the average kinetic energy of the molecules will change . thus , heat transfer results in a change in the system 's temperature as long as the system is not undergoing a phase change . the change in temperature resulting from heat transferred to or from a system depends on how many molecules are in the system . we can use a thermometer to measure the change in a system 's temperature . how can we use the change in temperature to calculate the heat transferred ? in order to figure out how the heat transferred to a system will change the temperature of the system , we need to know at least $ 2 $ things : the number of molecules in the system the heat capacity of the system the heat capacity tells us how much energy is needed to change the temperature of a given substance assuming that no phase changes are occurring . there are two main ways that heat capacity is reported . the specific heat capacity ( also called specific heat ) , represented by the symbol $ \text c $ or $ \text c $ , is how much energy is needed to increase the temperature of one gram of a substance by $ 1~^ { \circ } \text c $ or $ 1\ , \text k $ . specific heat capacity usually has units of $ \dfrac { \text j } { \text { grams } \cdot\text k } $ . the molar heat capacity , $ \text c_\text m $ or $ \text c_ { \text { mol } } $ , measures the amount of thermal energy it takes to raise the temperature of one mole of a substance by $ 1~^ { \circ } \text c $ or $ 1\ , \text k $ , and it usually has units of $ \dfrac { \text j } { \text { mol } \cdot\text k } $ . for example , the heat capacity of lead might be given as the specific heat capacity , $ 0.129\ , \dfrac { \text j } { \text { g } \cdot\text k } $ , or the molar heat capacity , $ 26.65\ , \dfrac { \text j } { \text { mol } \cdot\text k } $ . calculating $ \text q $ using the heat capacity we can use the heat capacity to determine the heat released or absorbed by a material using the following formula : $ \text q = \text { m } \times \text c \times \delta \text t $ where $ \text { m } $ is the mass of the substance ( in grams ) , $ \text { c } $ is the specific heat capacity , and $ \delta \text t $ is the change in temperature during the heat transfer . note that both mass and specific heat capacity can only have positive values , so the sign of $ \text q $ will depend on the sign of $ \delta \text t $ . we can calculate $ \delta \text t $ using the following equation : $ \delta \text t=\text t_ { \text { final } } -\text t_ { \text { initial } } $ where $ \text t_ { \text { final } } $ and $ \text t_ { \text { initial } } $ can have units of either $ ~^ { \circ } \text c $ or $ \text k $ . based on this equation , if $ \text q $ is positive ( energy of the system increases ) , then our system increases in temperature and $ \text t_ { \text { final } } & gt ; \text t_ { \text { initial } } $ . if $ \text q $ is negative ( energy of the system decreases ) , then our system 's temperature decreases and $ \text t_ { \text { final } } & lt ; \text t_ { \text { initial } } $ . example problem : cooling a cup of tea let 's say that we have $ 250\ , \text { ml } $ of hot tea which we would like to cool down before we try to drink it . the tea is currently at $ 370\ , \text k $ , and we 'd like to cool it down to $ 350\ , \text k $ . how much thermal energy has to be transferred from the tea to the surroundings to cool the tea ? we are going to assume that the tea is mostly water , so we can use the density and heat capacity of water in our calculations . the specific heat capacity of water is $ 4.18\ , \dfrac { \text j } { \text g \cdot \text k } $ , and the density of water is $ 1.00\ , \dfrac { \text g } { \text { ml } } $ . we can calculate the energy transferred in the process of cooling the tea using the following steps : 1 . calculate the mass of the substance we can calculate the mass of the tea/water using the volume and density of water : $ \text m=250\ , \cancel { \text { ml } } \times 1.00\ , \dfrac { \text g } { \cancel { \text { ml } } } =250\ , \text g $ 2 . calculate the change in temperature , $ \delta \text t $ we can calculate the change in temperature , $ \delta \text t $ , from the initial and final temperatures : $ \begin { align } \delta \text t & amp ; =\text t_ { \text { final } } -\text t_ { \text { initial } } \ \ & amp ; =350\ , \text k-370\ , \text k\ \ & amp ; =-20\ , \text k\end { align } $ since the temperature of the tea is decreasing and $ \delta \text t $ is negative , we would expect $ \text q $ to also be negative since our system is losing thermal energy . 3 . solve for $ \text q $ now we can solve for the heat transferred from the hot tea using the equation for heat : $ \begin { align } \text q & amp ; = \text { m } \times \text c \times \delta \text t\ & amp ; =250\ , \cancel { \text g } \times4.18\ , \dfrac { \text j } { \cancel { \text g } \cdot \cancel { \text k } } \times -20\ , \cancel { \text k } \ & amp ; =-21000\ , \text j\end { align } $ thus , we calculated that the tea will transfer $ 21000\ , \text j $ of energy to the surroundings when it cools down from $ 370\ , \text k $ to $ 350\ , \text k $ . conclusions in thermodynamics , heat and temperature are closely related concepts with precise definitions . heat , $ \text q $ , is thermal energy transferred from a hotter system to a cooler system that are in contact . temperature is a measure of the average kinetic energy of the atoms or molecules in the system . the zeroth law of thermodynamics says that no heat is transferred between two objects in thermal equilibrium ; therefore , they are the same temperature . we can calculate the heat released or absorbed using the specific heat capacity $ \text c $ , the mass of the substance $ \text m $ , and the change in temperature $ \delta \text t $ in the following equation : $ \text q = \text { m } \times \text c \times \delta \text t $
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when the two systems in contact are at the same temperature , we say they are in thermal equilibrium . zeroth law of thermodynamics : defining thermal equilibrium the zeroth law of thermodynamics defines thermal equilibrium within an isolated system . the zeroth law says when two objects at thermal equilibrium are in contact , there is no net heat transfer between the objects ; therefore , they are the same temperature .
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why the zeroth law of thermodynamics is called so , is it the most basic law ?
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key points heat , $ \text q $ , is thermal energy transferred from a hotter system to a cooler system that are in contact . temperature is a measure of the average kinetic energy of the atoms or molecules in the system . the zeroth law of thermodynamics says that no heat is transferred between two objects in thermal equilibrium ; therefore , they are the same temperature . we can calculate the heat released or absorbed using the specific heat capacity $ \text c $ , the mass of the substance $ \text m $ , and the change in temperature $ \delta \text t $ in the equation : $ \text q = \text { m } \times \text c \times \delta \text t $ heat in thermodynamics what contains more heat , a cup of coffee or a glass of iced tea ? in chemistry class , that would be a trick question ( sorry ! ) . in thermodynamics , heat has a very specific meaning that is different from how we might use the word in everyday speech . scientists define heat as thermal energy transferred between two systems at different temperatures that come in contact . heat is written with the symbol q or q , and it has units of joules ( $ \text j $ ) . heat is sometimes called a process quantity , because it is defined in the context of a process by which energy can be transferred . we do n't talk about a cup of coffee containing heat , but we can talk about the heat transferred from the cup of hot coffee to your hand . heat is also an extensive property , so the change in temperature resulting from heat transferred to a system depends on how many molecules are in the system . relationship between heat and temperature heat and temperature are two different but closely related concepts . note that they have different units : temperature typically has units of degrees celsius ( $ ^\circ\text c $ ) or kelvin ( $ \text k $ ) , and heat has units of energy , joules ( $ \text j $ ) . temperature is a measure of the average kinetic energy of the atoms or molecules in the system . the water molecules in a cup of hot coffee have a higher average kinetic energy than the water molecules in a cup of iced tea , which also means they are moving at a higher velocity . temperature is also an intensive property , which means that the temperature does n't change no matter how much of a substance you have ( as long as it is all at the same temperature ! ) . this is why chemists can use the melting point to help identify a pure substance $ - $ the temperature at which it melts is a property of the substance with no dependence on the mass of a sample . on an atomic level , the molecules in each object are constantly in motion and colliding with each other . every time molecules collide , kinetic energy can be transferred . when the two systems are in contact , heat will be transferred through molecular collisions from the hotter system to the cooler system . the thermal energy will flow in that direction until the two objects are at the same temperature . when the two systems in contact are at the same temperature , we say they are in thermal equilibrium . zeroth law of thermodynamics : defining thermal equilibrium the zeroth law of thermodynamics defines thermal equilibrium within an isolated system . the zeroth law says when two objects at thermal equilibrium are in contact , there is no net heat transfer between the objects ; therefore , they are the same temperature . another way to state the zeroth law is to say that if two objects are both separately in thermal equilibrium with a third object , then they are in thermal equilibrium with each other . the zeroth law allows us to measure the temperature of objects . any time we use a thermometer , we are using the zeroth law of thermodynamics . let 's say we are measuring the temperature of a water bath . in order to make sure the reading is accurate , we usually want to wait for the temperature reading to stay constant . we are waiting for the thermometer and the water to reach thermal equilibrium ! at thermal equilibrium , the temperature of the thermometer bulb and the water bath will be the same , and there should be no net heat transfer from one object to the other ( assuming no other loss of heat to the surroundings ) . heat capacity : converting between heat and change in temperature how can we measure heat ? here are some things we know about heat so far : when a system absorbs or loses heat , the average kinetic energy of the molecules will change . thus , heat transfer results in a change in the system 's temperature as long as the system is not undergoing a phase change . the change in temperature resulting from heat transferred to or from a system depends on how many molecules are in the system . we can use a thermometer to measure the change in a system 's temperature . how can we use the change in temperature to calculate the heat transferred ? in order to figure out how the heat transferred to a system will change the temperature of the system , we need to know at least $ 2 $ things : the number of molecules in the system the heat capacity of the system the heat capacity tells us how much energy is needed to change the temperature of a given substance assuming that no phase changes are occurring . there are two main ways that heat capacity is reported . the specific heat capacity ( also called specific heat ) , represented by the symbol $ \text c $ or $ \text c $ , is how much energy is needed to increase the temperature of one gram of a substance by $ 1~^ { \circ } \text c $ or $ 1\ , \text k $ . specific heat capacity usually has units of $ \dfrac { \text j } { \text { grams } \cdot\text k } $ . the molar heat capacity , $ \text c_\text m $ or $ \text c_ { \text { mol } } $ , measures the amount of thermal energy it takes to raise the temperature of one mole of a substance by $ 1~^ { \circ } \text c $ or $ 1\ , \text k $ , and it usually has units of $ \dfrac { \text j } { \text { mol } \cdot\text k } $ . for example , the heat capacity of lead might be given as the specific heat capacity , $ 0.129\ , \dfrac { \text j } { \text { g } \cdot\text k } $ , or the molar heat capacity , $ 26.65\ , \dfrac { \text j } { \text { mol } \cdot\text k } $ . calculating $ \text q $ using the heat capacity we can use the heat capacity to determine the heat released or absorbed by a material using the following formula : $ \text q = \text { m } \times \text c \times \delta \text t $ where $ \text { m } $ is the mass of the substance ( in grams ) , $ \text { c } $ is the specific heat capacity , and $ \delta \text t $ is the change in temperature during the heat transfer . note that both mass and specific heat capacity can only have positive values , so the sign of $ \text q $ will depend on the sign of $ \delta \text t $ . we can calculate $ \delta \text t $ using the following equation : $ \delta \text t=\text t_ { \text { final } } -\text t_ { \text { initial } } $ where $ \text t_ { \text { final } } $ and $ \text t_ { \text { initial } } $ can have units of either $ ~^ { \circ } \text c $ or $ \text k $ . based on this equation , if $ \text q $ is positive ( energy of the system increases ) , then our system increases in temperature and $ \text t_ { \text { final } } & gt ; \text t_ { \text { initial } } $ . if $ \text q $ is negative ( energy of the system decreases ) , then our system 's temperature decreases and $ \text t_ { \text { final } } & lt ; \text t_ { \text { initial } } $ . example problem : cooling a cup of tea let 's say that we have $ 250\ , \text { ml } $ of hot tea which we would like to cool down before we try to drink it . the tea is currently at $ 370\ , \text k $ , and we 'd like to cool it down to $ 350\ , \text k $ . how much thermal energy has to be transferred from the tea to the surroundings to cool the tea ? we are going to assume that the tea is mostly water , so we can use the density and heat capacity of water in our calculations . the specific heat capacity of water is $ 4.18\ , \dfrac { \text j } { \text g \cdot \text k } $ , and the density of water is $ 1.00\ , \dfrac { \text g } { \text { ml } } $ . we can calculate the energy transferred in the process of cooling the tea using the following steps : 1 . calculate the mass of the substance we can calculate the mass of the tea/water using the volume and density of water : $ \text m=250\ , \cancel { \text { ml } } \times 1.00\ , \dfrac { \text g } { \cancel { \text { ml } } } =250\ , \text g $ 2 . calculate the change in temperature , $ \delta \text t $ we can calculate the change in temperature , $ \delta \text t $ , from the initial and final temperatures : $ \begin { align } \delta \text t & amp ; =\text t_ { \text { final } } -\text t_ { \text { initial } } \ \ & amp ; =350\ , \text k-370\ , \text k\ \ & amp ; =-20\ , \text k\end { align } $ since the temperature of the tea is decreasing and $ \delta \text t $ is negative , we would expect $ \text q $ to also be negative since our system is losing thermal energy . 3 . solve for $ \text q $ now we can solve for the heat transferred from the hot tea using the equation for heat : $ \begin { align } \text q & amp ; = \text { m } \times \text c \times \delta \text t\ & amp ; =250\ , \cancel { \text g } \times4.18\ , \dfrac { \text j } { \cancel { \text g } \cdot \cancel { \text k } } \times -20\ , \cancel { \text k } \ & amp ; =-21000\ , \text j\end { align } $ thus , we calculated that the tea will transfer $ 21000\ , \text j $ of energy to the surroundings when it cools down from $ 370\ , \text k $ to $ 350\ , \text k $ . conclusions in thermodynamics , heat and temperature are closely related concepts with precise definitions . heat , $ \text q $ , is thermal energy transferred from a hotter system to a cooler system that are in contact . temperature is a measure of the average kinetic energy of the atoms or molecules in the system . the zeroth law of thermodynamics says that no heat is transferred between two objects in thermal equilibrium ; therefore , they are the same temperature . we can calculate the heat released or absorbed using the specific heat capacity $ \text c $ , the mass of the substance $ \text m $ , and the change in temperature $ \delta \text t $ in the following equation : $ \text q = \text { m } \times \text c \times \delta \text t $
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at thermal equilibrium , the temperature of the thermometer bulb and the water bath will be the same , and there should be no net heat transfer from one object to the other ( assuming no other loss of heat to the surroundings ) . heat capacity : converting between heat and change in temperature how can we measure heat ? here are some things we know about heat so far : when a system absorbs or loses heat , the average kinetic energy of the molecules will change .
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should n't it read , `` as ice melts , heat is transferred from the surroundings to the ice '' or something similar ?
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key points heat , $ \text q $ , is thermal energy transferred from a hotter system to a cooler system that are in contact . temperature is a measure of the average kinetic energy of the atoms or molecules in the system . the zeroth law of thermodynamics says that no heat is transferred between two objects in thermal equilibrium ; therefore , they are the same temperature . we can calculate the heat released or absorbed using the specific heat capacity $ \text c $ , the mass of the substance $ \text m $ , and the change in temperature $ \delta \text t $ in the equation : $ \text q = \text { m } \times \text c \times \delta \text t $ heat in thermodynamics what contains more heat , a cup of coffee or a glass of iced tea ? in chemistry class , that would be a trick question ( sorry ! ) . in thermodynamics , heat has a very specific meaning that is different from how we might use the word in everyday speech . scientists define heat as thermal energy transferred between two systems at different temperatures that come in contact . heat is written with the symbol q or q , and it has units of joules ( $ \text j $ ) . heat is sometimes called a process quantity , because it is defined in the context of a process by which energy can be transferred . we do n't talk about a cup of coffee containing heat , but we can talk about the heat transferred from the cup of hot coffee to your hand . heat is also an extensive property , so the change in temperature resulting from heat transferred to a system depends on how many molecules are in the system . relationship between heat and temperature heat and temperature are two different but closely related concepts . note that they have different units : temperature typically has units of degrees celsius ( $ ^\circ\text c $ ) or kelvin ( $ \text k $ ) , and heat has units of energy , joules ( $ \text j $ ) . temperature is a measure of the average kinetic energy of the atoms or molecules in the system . the water molecules in a cup of hot coffee have a higher average kinetic energy than the water molecules in a cup of iced tea , which also means they are moving at a higher velocity . temperature is also an intensive property , which means that the temperature does n't change no matter how much of a substance you have ( as long as it is all at the same temperature ! ) . this is why chemists can use the melting point to help identify a pure substance $ - $ the temperature at which it melts is a property of the substance with no dependence on the mass of a sample . on an atomic level , the molecules in each object are constantly in motion and colliding with each other . every time molecules collide , kinetic energy can be transferred . when the two systems are in contact , heat will be transferred through molecular collisions from the hotter system to the cooler system . the thermal energy will flow in that direction until the two objects are at the same temperature . when the two systems in contact are at the same temperature , we say they are in thermal equilibrium . zeroth law of thermodynamics : defining thermal equilibrium the zeroth law of thermodynamics defines thermal equilibrium within an isolated system . the zeroth law says when two objects at thermal equilibrium are in contact , there is no net heat transfer between the objects ; therefore , they are the same temperature . another way to state the zeroth law is to say that if two objects are both separately in thermal equilibrium with a third object , then they are in thermal equilibrium with each other . the zeroth law allows us to measure the temperature of objects . any time we use a thermometer , we are using the zeroth law of thermodynamics . let 's say we are measuring the temperature of a water bath . in order to make sure the reading is accurate , we usually want to wait for the temperature reading to stay constant . we are waiting for the thermometer and the water to reach thermal equilibrium ! at thermal equilibrium , the temperature of the thermometer bulb and the water bath will be the same , and there should be no net heat transfer from one object to the other ( assuming no other loss of heat to the surroundings ) . heat capacity : converting between heat and change in temperature how can we measure heat ? here are some things we know about heat so far : when a system absorbs or loses heat , the average kinetic energy of the molecules will change . thus , heat transfer results in a change in the system 's temperature as long as the system is not undergoing a phase change . the change in temperature resulting from heat transferred to or from a system depends on how many molecules are in the system . we can use a thermometer to measure the change in a system 's temperature . how can we use the change in temperature to calculate the heat transferred ? in order to figure out how the heat transferred to a system will change the temperature of the system , we need to know at least $ 2 $ things : the number of molecules in the system the heat capacity of the system the heat capacity tells us how much energy is needed to change the temperature of a given substance assuming that no phase changes are occurring . there are two main ways that heat capacity is reported . the specific heat capacity ( also called specific heat ) , represented by the symbol $ \text c $ or $ \text c $ , is how much energy is needed to increase the temperature of one gram of a substance by $ 1~^ { \circ } \text c $ or $ 1\ , \text k $ . specific heat capacity usually has units of $ \dfrac { \text j } { \text { grams } \cdot\text k } $ . the molar heat capacity , $ \text c_\text m $ or $ \text c_ { \text { mol } } $ , measures the amount of thermal energy it takes to raise the temperature of one mole of a substance by $ 1~^ { \circ } \text c $ or $ 1\ , \text k $ , and it usually has units of $ \dfrac { \text j } { \text { mol } \cdot\text k } $ . for example , the heat capacity of lead might be given as the specific heat capacity , $ 0.129\ , \dfrac { \text j } { \text { g } \cdot\text k } $ , or the molar heat capacity , $ 26.65\ , \dfrac { \text j } { \text { mol } \cdot\text k } $ . calculating $ \text q $ using the heat capacity we can use the heat capacity to determine the heat released or absorbed by a material using the following formula : $ \text q = \text { m } \times \text c \times \delta \text t $ where $ \text { m } $ is the mass of the substance ( in grams ) , $ \text { c } $ is the specific heat capacity , and $ \delta \text t $ is the change in temperature during the heat transfer . note that both mass and specific heat capacity can only have positive values , so the sign of $ \text q $ will depend on the sign of $ \delta \text t $ . we can calculate $ \delta \text t $ using the following equation : $ \delta \text t=\text t_ { \text { final } } -\text t_ { \text { initial } } $ where $ \text t_ { \text { final } } $ and $ \text t_ { \text { initial } } $ can have units of either $ ~^ { \circ } \text c $ or $ \text k $ . based on this equation , if $ \text q $ is positive ( energy of the system increases ) , then our system increases in temperature and $ \text t_ { \text { final } } & gt ; \text t_ { \text { initial } } $ . if $ \text q $ is negative ( energy of the system decreases ) , then our system 's temperature decreases and $ \text t_ { \text { final } } & lt ; \text t_ { \text { initial } } $ . example problem : cooling a cup of tea let 's say that we have $ 250\ , \text { ml } $ of hot tea which we would like to cool down before we try to drink it . the tea is currently at $ 370\ , \text k $ , and we 'd like to cool it down to $ 350\ , \text k $ . how much thermal energy has to be transferred from the tea to the surroundings to cool the tea ? we are going to assume that the tea is mostly water , so we can use the density and heat capacity of water in our calculations . the specific heat capacity of water is $ 4.18\ , \dfrac { \text j } { \text g \cdot \text k } $ , and the density of water is $ 1.00\ , \dfrac { \text g } { \text { ml } } $ . we can calculate the energy transferred in the process of cooling the tea using the following steps : 1 . calculate the mass of the substance we can calculate the mass of the tea/water using the volume and density of water : $ \text m=250\ , \cancel { \text { ml } } \times 1.00\ , \dfrac { \text g } { \cancel { \text { ml } } } =250\ , \text g $ 2 . calculate the change in temperature , $ \delta \text t $ we can calculate the change in temperature , $ \delta \text t $ , from the initial and final temperatures : $ \begin { align } \delta \text t & amp ; =\text t_ { \text { final } } -\text t_ { \text { initial } } \ \ & amp ; =350\ , \text k-370\ , \text k\ \ & amp ; =-20\ , \text k\end { align } $ since the temperature of the tea is decreasing and $ \delta \text t $ is negative , we would expect $ \text q $ to also be negative since our system is losing thermal energy . 3 . solve for $ \text q $ now we can solve for the heat transferred from the hot tea using the equation for heat : $ \begin { align } \text q & amp ; = \text { m } \times \text c \times \delta \text t\ & amp ; =250\ , \cancel { \text g } \times4.18\ , \dfrac { \text j } { \cancel { \text g } \cdot \cancel { \text k } } \times -20\ , \cancel { \text k } \ & amp ; =-21000\ , \text j\end { align } $ thus , we calculated that the tea will transfer $ 21000\ , \text j $ of energy to the surroundings when it cools down from $ 370\ , \text k $ to $ 350\ , \text k $ . conclusions in thermodynamics , heat and temperature are closely related concepts with precise definitions . heat , $ \text q $ , is thermal energy transferred from a hotter system to a cooler system that are in contact . temperature is a measure of the average kinetic energy of the atoms or molecules in the system . the zeroth law of thermodynamics says that no heat is transferred between two objects in thermal equilibrium ; therefore , they are the same temperature . we can calculate the heat released or absorbed using the specific heat capacity $ \text c $ , the mass of the substance $ \text m $ , and the change in temperature $ \delta \text t $ in the following equation : $ \text q = \text { m } \times \text c \times \delta \text t $
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at thermal equilibrium , the temperature of the thermometer bulb and the water bath will be the same , and there should be no net heat transfer from one object to the other ( assuming no other loss of heat to the surroundings ) . heat capacity : converting between heat and change in temperature how can we measure heat ? here are some things we know about heat so far : when a system absorbs or loses heat , the average kinetic energy of the molecules will change .
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why is it often not possible to directly measure the heat energy change of the reactants and products ?
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key points heat , $ \text q $ , is thermal energy transferred from a hotter system to a cooler system that are in contact . temperature is a measure of the average kinetic energy of the atoms or molecules in the system . the zeroth law of thermodynamics says that no heat is transferred between two objects in thermal equilibrium ; therefore , they are the same temperature . we can calculate the heat released or absorbed using the specific heat capacity $ \text c $ , the mass of the substance $ \text m $ , and the change in temperature $ \delta \text t $ in the equation : $ \text q = \text { m } \times \text c \times \delta \text t $ heat in thermodynamics what contains more heat , a cup of coffee or a glass of iced tea ? in chemistry class , that would be a trick question ( sorry ! ) . in thermodynamics , heat has a very specific meaning that is different from how we might use the word in everyday speech . scientists define heat as thermal energy transferred between two systems at different temperatures that come in contact . heat is written with the symbol q or q , and it has units of joules ( $ \text j $ ) . heat is sometimes called a process quantity , because it is defined in the context of a process by which energy can be transferred . we do n't talk about a cup of coffee containing heat , but we can talk about the heat transferred from the cup of hot coffee to your hand . heat is also an extensive property , so the change in temperature resulting from heat transferred to a system depends on how many molecules are in the system . relationship between heat and temperature heat and temperature are two different but closely related concepts . note that they have different units : temperature typically has units of degrees celsius ( $ ^\circ\text c $ ) or kelvin ( $ \text k $ ) , and heat has units of energy , joules ( $ \text j $ ) . temperature is a measure of the average kinetic energy of the atoms or molecules in the system . the water molecules in a cup of hot coffee have a higher average kinetic energy than the water molecules in a cup of iced tea , which also means they are moving at a higher velocity . temperature is also an intensive property , which means that the temperature does n't change no matter how much of a substance you have ( as long as it is all at the same temperature ! ) . this is why chemists can use the melting point to help identify a pure substance $ - $ the temperature at which it melts is a property of the substance with no dependence on the mass of a sample . on an atomic level , the molecules in each object are constantly in motion and colliding with each other . every time molecules collide , kinetic energy can be transferred . when the two systems are in contact , heat will be transferred through molecular collisions from the hotter system to the cooler system . the thermal energy will flow in that direction until the two objects are at the same temperature . when the two systems in contact are at the same temperature , we say they are in thermal equilibrium . zeroth law of thermodynamics : defining thermal equilibrium the zeroth law of thermodynamics defines thermal equilibrium within an isolated system . the zeroth law says when two objects at thermal equilibrium are in contact , there is no net heat transfer between the objects ; therefore , they are the same temperature . another way to state the zeroth law is to say that if two objects are both separately in thermal equilibrium with a third object , then they are in thermal equilibrium with each other . the zeroth law allows us to measure the temperature of objects . any time we use a thermometer , we are using the zeroth law of thermodynamics . let 's say we are measuring the temperature of a water bath . in order to make sure the reading is accurate , we usually want to wait for the temperature reading to stay constant . we are waiting for the thermometer and the water to reach thermal equilibrium ! at thermal equilibrium , the temperature of the thermometer bulb and the water bath will be the same , and there should be no net heat transfer from one object to the other ( assuming no other loss of heat to the surroundings ) . heat capacity : converting between heat and change in temperature how can we measure heat ? here are some things we know about heat so far : when a system absorbs or loses heat , the average kinetic energy of the molecules will change . thus , heat transfer results in a change in the system 's temperature as long as the system is not undergoing a phase change . the change in temperature resulting from heat transferred to or from a system depends on how many molecules are in the system . we can use a thermometer to measure the change in a system 's temperature . how can we use the change in temperature to calculate the heat transferred ? in order to figure out how the heat transferred to a system will change the temperature of the system , we need to know at least $ 2 $ things : the number of molecules in the system the heat capacity of the system the heat capacity tells us how much energy is needed to change the temperature of a given substance assuming that no phase changes are occurring . there are two main ways that heat capacity is reported . the specific heat capacity ( also called specific heat ) , represented by the symbol $ \text c $ or $ \text c $ , is how much energy is needed to increase the temperature of one gram of a substance by $ 1~^ { \circ } \text c $ or $ 1\ , \text k $ . specific heat capacity usually has units of $ \dfrac { \text j } { \text { grams } \cdot\text k } $ . the molar heat capacity , $ \text c_\text m $ or $ \text c_ { \text { mol } } $ , measures the amount of thermal energy it takes to raise the temperature of one mole of a substance by $ 1~^ { \circ } \text c $ or $ 1\ , \text k $ , and it usually has units of $ \dfrac { \text j } { \text { mol } \cdot\text k } $ . for example , the heat capacity of lead might be given as the specific heat capacity , $ 0.129\ , \dfrac { \text j } { \text { g } \cdot\text k } $ , or the molar heat capacity , $ 26.65\ , \dfrac { \text j } { \text { mol } \cdot\text k } $ . calculating $ \text q $ using the heat capacity we can use the heat capacity to determine the heat released or absorbed by a material using the following formula : $ \text q = \text { m } \times \text c \times \delta \text t $ where $ \text { m } $ is the mass of the substance ( in grams ) , $ \text { c } $ is the specific heat capacity , and $ \delta \text t $ is the change in temperature during the heat transfer . note that both mass and specific heat capacity can only have positive values , so the sign of $ \text q $ will depend on the sign of $ \delta \text t $ . we can calculate $ \delta \text t $ using the following equation : $ \delta \text t=\text t_ { \text { final } } -\text t_ { \text { initial } } $ where $ \text t_ { \text { final } } $ and $ \text t_ { \text { initial } } $ can have units of either $ ~^ { \circ } \text c $ or $ \text k $ . based on this equation , if $ \text q $ is positive ( energy of the system increases ) , then our system increases in temperature and $ \text t_ { \text { final } } & gt ; \text t_ { \text { initial } } $ . if $ \text q $ is negative ( energy of the system decreases ) , then our system 's temperature decreases and $ \text t_ { \text { final } } & lt ; \text t_ { \text { initial } } $ . example problem : cooling a cup of tea let 's say that we have $ 250\ , \text { ml } $ of hot tea which we would like to cool down before we try to drink it . the tea is currently at $ 370\ , \text k $ , and we 'd like to cool it down to $ 350\ , \text k $ . how much thermal energy has to be transferred from the tea to the surroundings to cool the tea ? we are going to assume that the tea is mostly water , so we can use the density and heat capacity of water in our calculations . the specific heat capacity of water is $ 4.18\ , \dfrac { \text j } { \text g \cdot \text k } $ , and the density of water is $ 1.00\ , \dfrac { \text g } { \text { ml } } $ . we can calculate the energy transferred in the process of cooling the tea using the following steps : 1 . calculate the mass of the substance we can calculate the mass of the tea/water using the volume and density of water : $ \text m=250\ , \cancel { \text { ml } } \times 1.00\ , \dfrac { \text g } { \cancel { \text { ml } } } =250\ , \text g $ 2 . calculate the change in temperature , $ \delta \text t $ we can calculate the change in temperature , $ \delta \text t $ , from the initial and final temperatures : $ \begin { align } \delta \text t & amp ; =\text t_ { \text { final } } -\text t_ { \text { initial } } \ \ & amp ; =350\ , \text k-370\ , \text k\ \ & amp ; =-20\ , \text k\end { align } $ since the temperature of the tea is decreasing and $ \delta \text t $ is negative , we would expect $ \text q $ to also be negative since our system is losing thermal energy . 3 . solve for $ \text q $ now we can solve for the heat transferred from the hot tea using the equation for heat : $ \begin { align } \text q & amp ; = \text { m } \times \text c \times \delta \text t\ & amp ; =250\ , \cancel { \text g } \times4.18\ , \dfrac { \text j } { \cancel { \text g } \cdot \cancel { \text k } } \times -20\ , \cancel { \text k } \ & amp ; =-21000\ , \text j\end { align } $ thus , we calculated that the tea will transfer $ 21000\ , \text j $ of energy to the surroundings when it cools down from $ 370\ , \text k $ to $ 350\ , \text k $ . conclusions in thermodynamics , heat and temperature are closely related concepts with precise definitions . heat , $ \text q $ , is thermal energy transferred from a hotter system to a cooler system that are in contact . temperature is a measure of the average kinetic energy of the atoms or molecules in the system . the zeroth law of thermodynamics says that no heat is transferred between two objects in thermal equilibrium ; therefore , they are the same temperature . we can calculate the heat released or absorbed using the specific heat capacity $ \text c $ , the mass of the substance $ \text m $ , and the change in temperature $ \delta \text t $ in the following equation : $ \text q = \text { m } \times \text c \times \delta \text t $
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at thermal equilibrium , the temperature of the thermometer bulb and the water bath will be the same , and there should be no net heat transfer from one object to the other ( assuming no other loss of heat to the surroundings ) . heat capacity : converting between heat and change in temperature how can we measure heat ? here are some things we know about heat so far : when a system absorbs or loses heat , the average kinetic energy of the molecules will change .
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difference between work and heat ?
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key points heat , $ \text q $ , is thermal energy transferred from a hotter system to a cooler system that are in contact . temperature is a measure of the average kinetic energy of the atoms or molecules in the system . the zeroth law of thermodynamics says that no heat is transferred between two objects in thermal equilibrium ; therefore , they are the same temperature . we can calculate the heat released or absorbed using the specific heat capacity $ \text c $ , the mass of the substance $ \text m $ , and the change in temperature $ \delta \text t $ in the equation : $ \text q = \text { m } \times \text c \times \delta \text t $ heat in thermodynamics what contains more heat , a cup of coffee or a glass of iced tea ? in chemistry class , that would be a trick question ( sorry ! ) . in thermodynamics , heat has a very specific meaning that is different from how we might use the word in everyday speech . scientists define heat as thermal energy transferred between two systems at different temperatures that come in contact . heat is written with the symbol q or q , and it has units of joules ( $ \text j $ ) . heat is sometimes called a process quantity , because it is defined in the context of a process by which energy can be transferred . we do n't talk about a cup of coffee containing heat , but we can talk about the heat transferred from the cup of hot coffee to your hand . heat is also an extensive property , so the change in temperature resulting from heat transferred to a system depends on how many molecules are in the system . relationship between heat and temperature heat and temperature are two different but closely related concepts . note that they have different units : temperature typically has units of degrees celsius ( $ ^\circ\text c $ ) or kelvin ( $ \text k $ ) , and heat has units of energy , joules ( $ \text j $ ) . temperature is a measure of the average kinetic energy of the atoms or molecules in the system . the water molecules in a cup of hot coffee have a higher average kinetic energy than the water molecules in a cup of iced tea , which also means they are moving at a higher velocity . temperature is also an intensive property , which means that the temperature does n't change no matter how much of a substance you have ( as long as it is all at the same temperature ! ) . this is why chemists can use the melting point to help identify a pure substance $ - $ the temperature at which it melts is a property of the substance with no dependence on the mass of a sample . on an atomic level , the molecules in each object are constantly in motion and colliding with each other . every time molecules collide , kinetic energy can be transferred . when the two systems are in contact , heat will be transferred through molecular collisions from the hotter system to the cooler system . the thermal energy will flow in that direction until the two objects are at the same temperature . when the two systems in contact are at the same temperature , we say they are in thermal equilibrium . zeroth law of thermodynamics : defining thermal equilibrium the zeroth law of thermodynamics defines thermal equilibrium within an isolated system . the zeroth law says when two objects at thermal equilibrium are in contact , there is no net heat transfer between the objects ; therefore , they are the same temperature . another way to state the zeroth law is to say that if two objects are both separately in thermal equilibrium with a third object , then they are in thermal equilibrium with each other . the zeroth law allows us to measure the temperature of objects . any time we use a thermometer , we are using the zeroth law of thermodynamics . let 's say we are measuring the temperature of a water bath . in order to make sure the reading is accurate , we usually want to wait for the temperature reading to stay constant . we are waiting for the thermometer and the water to reach thermal equilibrium ! at thermal equilibrium , the temperature of the thermometer bulb and the water bath will be the same , and there should be no net heat transfer from one object to the other ( assuming no other loss of heat to the surroundings ) . heat capacity : converting between heat and change in temperature how can we measure heat ? here are some things we know about heat so far : when a system absorbs or loses heat , the average kinetic energy of the molecules will change . thus , heat transfer results in a change in the system 's temperature as long as the system is not undergoing a phase change . the change in temperature resulting from heat transferred to or from a system depends on how many molecules are in the system . we can use a thermometer to measure the change in a system 's temperature . how can we use the change in temperature to calculate the heat transferred ? in order to figure out how the heat transferred to a system will change the temperature of the system , we need to know at least $ 2 $ things : the number of molecules in the system the heat capacity of the system the heat capacity tells us how much energy is needed to change the temperature of a given substance assuming that no phase changes are occurring . there are two main ways that heat capacity is reported . the specific heat capacity ( also called specific heat ) , represented by the symbol $ \text c $ or $ \text c $ , is how much energy is needed to increase the temperature of one gram of a substance by $ 1~^ { \circ } \text c $ or $ 1\ , \text k $ . specific heat capacity usually has units of $ \dfrac { \text j } { \text { grams } \cdot\text k } $ . the molar heat capacity , $ \text c_\text m $ or $ \text c_ { \text { mol } } $ , measures the amount of thermal energy it takes to raise the temperature of one mole of a substance by $ 1~^ { \circ } \text c $ or $ 1\ , \text k $ , and it usually has units of $ \dfrac { \text j } { \text { mol } \cdot\text k } $ . for example , the heat capacity of lead might be given as the specific heat capacity , $ 0.129\ , \dfrac { \text j } { \text { g } \cdot\text k } $ , or the molar heat capacity , $ 26.65\ , \dfrac { \text j } { \text { mol } \cdot\text k } $ . calculating $ \text q $ using the heat capacity we can use the heat capacity to determine the heat released or absorbed by a material using the following formula : $ \text q = \text { m } \times \text c \times \delta \text t $ where $ \text { m } $ is the mass of the substance ( in grams ) , $ \text { c } $ is the specific heat capacity , and $ \delta \text t $ is the change in temperature during the heat transfer . note that both mass and specific heat capacity can only have positive values , so the sign of $ \text q $ will depend on the sign of $ \delta \text t $ . we can calculate $ \delta \text t $ using the following equation : $ \delta \text t=\text t_ { \text { final } } -\text t_ { \text { initial } } $ where $ \text t_ { \text { final } } $ and $ \text t_ { \text { initial } } $ can have units of either $ ~^ { \circ } \text c $ or $ \text k $ . based on this equation , if $ \text q $ is positive ( energy of the system increases ) , then our system increases in temperature and $ \text t_ { \text { final } } & gt ; \text t_ { \text { initial } } $ . if $ \text q $ is negative ( energy of the system decreases ) , then our system 's temperature decreases and $ \text t_ { \text { final } } & lt ; \text t_ { \text { initial } } $ . example problem : cooling a cup of tea let 's say that we have $ 250\ , \text { ml } $ of hot tea which we would like to cool down before we try to drink it . the tea is currently at $ 370\ , \text k $ , and we 'd like to cool it down to $ 350\ , \text k $ . how much thermal energy has to be transferred from the tea to the surroundings to cool the tea ? we are going to assume that the tea is mostly water , so we can use the density and heat capacity of water in our calculations . the specific heat capacity of water is $ 4.18\ , \dfrac { \text j } { \text g \cdot \text k } $ , and the density of water is $ 1.00\ , \dfrac { \text g } { \text { ml } } $ . we can calculate the energy transferred in the process of cooling the tea using the following steps : 1 . calculate the mass of the substance we can calculate the mass of the tea/water using the volume and density of water : $ \text m=250\ , \cancel { \text { ml } } \times 1.00\ , \dfrac { \text g } { \cancel { \text { ml } } } =250\ , \text g $ 2 . calculate the change in temperature , $ \delta \text t $ we can calculate the change in temperature , $ \delta \text t $ , from the initial and final temperatures : $ \begin { align } \delta \text t & amp ; =\text t_ { \text { final } } -\text t_ { \text { initial } } \ \ & amp ; =350\ , \text k-370\ , \text k\ \ & amp ; =-20\ , \text k\end { align } $ since the temperature of the tea is decreasing and $ \delta \text t $ is negative , we would expect $ \text q $ to also be negative since our system is losing thermal energy . 3 . solve for $ \text q $ now we can solve for the heat transferred from the hot tea using the equation for heat : $ \begin { align } \text q & amp ; = \text { m } \times \text c \times \delta \text t\ & amp ; =250\ , \cancel { \text g } \times4.18\ , \dfrac { \text j } { \cancel { \text g } \cdot \cancel { \text k } } \times -20\ , \cancel { \text k } \ & amp ; =-21000\ , \text j\end { align } $ thus , we calculated that the tea will transfer $ 21000\ , \text j $ of energy to the surroundings when it cools down from $ 370\ , \text k $ to $ 350\ , \text k $ . conclusions in thermodynamics , heat and temperature are closely related concepts with precise definitions . heat , $ \text q $ , is thermal energy transferred from a hotter system to a cooler system that are in contact . temperature is a measure of the average kinetic energy of the atoms or molecules in the system . the zeroth law of thermodynamics says that no heat is transferred between two objects in thermal equilibrium ; therefore , they are the same temperature . we can calculate the heat released or absorbed using the specific heat capacity $ \text c $ , the mass of the substance $ \text m $ , and the change in temperature $ \delta \text t $ in the following equation : $ \text q = \text { m } \times \text c \times \delta \text t $
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in thermodynamics , heat has a very specific meaning that is different from how we might use the word in everyday speech . scientists define heat as thermal energy transferred between two systems at different temperatures that come in contact . heat is written with the symbol q or q , and it has units of joules ( $ \text j $ ) .
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is thermal energy and heat are the same thing ?
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key points heat , $ \text q $ , is thermal energy transferred from a hotter system to a cooler system that are in contact . temperature is a measure of the average kinetic energy of the atoms or molecules in the system . the zeroth law of thermodynamics says that no heat is transferred between two objects in thermal equilibrium ; therefore , they are the same temperature . we can calculate the heat released or absorbed using the specific heat capacity $ \text c $ , the mass of the substance $ \text m $ , and the change in temperature $ \delta \text t $ in the equation : $ \text q = \text { m } \times \text c \times \delta \text t $ heat in thermodynamics what contains more heat , a cup of coffee or a glass of iced tea ? in chemistry class , that would be a trick question ( sorry ! ) . in thermodynamics , heat has a very specific meaning that is different from how we might use the word in everyday speech . scientists define heat as thermal energy transferred between two systems at different temperatures that come in contact . heat is written with the symbol q or q , and it has units of joules ( $ \text j $ ) . heat is sometimes called a process quantity , because it is defined in the context of a process by which energy can be transferred . we do n't talk about a cup of coffee containing heat , but we can talk about the heat transferred from the cup of hot coffee to your hand . heat is also an extensive property , so the change in temperature resulting from heat transferred to a system depends on how many molecules are in the system . relationship between heat and temperature heat and temperature are two different but closely related concepts . note that they have different units : temperature typically has units of degrees celsius ( $ ^\circ\text c $ ) or kelvin ( $ \text k $ ) , and heat has units of energy , joules ( $ \text j $ ) . temperature is a measure of the average kinetic energy of the atoms or molecules in the system . the water molecules in a cup of hot coffee have a higher average kinetic energy than the water molecules in a cup of iced tea , which also means they are moving at a higher velocity . temperature is also an intensive property , which means that the temperature does n't change no matter how much of a substance you have ( as long as it is all at the same temperature ! ) . this is why chemists can use the melting point to help identify a pure substance $ - $ the temperature at which it melts is a property of the substance with no dependence on the mass of a sample . on an atomic level , the molecules in each object are constantly in motion and colliding with each other . every time molecules collide , kinetic energy can be transferred . when the two systems are in contact , heat will be transferred through molecular collisions from the hotter system to the cooler system . the thermal energy will flow in that direction until the two objects are at the same temperature . when the two systems in contact are at the same temperature , we say they are in thermal equilibrium . zeroth law of thermodynamics : defining thermal equilibrium the zeroth law of thermodynamics defines thermal equilibrium within an isolated system . the zeroth law says when two objects at thermal equilibrium are in contact , there is no net heat transfer between the objects ; therefore , they are the same temperature . another way to state the zeroth law is to say that if two objects are both separately in thermal equilibrium with a third object , then they are in thermal equilibrium with each other . the zeroth law allows us to measure the temperature of objects . any time we use a thermometer , we are using the zeroth law of thermodynamics . let 's say we are measuring the temperature of a water bath . in order to make sure the reading is accurate , we usually want to wait for the temperature reading to stay constant . we are waiting for the thermometer and the water to reach thermal equilibrium ! at thermal equilibrium , the temperature of the thermometer bulb and the water bath will be the same , and there should be no net heat transfer from one object to the other ( assuming no other loss of heat to the surroundings ) . heat capacity : converting between heat and change in temperature how can we measure heat ? here are some things we know about heat so far : when a system absorbs or loses heat , the average kinetic energy of the molecules will change . thus , heat transfer results in a change in the system 's temperature as long as the system is not undergoing a phase change . the change in temperature resulting from heat transferred to or from a system depends on how many molecules are in the system . we can use a thermometer to measure the change in a system 's temperature . how can we use the change in temperature to calculate the heat transferred ? in order to figure out how the heat transferred to a system will change the temperature of the system , we need to know at least $ 2 $ things : the number of molecules in the system the heat capacity of the system the heat capacity tells us how much energy is needed to change the temperature of a given substance assuming that no phase changes are occurring . there are two main ways that heat capacity is reported . the specific heat capacity ( also called specific heat ) , represented by the symbol $ \text c $ or $ \text c $ , is how much energy is needed to increase the temperature of one gram of a substance by $ 1~^ { \circ } \text c $ or $ 1\ , \text k $ . specific heat capacity usually has units of $ \dfrac { \text j } { \text { grams } \cdot\text k } $ . the molar heat capacity , $ \text c_\text m $ or $ \text c_ { \text { mol } } $ , measures the amount of thermal energy it takes to raise the temperature of one mole of a substance by $ 1~^ { \circ } \text c $ or $ 1\ , \text k $ , and it usually has units of $ \dfrac { \text j } { \text { mol } \cdot\text k } $ . for example , the heat capacity of lead might be given as the specific heat capacity , $ 0.129\ , \dfrac { \text j } { \text { g } \cdot\text k } $ , or the molar heat capacity , $ 26.65\ , \dfrac { \text j } { \text { mol } \cdot\text k } $ . calculating $ \text q $ using the heat capacity we can use the heat capacity to determine the heat released or absorbed by a material using the following formula : $ \text q = \text { m } \times \text c \times \delta \text t $ where $ \text { m } $ is the mass of the substance ( in grams ) , $ \text { c } $ is the specific heat capacity , and $ \delta \text t $ is the change in temperature during the heat transfer . note that both mass and specific heat capacity can only have positive values , so the sign of $ \text q $ will depend on the sign of $ \delta \text t $ . we can calculate $ \delta \text t $ using the following equation : $ \delta \text t=\text t_ { \text { final } } -\text t_ { \text { initial } } $ where $ \text t_ { \text { final } } $ and $ \text t_ { \text { initial } } $ can have units of either $ ~^ { \circ } \text c $ or $ \text k $ . based on this equation , if $ \text q $ is positive ( energy of the system increases ) , then our system increases in temperature and $ \text t_ { \text { final } } & gt ; \text t_ { \text { initial } } $ . if $ \text q $ is negative ( energy of the system decreases ) , then our system 's temperature decreases and $ \text t_ { \text { final } } & lt ; \text t_ { \text { initial } } $ . example problem : cooling a cup of tea let 's say that we have $ 250\ , \text { ml } $ of hot tea which we would like to cool down before we try to drink it . the tea is currently at $ 370\ , \text k $ , and we 'd like to cool it down to $ 350\ , \text k $ . how much thermal energy has to be transferred from the tea to the surroundings to cool the tea ? we are going to assume that the tea is mostly water , so we can use the density and heat capacity of water in our calculations . the specific heat capacity of water is $ 4.18\ , \dfrac { \text j } { \text g \cdot \text k } $ , and the density of water is $ 1.00\ , \dfrac { \text g } { \text { ml } } $ . we can calculate the energy transferred in the process of cooling the tea using the following steps : 1 . calculate the mass of the substance we can calculate the mass of the tea/water using the volume and density of water : $ \text m=250\ , \cancel { \text { ml } } \times 1.00\ , \dfrac { \text g } { \cancel { \text { ml } } } =250\ , \text g $ 2 . calculate the change in temperature , $ \delta \text t $ we can calculate the change in temperature , $ \delta \text t $ , from the initial and final temperatures : $ \begin { align } \delta \text t & amp ; =\text t_ { \text { final } } -\text t_ { \text { initial } } \ \ & amp ; =350\ , \text k-370\ , \text k\ \ & amp ; =-20\ , \text k\end { align } $ since the temperature of the tea is decreasing and $ \delta \text t $ is negative , we would expect $ \text q $ to also be negative since our system is losing thermal energy . 3 . solve for $ \text q $ now we can solve for the heat transferred from the hot tea using the equation for heat : $ \begin { align } \text q & amp ; = \text { m } \times \text c \times \delta \text t\ & amp ; =250\ , \cancel { \text g } \times4.18\ , \dfrac { \text j } { \cancel { \text g } \cdot \cancel { \text k } } \times -20\ , \cancel { \text k } \ & amp ; =-21000\ , \text j\end { align } $ thus , we calculated that the tea will transfer $ 21000\ , \text j $ of energy to the surroundings when it cools down from $ 370\ , \text k $ to $ 350\ , \text k $ . conclusions in thermodynamics , heat and temperature are closely related concepts with precise definitions . heat , $ \text q $ , is thermal energy transferred from a hotter system to a cooler system that are in contact . temperature is a measure of the average kinetic energy of the atoms or molecules in the system . the zeroth law of thermodynamics says that no heat is transferred between two objects in thermal equilibrium ; therefore , they are the same temperature . we can calculate the heat released or absorbed using the specific heat capacity $ \text c $ , the mass of the substance $ \text m $ , and the change in temperature $ \delta \text t $ in the following equation : $ \text q = \text { m } \times \text c \times \delta \text t $
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on an atomic level , the molecules in each object are constantly in motion and colliding with each other . every time molecules collide , kinetic energy can be transferred . when the two systems are in contact , heat will be transferred through molecular collisions from the hotter system to the cooler system .
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what does it mean , `` average kinetic energy per molecule '' ?
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key points heat , $ \text q $ , is thermal energy transferred from a hotter system to a cooler system that are in contact . temperature is a measure of the average kinetic energy of the atoms or molecules in the system . the zeroth law of thermodynamics says that no heat is transferred between two objects in thermal equilibrium ; therefore , they are the same temperature . we can calculate the heat released or absorbed using the specific heat capacity $ \text c $ , the mass of the substance $ \text m $ , and the change in temperature $ \delta \text t $ in the equation : $ \text q = \text { m } \times \text c \times \delta \text t $ heat in thermodynamics what contains more heat , a cup of coffee or a glass of iced tea ? in chemistry class , that would be a trick question ( sorry ! ) . in thermodynamics , heat has a very specific meaning that is different from how we might use the word in everyday speech . scientists define heat as thermal energy transferred between two systems at different temperatures that come in contact . heat is written with the symbol q or q , and it has units of joules ( $ \text j $ ) . heat is sometimes called a process quantity , because it is defined in the context of a process by which energy can be transferred . we do n't talk about a cup of coffee containing heat , but we can talk about the heat transferred from the cup of hot coffee to your hand . heat is also an extensive property , so the change in temperature resulting from heat transferred to a system depends on how many molecules are in the system . relationship between heat and temperature heat and temperature are two different but closely related concepts . note that they have different units : temperature typically has units of degrees celsius ( $ ^\circ\text c $ ) or kelvin ( $ \text k $ ) , and heat has units of energy , joules ( $ \text j $ ) . temperature is a measure of the average kinetic energy of the atoms or molecules in the system . the water molecules in a cup of hot coffee have a higher average kinetic energy than the water molecules in a cup of iced tea , which also means they are moving at a higher velocity . temperature is also an intensive property , which means that the temperature does n't change no matter how much of a substance you have ( as long as it is all at the same temperature ! ) . this is why chemists can use the melting point to help identify a pure substance $ - $ the temperature at which it melts is a property of the substance with no dependence on the mass of a sample . on an atomic level , the molecules in each object are constantly in motion and colliding with each other . every time molecules collide , kinetic energy can be transferred . when the two systems are in contact , heat will be transferred through molecular collisions from the hotter system to the cooler system . the thermal energy will flow in that direction until the two objects are at the same temperature . when the two systems in contact are at the same temperature , we say they are in thermal equilibrium . zeroth law of thermodynamics : defining thermal equilibrium the zeroth law of thermodynamics defines thermal equilibrium within an isolated system . the zeroth law says when two objects at thermal equilibrium are in contact , there is no net heat transfer between the objects ; therefore , they are the same temperature . another way to state the zeroth law is to say that if two objects are both separately in thermal equilibrium with a third object , then they are in thermal equilibrium with each other . the zeroth law allows us to measure the temperature of objects . any time we use a thermometer , we are using the zeroth law of thermodynamics . let 's say we are measuring the temperature of a water bath . in order to make sure the reading is accurate , we usually want to wait for the temperature reading to stay constant . we are waiting for the thermometer and the water to reach thermal equilibrium ! at thermal equilibrium , the temperature of the thermometer bulb and the water bath will be the same , and there should be no net heat transfer from one object to the other ( assuming no other loss of heat to the surroundings ) . heat capacity : converting between heat and change in temperature how can we measure heat ? here are some things we know about heat so far : when a system absorbs or loses heat , the average kinetic energy of the molecules will change . thus , heat transfer results in a change in the system 's temperature as long as the system is not undergoing a phase change . the change in temperature resulting from heat transferred to or from a system depends on how many molecules are in the system . we can use a thermometer to measure the change in a system 's temperature . how can we use the change in temperature to calculate the heat transferred ? in order to figure out how the heat transferred to a system will change the temperature of the system , we need to know at least $ 2 $ things : the number of molecules in the system the heat capacity of the system the heat capacity tells us how much energy is needed to change the temperature of a given substance assuming that no phase changes are occurring . there are two main ways that heat capacity is reported . the specific heat capacity ( also called specific heat ) , represented by the symbol $ \text c $ or $ \text c $ , is how much energy is needed to increase the temperature of one gram of a substance by $ 1~^ { \circ } \text c $ or $ 1\ , \text k $ . specific heat capacity usually has units of $ \dfrac { \text j } { \text { grams } \cdot\text k } $ . the molar heat capacity , $ \text c_\text m $ or $ \text c_ { \text { mol } } $ , measures the amount of thermal energy it takes to raise the temperature of one mole of a substance by $ 1~^ { \circ } \text c $ or $ 1\ , \text k $ , and it usually has units of $ \dfrac { \text j } { \text { mol } \cdot\text k } $ . for example , the heat capacity of lead might be given as the specific heat capacity , $ 0.129\ , \dfrac { \text j } { \text { g } \cdot\text k } $ , or the molar heat capacity , $ 26.65\ , \dfrac { \text j } { \text { mol } \cdot\text k } $ . calculating $ \text q $ using the heat capacity we can use the heat capacity to determine the heat released or absorbed by a material using the following formula : $ \text q = \text { m } \times \text c \times \delta \text t $ where $ \text { m } $ is the mass of the substance ( in grams ) , $ \text { c } $ is the specific heat capacity , and $ \delta \text t $ is the change in temperature during the heat transfer . note that both mass and specific heat capacity can only have positive values , so the sign of $ \text q $ will depend on the sign of $ \delta \text t $ . we can calculate $ \delta \text t $ using the following equation : $ \delta \text t=\text t_ { \text { final } } -\text t_ { \text { initial } } $ where $ \text t_ { \text { final } } $ and $ \text t_ { \text { initial } } $ can have units of either $ ~^ { \circ } \text c $ or $ \text k $ . based on this equation , if $ \text q $ is positive ( energy of the system increases ) , then our system increases in temperature and $ \text t_ { \text { final } } & gt ; \text t_ { \text { initial } } $ . if $ \text q $ is negative ( energy of the system decreases ) , then our system 's temperature decreases and $ \text t_ { \text { final } } & lt ; \text t_ { \text { initial } } $ . example problem : cooling a cup of tea let 's say that we have $ 250\ , \text { ml } $ of hot tea which we would like to cool down before we try to drink it . the tea is currently at $ 370\ , \text k $ , and we 'd like to cool it down to $ 350\ , \text k $ . how much thermal energy has to be transferred from the tea to the surroundings to cool the tea ? we are going to assume that the tea is mostly water , so we can use the density and heat capacity of water in our calculations . the specific heat capacity of water is $ 4.18\ , \dfrac { \text j } { \text g \cdot \text k } $ , and the density of water is $ 1.00\ , \dfrac { \text g } { \text { ml } } $ . we can calculate the energy transferred in the process of cooling the tea using the following steps : 1 . calculate the mass of the substance we can calculate the mass of the tea/water using the volume and density of water : $ \text m=250\ , \cancel { \text { ml } } \times 1.00\ , \dfrac { \text g } { \cancel { \text { ml } } } =250\ , \text g $ 2 . calculate the change in temperature , $ \delta \text t $ we can calculate the change in temperature , $ \delta \text t $ , from the initial and final temperatures : $ \begin { align } \delta \text t & amp ; =\text t_ { \text { final } } -\text t_ { \text { initial } } \ \ & amp ; =350\ , \text k-370\ , \text k\ \ & amp ; =-20\ , \text k\end { align } $ since the temperature of the tea is decreasing and $ \delta \text t $ is negative , we would expect $ \text q $ to also be negative since our system is losing thermal energy . 3 . solve for $ \text q $ now we can solve for the heat transferred from the hot tea using the equation for heat : $ \begin { align } \text q & amp ; = \text { m } \times \text c \times \delta \text t\ & amp ; =250\ , \cancel { \text g } \times4.18\ , \dfrac { \text j } { \cancel { \text g } \cdot \cancel { \text k } } \times -20\ , \cancel { \text k } \ & amp ; =-21000\ , \text j\end { align } $ thus , we calculated that the tea will transfer $ 21000\ , \text j $ of energy to the surroundings when it cools down from $ 370\ , \text k $ to $ 350\ , \text k $ . conclusions in thermodynamics , heat and temperature are closely related concepts with precise definitions . heat , $ \text q $ , is thermal energy transferred from a hotter system to a cooler system that are in contact . temperature is a measure of the average kinetic energy of the atoms or molecules in the system . the zeroth law of thermodynamics says that no heat is transferred between two objects in thermal equilibrium ; therefore , they are the same temperature . we can calculate the heat released or absorbed using the specific heat capacity $ \text c $ , the mass of the substance $ \text m $ , and the change in temperature $ \delta \text t $ in the following equation : $ \text q = \text { m } \times \text c \times \delta \text t $
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temperature is a measure of the average kinetic energy of the atoms or molecules in the system . the zeroth law of thermodynamics says that no heat is transferred between two objects in thermal equilibrium ; therefore , they are the same temperature . we can calculate the heat released or absorbed using the specific heat capacity $ \text c $ , the mass of the substance $ \text m $ , and the change in temperature $ \delta \text t $ in the equation : $ \text q = \text { m } \times \text c \times \delta \text t $ heat in thermodynamics what contains more heat , a cup of coffee or a glass of iced tea ?
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may i know why there 's no more heat transferred when both objects have achieved thermal equilibrium ?
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key points heat , $ \text q $ , is thermal energy transferred from a hotter system to a cooler system that are in contact . temperature is a measure of the average kinetic energy of the atoms or molecules in the system . the zeroth law of thermodynamics says that no heat is transferred between two objects in thermal equilibrium ; therefore , they are the same temperature . we can calculate the heat released or absorbed using the specific heat capacity $ \text c $ , the mass of the substance $ \text m $ , and the change in temperature $ \delta \text t $ in the equation : $ \text q = \text { m } \times \text c \times \delta \text t $ heat in thermodynamics what contains more heat , a cup of coffee or a glass of iced tea ? in chemistry class , that would be a trick question ( sorry ! ) . in thermodynamics , heat has a very specific meaning that is different from how we might use the word in everyday speech . scientists define heat as thermal energy transferred between two systems at different temperatures that come in contact . heat is written with the symbol q or q , and it has units of joules ( $ \text j $ ) . heat is sometimes called a process quantity , because it is defined in the context of a process by which energy can be transferred . we do n't talk about a cup of coffee containing heat , but we can talk about the heat transferred from the cup of hot coffee to your hand . heat is also an extensive property , so the change in temperature resulting from heat transferred to a system depends on how many molecules are in the system . relationship between heat and temperature heat and temperature are two different but closely related concepts . note that they have different units : temperature typically has units of degrees celsius ( $ ^\circ\text c $ ) or kelvin ( $ \text k $ ) , and heat has units of energy , joules ( $ \text j $ ) . temperature is a measure of the average kinetic energy of the atoms or molecules in the system . the water molecules in a cup of hot coffee have a higher average kinetic energy than the water molecules in a cup of iced tea , which also means they are moving at a higher velocity . temperature is also an intensive property , which means that the temperature does n't change no matter how much of a substance you have ( as long as it is all at the same temperature ! ) . this is why chemists can use the melting point to help identify a pure substance $ - $ the temperature at which it melts is a property of the substance with no dependence on the mass of a sample . on an atomic level , the molecules in each object are constantly in motion and colliding with each other . every time molecules collide , kinetic energy can be transferred . when the two systems are in contact , heat will be transferred through molecular collisions from the hotter system to the cooler system . the thermal energy will flow in that direction until the two objects are at the same temperature . when the two systems in contact are at the same temperature , we say they are in thermal equilibrium . zeroth law of thermodynamics : defining thermal equilibrium the zeroth law of thermodynamics defines thermal equilibrium within an isolated system . the zeroth law says when two objects at thermal equilibrium are in contact , there is no net heat transfer between the objects ; therefore , they are the same temperature . another way to state the zeroth law is to say that if two objects are both separately in thermal equilibrium with a third object , then they are in thermal equilibrium with each other . the zeroth law allows us to measure the temperature of objects . any time we use a thermometer , we are using the zeroth law of thermodynamics . let 's say we are measuring the temperature of a water bath . in order to make sure the reading is accurate , we usually want to wait for the temperature reading to stay constant . we are waiting for the thermometer and the water to reach thermal equilibrium ! at thermal equilibrium , the temperature of the thermometer bulb and the water bath will be the same , and there should be no net heat transfer from one object to the other ( assuming no other loss of heat to the surroundings ) . heat capacity : converting between heat and change in temperature how can we measure heat ? here are some things we know about heat so far : when a system absorbs or loses heat , the average kinetic energy of the molecules will change . thus , heat transfer results in a change in the system 's temperature as long as the system is not undergoing a phase change . the change in temperature resulting from heat transferred to or from a system depends on how many molecules are in the system . we can use a thermometer to measure the change in a system 's temperature . how can we use the change in temperature to calculate the heat transferred ? in order to figure out how the heat transferred to a system will change the temperature of the system , we need to know at least $ 2 $ things : the number of molecules in the system the heat capacity of the system the heat capacity tells us how much energy is needed to change the temperature of a given substance assuming that no phase changes are occurring . there are two main ways that heat capacity is reported . the specific heat capacity ( also called specific heat ) , represented by the symbol $ \text c $ or $ \text c $ , is how much energy is needed to increase the temperature of one gram of a substance by $ 1~^ { \circ } \text c $ or $ 1\ , \text k $ . specific heat capacity usually has units of $ \dfrac { \text j } { \text { grams } \cdot\text k } $ . the molar heat capacity , $ \text c_\text m $ or $ \text c_ { \text { mol } } $ , measures the amount of thermal energy it takes to raise the temperature of one mole of a substance by $ 1~^ { \circ } \text c $ or $ 1\ , \text k $ , and it usually has units of $ \dfrac { \text j } { \text { mol } \cdot\text k } $ . for example , the heat capacity of lead might be given as the specific heat capacity , $ 0.129\ , \dfrac { \text j } { \text { g } \cdot\text k } $ , or the molar heat capacity , $ 26.65\ , \dfrac { \text j } { \text { mol } \cdot\text k } $ . calculating $ \text q $ using the heat capacity we can use the heat capacity to determine the heat released or absorbed by a material using the following formula : $ \text q = \text { m } \times \text c \times \delta \text t $ where $ \text { m } $ is the mass of the substance ( in grams ) , $ \text { c } $ is the specific heat capacity , and $ \delta \text t $ is the change in temperature during the heat transfer . note that both mass and specific heat capacity can only have positive values , so the sign of $ \text q $ will depend on the sign of $ \delta \text t $ . we can calculate $ \delta \text t $ using the following equation : $ \delta \text t=\text t_ { \text { final } } -\text t_ { \text { initial } } $ where $ \text t_ { \text { final } } $ and $ \text t_ { \text { initial } } $ can have units of either $ ~^ { \circ } \text c $ or $ \text k $ . based on this equation , if $ \text q $ is positive ( energy of the system increases ) , then our system increases in temperature and $ \text t_ { \text { final } } & gt ; \text t_ { \text { initial } } $ . if $ \text q $ is negative ( energy of the system decreases ) , then our system 's temperature decreases and $ \text t_ { \text { final } } & lt ; \text t_ { \text { initial } } $ . example problem : cooling a cup of tea let 's say that we have $ 250\ , \text { ml } $ of hot tea which we would like to cool down before we try to drink it . the tea is currently at $ 370\ , \text k $ , and we 'd like to cool it down to $ 350\ , \text k $ . how much thermal energy has to be transferred from the tea to the surroundings to cool the tea ? we are going to assume that the tea is mostly water , so we can use the density and heat capacity of water in our calculations . the specific heat capacity of water is $ 4.18\ , \dfrac { \text j } { \text g \cdot \text k } $ , and the density of water is $ 1.00\ , \dfrac { \text g } { \text { ml } } $ . we can calculate the energy transferred in the process of cooling the tea using the following steps : 1 . calculate the mass of the substance we can calculate the mass of the tea/water using the volume and density of water : $ \text m=250\ , \cancel { \text { ml } } \times 1.00\ , \dfrac { \text g } { \cancel { \text { ml } } } =250\ , \text g $ 2 . calculate the change in temperature , $ \delta \text t $ we can calculate the change in temperature , $ \delta \text t $ , from the initial and final temperatures : $ \begin { align } \delta \text t & amp ; =\text t_ { \text { final } } -\text t_ { \text { initial } } \ \ & amp ; =350\ , \text k-370\ , \text k\ \ & amp ; =-20\ , \text k\end { align } $ since the temperature of the tea is decreasing and $ \delta \text t $ is negative , we would expect $ \text q $ to also be negative since our system is losing thermal energy . 3 . solve for $ \text q $ now we can solve for the heat transferred from the hot tea using the equation for heat : $ \begin { align } \text q & amp ; = \text { m } \times \text c \times \delta \text t\ & amp ; =250\ , \cancel { \text g } \times4.18\ , \dfrac { \text j } { \cancel { \text g } \cdot \cancel { \text k } } \times -20\ , \cancel { \text k } \ & amp ; =-21000\ , \text j\end { align } $ thus , we calculated that the tea will transfer $ 21000\ , \text j $ of energy to the surroundings when it cools down from $ 370\ , \text k $ to $ 350\ , \text k $ . conclusions in thermodynamics , heat and temperature are closely related concepts with precise definitions . heat , $ \text q $ , is thermal energy transferred from a hotter system to a cooler system that are in contact . temperature is a measure of the average kinetic energy of the atoms or molecules in the system . the zeroth law of thermodynamics says that no heat is transferred between two objects in thermal equilibrium ; therefore , they are the same temperature . we can calculate the heat released or absorbed using the specific heat capacity $ \text c $ , the mass of the substance $ \text m $ , and the change in temperature $ \delta \text t $ in the following equation : $ \text q = \text { m } \times \text c \times \delta \text t $
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at thermal equilibrium , the temperature of the thermometer bulb and the water bath will be the same , and there should be no net heat transfer from one object to the other ( assuming no other loss of heat to the surroundings ) . heat capacity : converting between heat and change in temperature how can we measure heat ? here are some things we know about heat so far : when a system absorbs or loses heat , the average kinetic energy of the molecules will change .
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in the section called heat capacity : converting between heat and change in temperature , in the last paragraph , what is k ?
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key points heat , $ \text q $ , is thermal energy transferred from a hotter system to a cooler system that are in contact . temperature is a measure of the average kinetic energy of the atoms or molecules in the system . the zeroth law of thermodynamics says that no heat is transferred between two objects in thermal equilibrium ; therefore , they are the same temperature . we can calculate the heat released or absorbed using the specific heat capacity $ \text c $ , the mass of the substance $ \text m $ , and the change in temperature $ \delta \text t $ in the equation : $ \text q = \text { m } \times \text c \times \delta \text t $ heat in thermodynamics what contains more heat , a cup of coffee or a glass of iced tea ? in chemistry class , that would be a trick question ( sorry ! ) . in thermodynamics , heat has a very specific meaning that is different from how we might use the word in everyday speech . scientists define heat as thermal energy transferred between two systems at different temperatures that come in contact . heat is written with the symbol q or q , and it has units of joules ( $ \text j $ ) . heat is sometimes called a process quantity , because it is defined in the context of a process by which energy can be transferred . we do n't talk about a cup of coffee containing heat , but we can talk about the heat transferred from the cup of hot coffee to your hand . heat is also an extensive property , so the change in temperature resulting from heat transferred to a system depends on how many molecules are in the system . relationship between heat and temperature heat and temperature are two different but closely related concepts . note that they have different units : temperature typically has units of degrees celsius ( $ ^\circ\text c $ ) or kelvin ( $ \text k $ ) , and heat has units of energy , joules ( $ \text j $ ) . temperature is a measure of the average kinetic energy of the atoms or molecules in the system . the water molecules in a cup of hot coffee have a higher average kinetic energy than the water molecules in a cup of iced tea , which also means they are moving at a higher velocity . temperature is also an intensive property , which means that the temperature does n't change no matter how much of a substance you have ( as long as it is all at the same temperature ! ) . this is why chemists can use the melting point to help identify a pure substance $ - $ the temperature at which it melts is a property of the substance with no dependence on the mass of a sample . on an atomic level , the molecules in each object are constantly in motion and colliding with each other . every time molecules collide , kinetic energy can be transferred . when the two systems are in contact , heat will be transferred through molecular collisions from the hotter system to the cooler system . the thermal energy will flow in that direction until the two objects are at the same temperature . when the two systems in contact are at the same temperature , we say they are in thermal equilibrium . zeroth law of thermodynamics : defining thermal equilibrium the zeroth law of thermodynamics defines thermal equilibrium within an isolated system . the zeroth law says when two objects at thermal equilibrium are in contact , there is no net heat transfer between the objects ; therefore , they are the same temperature . another way to state the zeroth law is to say that if two objects are both separately in thermal equilibrium with a third object , then they are in thermal equilibrium with each other . the zeroth law allows us to measure the temperature of objects . any time we use a thermometer , we are using the zeroth law of thermodynamics . let 's say we are measuring the temperature of a water bath . in order to make sure the reading is accurate , we usually want to wait for the temperature reading to stay constant . we are waiting for the thermometer and the water to reach thermal equilibrium ! at thermal equilibrium , the temperature of the thermometer bulb and the water bath will be the same , and there should be no net heat transfer from one object to the other ( assuming no other loss of heat to the surroundings ) . heat capacity : converting between heat and change in temperature how can we measure heat ? here are some things we know about heat so far : when a system absorbs or loses heat , the average kinetic energy of the molecules will change . thus , heat transfer results in a change in the system 's temperature as long as the system is not undergoing a phase change . the change in temperature resulting from heat transferred to or from a system depends on how many molecules are in the system . we can use a thermometer to measure the change in a system 's temperature . how can we use the change in temperature to calculate the heat transferred ? in order to figure out how the heat transferred to a system will change the temperature of the system , we need to know at least $ 2 $ things : the number of molecules in the system the heat capacity of the system the heat capacity tells us how much energy is needed to change the temperature of a given substance assuming that no phase changes are occurring . there are two main ways that heat capacity is reported . the specific heat capacity ( also called specific heat ) , represented by the symbol $ \text c $ or $ \text c $ , is how much energy is needed to increase the temperature of one gram of a substance by $ 1~^ { \circ } \text c $ or $ 1\ , \text k $ . specific heat capacity usually has units of $ \dfrac { \text j } { \text { grams } \cdot\text k } $ . the molar heat capacity , $ \text c_\text m $ or $ \text c_ { \text { mol } } $ , measures the amount of thermal energy it takes to raise the temperature of one mole of a substance by $ 1~^ { \circ } \text c $ or $ 1\ , \text k $ , and it usually has units of $ \dfrac { \text j } { \text { mol } \cdot\text k } $ . for example , the heat capacity of lead might be given as the specific heat capacity , $ 0.129\ , \dfrac { \text j } { \text { g } \cdot\text k } $ , or the molar heat capacity , $ 26.65\ , \dfrac { \text j } { \text { mol } \cdot\text k } $ . calculating $ \text q $ using the heat capacity we can use the heat capacity to determine the heat released or absorbed by a material using the following formula : $ \text q = \text { m } \times \text c \times \delta \text t $ where $ \text { m } $ is the mass of the substance ( in grams ) , $ \text { c } $ is the specific heat capacity , and $ \delta \text t $ is the change in temperature during the heat transfer . note that both mass and specific heat capacity can only have positive values , so the sign of $ \text q $ will depend on the sign of $ \delta \text t $ . we can calculate $ \delta \text t $ using the following equation : $ \delta \text t=\text t_ { \text { final } } -\text t_ { \text { initial } } $ where $ \text t_ { \text { final } } $ and $ \text t_ { \text { initial } } $ can have units of either $ ~^ { \circ } \text c $ or $ \text k $ . based on this equation , if $ \text q $ is positive ( energy of the system increases ) , then our system increases in temperature and $ \text t_ { \text { final } } & gt ; \text t_ { \text { initial } } $ . if $ \text q $ is negative ( energy of the system decreases ) , then our system 's temperature decreases and $ \text t_ { \text { final } } & lt ; \text t_ { \text { initial } } $ . example problem : cooling a cup of tea let 's say that we have $ 250\ , \text { ml } $ of hot tea which we would like to cool down before we try to drink it . the tea is currently at $ 370\ , \text k $ , and we 'd like to cool it down to $ 350\ , \text k $ . how much thermal energy has to be transferred from the tea to the surroundings to cool the tea ? we are going to assume that the tea is mostly water , so we can use the density and heat capacity of water in our calculations . the specific heat capacity of water is $ 4.18\ , \dfrac { \text j } { \text g \cdot \text k } $ , and the density of water is $ 1.00\ , \dfrac { \text g } { \text { ml } } $ . we can calculate the energy transferred in the process of cooling the tea using the following steps : 1 . calculate the mass of the substance we can calculate the mass of the tea/water using the volume and density of water : $ \text m=250\ , \cancel { \text { ml } } \times 1.00\ , \dfrac { \text g } { \cancel { \text { ml } } } =250\ , \text g $ 2 . calculate the change in temperature , $ \delta \text t $ we can calculate the change in temperature , $ \delta \text t $ , from the initial and final temperatures : $ \begin { align } \delta \text t & amp ; =\text t_ { \text { final } } -\text t_ { \text { initial } } \ \ & amp ; =350\ , \text k-370\ , \text k\ \ & amp ; =-20\ , \text k\end { align } $ since the temperature of the tea is decreasing and $ \delta \text t $ is negative , we would expect $ \text q $ to also be negative since our system is losing thermal energy . 3 . solve for $ \text q $ now we can solve for the heat transferred from the hot tea using the equation for heat : $ \begin { align } \text q & amp ; = \text { m } \times \text c \times \delta \text t\ & amp ; =250\ , \cancel { \text g } \times4.18\ , \dfrac { \text j } { \cancel { \text g } \cdot \cancel { \text k } } \times -20\ , \cancel { \text k } \ & amp ; =-21000\ , \text j\end { align } $ thus , we calculated that the tea will transfer $ 21000\ , \text j $ of energy to the surroundings when it cools down from $ 370\ , \text k $ to $ 350\ , \text k $ . conclusions in thermodynamics , heat and temperature are closely related concepts with precise definitions . heat , $ \text q $ , is thermal energy transferred from a hotter system to a cooler system that are in contact . temperature is a measure of the average kinetic energy of the atoms or molecules in the system . the zeroth law of thermodynamics says that no heat is transferred between two objects in thermal equilibrium ; therefore , they are the same temperature . we can calculate the heat released or absorbed using the specific heat capacity $ \text c $ , the mass of the substance $ \text m $ , and the change in temperature $ \delta \text t $ in the following equation : $ \text q = \text { m } \times \text c \times \delta \text t $
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at thermal equilibrium , the temperature of the thermometer bulb and the water bath will be the same , and there should be no net heat transfer from one object to the other ( assuming no other loss of heat to the surroundings ) . heat capacity : converting between heat and change in temperature how can we measure heat ? here are some things we know about heat so far : when a system absorbs or loses heat , the average kinetic energy of the molecules will change .
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why there is no change of dalta h for the change of temperature , when the change of heat capacity for the reaction is zero ?
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key points heat , $ \text q $ , is thermal energy transferred from a hotter system to a cooler system that are in contact . temperature is a measure of the average kinetic energy of the atoms or molecules in the system . the zeroth law of thermodynamics says that no heat is transferred between two objects in thermal equilibrium ; therefore , they are the same temperature . we can calculate the heat released or absorbed using the specific heat capacity $ \text c $ , the mass of the substance $ \text m $ , and the change in temperature $ \delta \text t $ in the equation : $ \text q = \text { m } \times \text c \times \delta \text t $ heat in thermodynamics what contains more heat , a cup of coffee or a glass of iced tea ? in chemistry class , that would be a trick question ( sorry ! ) . in thermodynamics , heat has a very specific meaning that is different from how we might use the word in everyday speech . scientists define heat as thermal energy transferred between two systems at different temperatures that come in contact . heat is written with the symbol q or q , and it has units of joules ( $ \text j $ ) . heat is sometimes called a process quantity , because it is defined in the context of a process by which energy can be transferred . we do n't talk about a cup of coffee containing heat , but we can talk about the heat transferred from the cup of hot coffee to your hand . heat is also an extensive property , so the change in temperature resulting from heat transferred to a system depends on how many molecules are in the system . relationship between heat and temperature heat and temperature are two different but closely related concepts . note that they have different units : temperature typically has units of degrees celsius ( $ ^\circ\text c $ ) or kelvin ( $ \text k $ ) , and heat has units of energy , joules ( $ \text j $ ) . temperature is a measure of the average kinetic energy of the atoms or molecules in the system . the water molecules in a cup of hot coffee have a higher average kinetic energy than the water molecules in a cup of iced tea , which also means they are moving at a higher velocity . temperature is also an intensive property , which means that the temperature does n't change no matter how much of a substance you have ( as long as it is all at the same temperature ! ) . this is why chemists can use the melting point to help identify a pure substance $ - $ the temperature at which it melts is a property of the substance with no dependence on the mass of a sample . on an atomic level , the molecules in each object are constantly in motion and colliding with each other . every time molecules collide , kinetic energy can be transferred . when the two systems are in contact , heat will be transferred through molecular collisions from the hotter system to the cooler system . the thermal energy will flow in that direction until the two objects are at the same temperature . when the two systems in contact are at the same temperature , we say they are in thermal equilibrium . zeroth law of thermodynamics : defining thermal equilibrium the zeroth law of thermodynamics defines thermal equilibrium within an isolated system . the zeroth law says when two objects at thermal equilibrium are in contact , there is no net heat transfer between the objects ; therefore , they are the same temperature . another way to state the zeroth law is to say that if two objects are both separately in thermal equilibrium with a third object , then they are in thermal equilibrium with each other . the zeroth law allows us to measure the temperature of objects . any time we use a thermometer , we are using the zeroth law of thermodynamics . let 's say we are measuring the temperature of a water bath . in order to make sure the reading is accurate , we usually want to wait for the temperature reading to stay constant . we are waiting for the thermometer and the water to reach thermal equilibrium ! at thermal equilibrium , the temperature of the thermometer bulb and the water bath will be the same , and there should be no net heat transfer from one object to the other ( assuming no other loss of heat to the surroundings ) . heat capacity : converting between heat and change in temperature how can we measure heat ? here are some things we know about heat so far : when a system absorbs or loses heat , the average kinetic energy of the molecules will change . thus , heat transfer results in a change in the system 's temperature as long as the system is not undergoing a phase change . the change in temperature resulting from heat transferred to or from a system depends on how many molecules are in the system . we can use a thermometer to measure the change in a system 's temperature . how can we use the change in temperature to calculate the heat transferred ? in order to figure out how the heat transferred to a system will change the temperature of the system , we need to know at least $ 2 $ things : the number of molecules in the system the heat capacity of the system the heat capacity tells us how much energy is needed to change the temperature of a given substance assuming that no phase changes are occurring . there are two main ways that heat capacity is reported . the specific heat capacity ( also called specific heat ) , represented by the symbol $ \text c $ or $ \text c $ , is how much energy is needed to increase the temperature of one gram of a substance by $ 1~^ { \circ } \text c $ or $ 1\ , \text k $ . specific heat capacity usually has units of $ \dfrac { \text j } { \text { grams } \cdot\text k } $ . the molar heat capacity , $ \text c_\text m $ or $ \text c_ { \text { mol } } $ , measures the amount of thermal energy it takes to raise the temperature of one mole of a substance by $ 1~^ { \circ } \text c $ or $ 1\ , \text k $ , and it usually has units of $ \dfrac { \text j } { \text { mol } \cdot\text k } $ . for example , the heat capacity of lead might be given as the specific heat capacity , $ 0.129\ , \dfrac { \text j } { \text { g } \cdot\text k } $ , or the molar heat capacity , $ 26.65\ , \dfrac { \text j } { \text { mol } \cdot\text k } $ . calculating $ \text q $ using the heat capacity we can use the heat capacity to determine the heat released or absorbed by a material using the following formula : $ \text q = \text { m } \times \text c \times \delta \text t $ where $ \text { m } $ is the mass of the substance ( in grams ) , $ \text { c } $ is the specific heat capacity , and $ \delta \text t $ is the change in temperature during the heat transfer . note that both mass and specific heat capacity can only have positive values , so the sign of $ \text q $ will depend on the sign of $ \delta \text t $ . we can calculate $ \delta \text t $ using the following equation : $ \delta \text t=\text t_ { \text { final } } -\text t_ { \text { initial } } $ where $ \text t_ { \text { final } } $ and $ \text t_ { \text { initial } } $ can have units of either $ ~^ { \circ } \text c $ or $ \text k $ . based on this equation , if $ \text q $ is positive ( energy of the system increases ) , then our system increases in temperature and $ \text t_ { \text { final } } & gt ; \text t_ { \text { initial } } $ . if $ \text q $ is negative ( energy of the system decreases ) , then our system 's temperature decreases and $ \text t_ { \text { final } } & lt ; \text t_ { \text { initial } } $ . example problem : cooling a cup of tea let 's say that we have $ 250\ , \text { ml } $ of hot tea which we would like to cool down before we try to drink it . the tea is currently at $ 370\ , \text k $ , and we 'd like to cool it down to $ 350\ , \text k $ . how much thermal energy has to be transferred from the tea to the surroundings to cool the tea ? we are going to assume that the tea is mostly water , so we can use the density and heat capacity of water in our calculations . the specific heat capacity of water is $ 4.18\ , \dfrac { \text j } { \text g \cdot \text k } $ , and the density of water is $ 1.00\ , \dfrac { \text g } { \text { ml } } $ . we can calculate the energy transferred in the process of cooling the tea using the following steps : 1 . calculate the mass of the substance we can calculate the mass of the tea/water using the volume and density of water : $ \text m=250\ , \cancel { \text { ml } } \times 1.00\ , \dfrac { \text g } { \cancel { \text { ml } } } =250\ , \text g $ 2 . calculate the change in temperature , $ \delta \text t $ we can calculate the change in temperature , $ \delta \text t $ , from the initial and final temperatures : $ \begin { align } \delta \text t & amp ; =\text t_ { \text { final } } -\text t_ { \text { initial } } \ \ & amp ; =350\ , \text k-370\ , \text k\ \ & amp ; =-20\ , \text k\end { align } $ since the temperature of the tea is decreasing and $ \delta \text t $ is negative , we would expect $ \text q $ to also be negative since our system is losing thermal energy . 3 . solve for $ \text q $ now we can solve for the heat transferred from the hot tea using the equation for heat : $ \begin { align } \text q & amp ; = \text { m } \times \text c \times \delta \text t\ & amp ; =250\ , \cancel { \text g } \times4.18\ , \dfrac { \text j } { \cancel { \text g } \cdot \cancel { \text k } } \times -20\ , \cancel { \text k } \ & amp ; =-21000\ , \text j\end { align } $ thus , we calculated that the tea will transfer $ 21000\ , \text j $ of energy to the surroundings when it cools down from $ 370\ , \text k $ to $ 350\ , \text k $ . conclusions in thermodynamics , heat and temperature are closely related concepts with precise definitions . heat , $ \text q $ , is thermal energy transferred from a hotter system to a cooler system that are in contact . temperature is a measure of the average kinetic energy of the atoms or molecules in the system . the zeroth law of thermodynamics says that no heat is transferred between two objects in thermal equilibrium ; therefore , they are the same temperature . we can calculate the heat released or absorbed using the specific heat capacity $ \text c $ , the mass of the substance $ \text m $ , and the change in temperature $ \delta \text t $ in the following equation : $ \text q = \text { m } \times \text c \times \delta \text t $
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the water molecules in a cup of hot coffee have a higher average kinetic energy than the water molecules in a cup of iced tea , which also means they are moving at a higher velocity . temperature is also an intensive property , which means that the temperature does n't change no matter how much of a substance you have ( as long as it is all at the same temperature ! ) . this is why chemists can use the melting point to help identify a pure substance $ - $ the temperature at which it melts is a property of the substance with no dependence on the mass of a sample .
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input temperature , output temperature , arithmetic average , or some other sort of average ?
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key points heat , $ \text q $ , is thermal energy transferred from a hotter system to a cooler system that are in contact . temperature is a measure of the average kinetic energy of the atoms or molecules in the system . the zeroth law of thermodynamics says that no heat is transferred between two objects in thermal equilibrium ; therefore , they are the same temperature . we can calculate the heat released or absorbed using the specific heat capacity $ \text c $ , the mass of the substance $ \text m $ , and the change in temperature $ \delta \text t $ in the equation : $ \text q = \text { m } \times \text c \times \delta \text t $ heat in thermodynamics what contains more heat , a cup of coffee or a glass of iced tea ? in chemistry class , that would be a trick question ( sorry ! ) . in thermodynamics , heat has a very specific meaning that is different from how we might use the word in everyday speech . scientists define heat as thermal energy transferred between two systems at different temperatures that come in contact . heat is written with the symbol q or q , and it has units of joules ( $ \text j $ ) . heat is sometimes called a process quantity , because it is defined in the context of a process by which energy can be transferred . we do n't talk about a cup of coffee containing heat , but we can talk about the heat transferred from the cup of hot coffee to your hand . heat is also an extensive property , so the change in temperature resulting from heat transferred to a system depends on how many molecules are in the system . relationship between heat and temperature heat and temperature are two different but closely related concepts . note that they have different units : temperature typically has units of degrees celsius ( $ ^\circ\text c $ ) or kelvin ( $ \text k $ ) , and heat has units of energy , joules ( $ \text j $ ) . temperature is a measure of the average kinetic energy of the atoms or molecules in the system . the water molecules in a cup of hot coffee have a higher average kinetic energy than the water molecules in a cup of iced tea , which also means they are moving at a higher velocity . temperature is also an intensive property , which means that the temperature does n't change no matter how much of a substance you have ( as long as it is all at the same temperature ! ) . this is why chemists can use the melting point to help identify a pure substance $ - $ the temperature at which it melts is a property of the substance with no dependence on the mass of a sample . on an atomic level , the molecules in each object are constantly in motion and colliding with each other . every time molecules collide , kinetic energy can be transferred . when the two systems are in contact , heat will be transferred through molecular collisions from the hotter system to the cooler system . the thermal energy will flow in that direction until the two objects are at the same temperature . when the two systems in contact are at the same temperature , we say they are in thermal equilibrium . zeroth law of thermodynamics : defining thermal equilibrium the zeroth law of thermodynamics defines thermal equilibrium within an isolated system . the zeroth law says when two objects at thermal equilibrium are in contact , there is no net heat transfer between the objects ; therefore , they are the same temperature . another way to state the zeroth law is to say that if two objects are both separately in thermal equilibrium with a third object , then they are in thermal equilibrium with each other . the zeroth law allows us to measure the temperature of objects . any time we use a thermometer , we are using the zeroth law of thermodynamics . let 's say we are measuring the temperature of a water bath . in order to make sure the reading is accurate , we usually want to wait for the temperature reading to stay constant . we are waiting for the thermometer and the water to reach thermal equilibrium ! at thermal equilibrium , the temperature of the thermometer bulb and the water bath will be the same , and there should be no net heat transfer from one object to the other ( assuming no other loss of heat to the surroundings ) . heat capacity : converting between heat and change in temperature how can we measure heat ? here are some things we know about heat so far : when a system absorbs or loses heat , the average kinetic energy of the molecules will change . thus , heat transfer results in a change in the system 's temperature as long as the system is not undergoing a phase change . the change in temperature resulting from heat transferred to or from a system depends on how many molecules are in the system . we can use a thermometer to measure the change in a system 's temperature . how can we use the change in temperature to calculate the heat transferred ? in order to figure out how the heat transferred to a system will change the temperature of the system , we need to know at least $ 2 $ things : the number of molecules in the system the heat capacity of the system the heat capacity tells us how much energy is needed to change the temperature of a given substance assuming that no phase changes are occurring . there are two main ways that heat capacity is reported . the specific heat capacity ( also called specific heat ) , represented by the symbol $ \text c $ or $ \text c $ , is how much energy is needed to increase the temperature of one gram of a substance by $ 1~^ { \circ } \text c $ or $ 1\ , \text k $ . specific heat capacity usually has units of $ \dfrac { \text j } { \text { grams } \cdot\text k } $ . the molar heat capacity , $ \text c_\text m $ or $ \text c_ { \text { mol } } $ , measures the amount of thermal energy it takes to raise the temperature of one mole of a substance by $ 1~^ { \circ } \text c $ or $ 1\ , \text k $ , and it usually has units of $ \dfrac { \text j } { \text { mol } \cdot\text k } $ . for example , the heat capacity of lead might be given as the specific heat capacity , $ 0.129\ , \dfrac { \text j } { \text { g } \cdot\text k } $ , or the molar heat capacity , $ 26.65\ , \dfrac { \text j } { \text { mol } \cdot\text k } $ . calculating $ \text q $ using the heat capacity we can use the heat capacity to determine the heat released or absorbed by a material using the following formula : $ \text q = \text { m } \times \text c \times \delta \text t $ where $ \text { m } $ is the mass of the substance ( in grams ) , $ \text { c } $ is the specific heat capacity , and $ \delta \text t $ is the change in temperature during the heat transfer . note that both mass and specific heat capacity can only have positive values , so the sign of $ \text q $ will depend on the sign of $ \delta \text t $ . we can calculate $ \delta \text t $ using the following equation : $ \delta \text t=\text t_ { \text { final } } -\text t_ { \text { initial } } $ where $ \text t_ { \text { final } } $ and $ \text t_ { \text { initial } } $ can have units of either $ ~^ { \circ } \text c $ or $ \text k $ . based on this equation , if $ \text q $ is positive ( energy of the system increases ) , then our system increases in temperature and $ \text t_ { \text { final } } & gt ; \text t_ { \text { initial } } $ . if $ \text q $ is negative ( energy of the system decreases ) , then our system 's temperature decreases and $ \text t_ { \text { final } } & lt ; \text t_ { \text { initial } } $ . example problem : cooling a cup of tea let 's say that we have $ 250\ , \text { ml } $ of hot tea which we would like to cool down before we try to drink it . the tea is currently at $ 370\ , \text k $ , and we 'd like to cool it down to $ 350\ , \text k $ . how much thermal energy has to be transferred from the tea to the surroundings to cool the tea ? we are going to assume that the tea is mostly water , so we can use the density and heat capacity of water in our calculations . the specific heat capacity of water is $ 4.18\ , \dfrac { \text j } { \text g \cdot \text k } $ , and the density of water is $ 1.00\ , \dfrac { \text g } { \text { ml } } $ . we can calculate the energy transferred in the process of cooling the tea using the following steps : 1 . calculate the mass of the substance we can calculate the mass of the tea/water using the volume and density of water : $ \text m=250\ , \cancel { \text { ml } } \times 1.00\ , \dfrac { \text g } { \cancel { \text { ml } } } =250\ , \text g $ 2 . calculate the change in temperature , $ \delta \text t $ we can calculate the change in temperature , $ \delta \text t $ , from the initial and final temperatures : $ \begin { align } \delta \text t & amp ; =\text t_ { \text { final } } -\text t_ { \text { initial } } \ \ & amp ; =350\ , \text k-370\ , \text k\ \ & amp ; =-20\ , \text k\end { align } $ since the temperature of the tea is decreasing and $ \delta \text t $ is negative , we would expect $ \text q $ to also be negative since our system is losing thermal energy . 3 . solve for $ \text q $ now we can solve for the heat transferred from the hot tea using the equation for heat : $ \begin { align } \text q & amp ; = \text { m } \times \text c \times \delta \text t\ & amp ; =250\ , \cancel { \text g } \times4.18\ , \dfrac { \text j } { \cancel { \text g } \cdot \cancel { \text k } } \times -20\ , \cancel { \text k } \ & amp ; =-21000\ , \text j\end { align } $ thus , we calculated that the tea will transfer $ 21000\ , \text j $ of energy to the surroundings when it cools down from $ 370\ , \text k $ to $ 350\ , \text k $ . conclusions in thermodynamics , heat and temperature are closely related concepts with precise definitions . heat , $ \text q $ , is thermal energy transferred from a hotter system to a cooler system that are in contact . temperature is a measure of the average kinetic energy of the atoms or molecules in the system . the zeroth law of thermodynamics says that no heat is transferred between two objects in thermal equilibrium ; therefore , they are the same temperature . we can calculate the heat released or absorbed using the specific heat capacity $ \text c $ , the mass of the substance $ \text m $ , and the change in temperature $ \delta \text t $ in the following equation : $ \text q = \text { m } \times \text c \times \delta \text t $
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at thermal equilibrium , the temperature of the thermometer bulb and the water bath will be the same , and there should be no net heat transfer from one object to the other ( assuming no other loss of heat to the surroundings ) . heat capacity : converting between heat and change in temperature how can we measure heat ? here are some things we know about heat so far : when a system absorbs or loses heat , the average kinetic energy of the molecules will change .
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how to get the time given the heat of fusion , work , specific heat and mass ?
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key points heat , $ \text q $ , is thermal energy transferred from a hotter system to a cooler system that are in contact . temperature is a measure of the average kinetic energy of the atoms or molecules in the system . the zeroth law of thermodynamics says that no heat is transferred between two objects in thermal equilibrium ; therefore , they are the same temperature . we can calculate the heat released or absorbed using the specific heat capacity $ \text c $ , the mass of the substance $ \text m $ , and the change in temperature $ \delta \text t $ in the equation : $ \text q = \text { m } \times \text c \times \delta \text t $ heat in thermodynamics what contains more heat , a cup of coffee or a glass of iced tea ? in chemistry class , that would be a trick question ( sorry ! ) . in thermodynamics , heat has a very specific meaning that is different from how we might use the word in everyday speech . scientists define heat as thermal energy transferred between two systems at different temperatures that come in contact . heat is written with the symbol q or q , and it has units of joules ( $ \text j $ ) . heat is sometimes called a process quantity , because it is defined in the context of a process by which energy can be transferred . we do n't talk about a cup of coffee containing heat , but we can talk about the heat transferred from the cup of hot coffee to your hand . heat is also an extensive property , so the change in temperature resulting from heat transferred to a system depends on how many molecules are in the system . relationship between heat and temperature heat and temperature are two different but closely related concepts . note that they have different units : temperature typically has units of degrees celsius ( $ ^\circ\text c $ ) or kelvin ( $ \text k $ ) , and heat has units of energy , joules ( $ \text j $ ) . temperature is a measure of the average kinetic energy of the atoms or molecules in the system . the water molecules in a cup of hot coffee have a higher average kinetic energy than the water molecules in a cup of iced tea , which also means they are moving at a higher velocity . temperature is also an intensive property , which means that the temperature does n't change no matter how much of a substance you have ( as long as it is all at the same temperature ! ) . this is why chemists can use the melting point to help identify a pure substance $ - $ the temperature at which it melts is a property of the substance with no dependence on the mass of a sample . on an atomic level , the molecules in each object are constantly in motion and colliding with each other . every time molecules collide , kinetic energy can be transferred . when the two systems are in contact , heat will be transferred through molecular collisions from the hotter system to the cooler system . the thermal energy will flow in that direction until the two objects are at the same temperature . when the two systems in contact are at the same temperature , we say they are in thermal equilibrium . zeroth law of thermodynamics : defining thermal equilibrium the zeroth law of thermodynamics defines thermal equilibrium within an isolated system . the zeroth law says when two objects at thermal equilibrium are in contact , there is no net heat transfer between the objects ; therefore , they are the same temperature . another way to state the zeroth law is to say that if two objects are both separately in thermal equilibrium with a third object , then they are in thermal equilibrium with each other . the zeroth law allows us to measure the temperature of objects . any time we use a thermometer , we are using the zeroth law of thermodynamics . let 's say we are measuring the temperature of a water bath . in order to make sure the reading is accurate , we usually want to wait for the temperature reading to stay constant . we are waiting for the thermometer and the water to reach thermal equilibrium ! at thermal equilibrium , the temperature of the thermometer bulb and the water bath will be the same , and there should be no net heat transfer from one object to the other ( assuming no other loss of heat to the surroundings ) . heat capacity : converting between heat and change in temperature how can we measure heat ? here are some things we know about heat so far : when a system absorbs or loses heat , the average kinetic energy of the molecules will change . thus , heat transfer results in a change in the system 's temperature as long as the system is not undergoing a phase change . the change in temperature resulting from heat transferred to or from a system depends on how many molecules are in the system . we can use a thermometer to measure the change in a system 's temperature . how can we use the change in temperature to calculate the heat transferred ? in order to figure out how the heat transferred to a system will change the temperature of the system , we need to know at least $ 2 $ things : the number of molecules in the system the heat capacity of the system the heat capacity tells us how much energy is needed to change the temperature of a given substance assuming that no phase changes are occurring . there are two main ways that heat capacity is reported . the specific heat capacity ( also called specific heat ) , represented by the symbol $ \text c $ or $ \text c $ , is how much energy is needed to increase the temperature of one gram of a substance by $ 1~^ { \circ } \text c $ or $ 1\ , \text k $ . specific heat capacity usually has units of $ \dfrac { \text j } { \text { grams } \cdot\text k } $ . the molar heat capacity , $ \text c_\text m $ or $ \text c_ { \text { mol } } $ , measures the amount of thermal energy it takes to raise the temperature of one mole of a substance by $ 1~^ { \circ } \text c $ or $ 1\ , \text k $ , and it usually has units of $ \dfrac { \text j } { \text { mol } \cdot\text k } $ . for example , the heat capacity of lead might be given as the specific heat capacity , $ 0.129\ , \dfrac { \text j } { \text { g } \cdot\text k } $ , or the molar heat capacity , $ 26.65\ , \dfrac { \text j } { \text { mol } \cdot\text k } $ . calculating $ \text q $ using the heat capacity we can use the heat capacity to determine the heat released or absorbed by a material using the following formula : $ \text q = \text { m } \times \text c \times \delta \text t $ where $ \text { m } $ is the mass of the substance ( in grams ) , $ \text { c } $ is the specific heat capacity , and $ \delta \text t $ is the change in temperature during the heat transfer . note that both mass and specific heat capacity can only have positive values , so the sign of $ \text q $ will depend on the sign of $ \delta \text t $ . we can calculate $ \delta \text t $ using the following equation : $ \delta \text t=\text t_ { \text { final } } -\text t_ { \text { initial } } $ where $ \text t_ { \text { final } } $ and $ \text t_ { \text { initial } } $ can have units of either $ ~^ { \circ } \text c $ or $ \text k $ . based on this equation , if $ \text q $ is positive ( energy of the system increases ) , then our system increases in temperature and $ \text t_ { \text { final } } & gt ; \text t_ { \text { initial } } $ . if $ \text q $ is negative ( energy of the system decreases ) , then our system 's temperature decreases and $ \text t_ { \text { final } } & lt ; \text t_ { \text { initial } } $ . example problem : cooling a cup of tea let 's say that we have $ 250\ , \text { ml } $ of hot tea which we would like to cool down before we try to drink it . the tea is currently at $ 370\ , \text k $ , and we 'd like to cool it down to $ 350\ , \text k $ . how much thermal energy has to be transferred from the tea to the surroundings to cool the tea ? we are going to assume that the tea is mostly water , so we can use the density and heat capacity of water in our calculations . the specific heat capacity of water is $ 4.18\ , \dfrac { \text j } { \text g \cdot \text k } $ , and the density of water is $ 1.00\ , \dfrac { \text g } { \text { ml } } $ . we can calculate the energy transferred in the process of cooling the tea using the following steps : 1 . calculate the mass of the substance we can calculate the mass of the tea/water using the volume and density of water : $ \text m=250\ , \cancel { \text { ml } } \times 1.00\ , \dfrac { \text g } { \cancel { \text { ml } } } =250\ , \text g $ 2 . calculate the change in temperature , $ \delta \text t $ we can calculate the change in temperature , $ \delta \text t $ , from the initial and final temperatures : $ \begin { align } \delta \text t & amp ; =\text t_ { \text { final } } -\text t_ { \text { initial } } \ \ & amp ; =350\ , \text k-370\ , \text k\ \ & amp ; =-20\ , \text k\end { align } $ since the temperature of the tea is decreasing and $ \delta \text t $ is negative , we would expect $ \text q $ to also be negative since our system is losing thermal energy . 3 . solve for $ \text q $ now we can solve for the heat transferred from the hot tea using the equation for heat : $ \begin { align } \text q & amp ; = \text { m } \times \text c \times \delta \text t\ & amp ; =250\ , \cancel { \text g } \times4.18\ , \dfrac { \text j } { \cancel { \text g } \cdot \cancel { \text k } } \times -20\ , \cancel { \text k } \ & amp ; =-21000\ , \text j\end { align } $ thus , we calculated that the tea will transfer $ 21000\ , \text j $ of energy to the surroundings when it cools down from $ 370\ , \text k $ to $ 350\ , \text k $ . conclusions in thermodynamics , heat and temperature are closely related concepts with precise definitions . heat , $ \text q $ , is thermal energy transferred from a hotter system to a cooler system that are in contact . temperature is a measure of the average kinetic energy of the atoms or molecules in the system . the zeroth law of thermodynamics says that no heat is transferred between two objects in thermal equilibrium ; therefore , they are the same temperature . we can calculate the heat released or absorbed using the specific heat capacity $ \text c $ , the mass of the substance $ \text m $ , and the change in temperature $ \delta \text t $ in the following equation : $ \text q = \text { m } \times \text c \times \delta \text t $
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the water molecules in a cup of hot coffee have a higher average kinetic energy than the water molecules in a cup of iced tea , which also means they are moving at a higher velocity . temperature is also an intensive property , which means that the temperature does n't change no matter how much of a substance you have ( as long as it is all at the same temperature ! ) . this is why chemists can use the melting point to help identify a pure substance $ - $ the temperature at which it melts is a property of the substance with no dependence on the mass of a sample .
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do the substances of same temperature and mass have same amount of internal energy ?
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key points heat , $ \text q $ , is thermal energy transferred from a hotter system to a cooler system that are in contact . temperature is a measure of the average kinetic energy of the atoms or molecules in the system . the zeroth law of thermodynamics says that no heat is transferred between two objects in thermal equilibrium ; therefore , they are the same temperature . we can calculate the heat released or absorbed using the specific heat capacity $ \text c $ , the mass of the substance $ \text m $ , and the change in temperature $ \delta \text t $ in the equation : $ \text q = \text { m } \times \text c \times \delta \text t $ heat in thermodynamics what contains more heat , a cup of coffee or a glass of iced tea ? in chemistry class , that would be a trick question ( sorry ! ) . in thermodynamics , heat has a very specific meaning that is different from how we might use the word in everyday speech . scientists define heat as thermal energy transferred between two systems at different temperatures that come in contact . heat is written with the symbol q or q , and it has units of joules ( $ \text j $ ) . heat is sometimes called a process quantity , because it is defined in the context of a process by which energy can be transferred . we do n't talk about a cup of coffee containing heat , but we can talk about the heat transferred from the cup of hot coffee to your hand . heat is also an extensive property , so the change in temperature resulting from heat transferred to a system depends on how many molecules are in the system . relationship between heat and temperature heat and temperature are two different but closely related concepts . note that they have different units : temperature typically has units of degrees celsius ( $ ^\circ\text c $ ) or kelvin ( $ \text k $ ) , and heat has units of energy , joules ( $ \text j $ ) . temperature is a measure of the average kinetic energy of the atoms or molecules in the system . the water molecules in a cup of hot coffee have a higher average kinetic energy than the water molecules in a cup of iced tea , which also means they are moving at a higher velocity . temperature is also an intensive property , which means that the temperature does n't change no matter how much of a substance you have ( as long as it is all at the same temperature ! ) . this is why chemists can use the melting point to help identify a pure substance $ - $ the temperature at which it melts is a property of the substance with no dependence on the mass of a sample . on an atomic level , the molecules in each object are constantly in motion and colliding with each other . every time molecules collide , kinetic energy can be transferred . when the two systems are in contact , heat will be transferred through molecular collisions from the hotter system to the cooler system . the thermal energy will flow in that direction until the two objects are at the same temperature . when the two systems in contact are at the same temperature , we say they are in thermal equilibrium . zeroth law of thermodynamics : defining thermal equilibrium the zeroth law of thermodynamics defines thermal equilibrium within an isolated system . the zeroth law says when two objects at thermal equilibrium are in contact , there is no net heat transfer between the objects ; therefore , they are the same temperature . another way to state the zeroth law is to say that if two objects are both separately in thermal equilibrium with a third object , then they are in thermal equilibrium with each other . the zeroth law allows us to measure the temperature of objects . any time we use a thermometer , we are using the zeroth law of thermodynamics . let 's say we are measuring the temperature of a water bath . in order to make sure the reading is accurate , we usually want to wait for the temperature reading to stay constant . we are waiting for the thermometer and the water to reach thermal equilibrium ! at thermal equilibrium , the temperature of the thermometer bulb and the water bath will be the same , and there should be no net heat transfer from one object to the other ( assuming no other loss of heat to the surroundings ) . heat capacity : converting between heat and change in temperature how can we measure heat ? here are some things we know about heat so far : when a system absorbs or loses heat , the average kinetic energy of the molecules will change . thus , heat transfer results in a change in the system 's temperature as long as the system is not undergoing a phase change . the change in temperature resulting from heat transferred to or from a system depends on how many molecules are in the system . we can use a thermometer to measure the change in a system 's temperature . how can we use the change in temperature to calculate the heat transferred ? in order to figure out how the heat transferred to a system will change the temperature of the system , we need to know at least $ 2 $ things : the number of molecules in the system the heat capacity of the system the heat capacity tells us how much energy is needed to change the temperature of a given substance assuming that no phase changes are occurring . there are two main ways that heat capacity is reported . the specific heat capacity ( also called specific heat ) , represented by the symbol $ \text c $ or $ \text c $ , is how much energy is needed to increase the temperature of one gram of a substance by $ 1~^ { \circ } \text c $ or $ 1\ , \text k $ . specific heat capacity usually has units of $ \dfrac { \text j } { \text { grams } \cdot\text k } $ . the molar heat capacity , $ \text c_\text m $ or $ \text c_ { \text { mol } } $ , measures the amount of thermal energy it takes to raise the temperature of one mole of a substance by $ 1~^ { \circ } \text c $ or $ 1\ , \text k $ , and it usually has units of $ \dfrac { \text j } { \text { mol } \cdot\text k } $ . for example , the heat capacity of lead might be given as the specific heat capacity , $ 0.129\ , \dfrac { \text j } { \text { g } \cdot\text k } $ , or the molar heat capacity , $ 26.65\ , \dfrac { \text j } { \text { mol } \cdot\text k } $ . calculating $ \text q $ using the heat capacity we can use the heat capacity to determine the heat released or absorbed by a material using the following formula : $ \text q = \text { m } \times \text c \times \delta \text t $ where $ \text { m } $ is the mass of the substance ( in grams ) , $ \text { c } $ is the specific heat capacity , and $ \delta \text t $ is the change in temperature during the heat transfer . note that both mass and specific heat capacity can only have positive values , so the sign of $ \text q $ will depend on the sign of $ \delta \text t $ . we can calculate $ \delta \text t $ using the following equation : $ \delta \text t=\text t_ { \text { final } } -\text t_ { \text { initial } } $ where $ \text t_ { \text { final } } $ and $ \text t_ { \text { initial } } $ can have units of either $ ~^ { \circ } \text c $ or $ \text k $ . based on this equation , if $ \text q $ is positive ( energy of the system increases ) , then our system increases in temperature and $ \text t_ { \text { final } } & gt ; \text t_ { \text { initial } } $ . if $ \text q $ is negative ( energy of the system decreases ) , then our system 's temperature decreases and $ \text t_ { \text { final } } & lt ; \text t_ { \text { initial } } $ . example problem : cooling a cup of tea let 's say that we have $ 250\ , \text { ml } $ of hot tea which we would like to cool down before we try to drink it . the tea is currently at $ 370\ , \text k $ , and we 'd like to cool it down to $ 350\ , \text k $ . how much thermal energy has to be transferred from the tea to the surroundings to cool the tea ? we are going to assume that the tea is mostly water , so we can use the density and heat capacity of water in our calculations . the specific heat capacity of water is $ 4.18\ , \dfrac { \text j } { \text g \cdot \text k } $ , and the density of water is $ 1.00\ , \dfrac { \text g } { \text { ml } } $ . we can calculate the energy transferred in the process of cooling the tea using the following steps : 1 . calculate the mass of the substance we can calculate the mass of the tea/water using the volume and density of water : $ \text m=250\ , \cancel { \text { ml } } \times 1.00\ , \dfrac { \text g } { \cancel { \text { ml } } } =250\ , \text g $ 2 . calculate the change in temperature , $ \delta \text t $ we can calculate the change in temperature , $ \delta \text t $ , from the initial and final temperatures : $ \begin { align } \delta \text t & amp ; =\text t_ { \text { final } } -\text t_ { \text { initial } } \ \ & amp ; =350\ , \text k-370\ , \text k\ \ & amp ; =-20\ , \text k\end { align } $ since the temperature of the tea is decreasing and $ \delta \text t $ is negative , we would expect $ \text q $ to also be negative since our system is losing thermal energy . 3 . solve for $ \text q $ now we can solve for the heat transferred from the hot tea using the equation for heat : $ \begin { align } \text q & amp ; = \text { m } \times \text c \times \delta \text t\ & amp ; =250\ , \cancel { \text g } \times4.18\ , \dfrac { \text j } { \cancel { \text g } \cdot \cancel { \text k } } \times -20\ , \cancel { \text k } \ & amp ; =-21000\ , \text j\end { align } $ thus , we calculated that the tea will transfer $ 21000\ , \text j $ of energy to the surroundings when it cools down from $ 370\ , \text k $ to $ 350\ , \text k $ . conclusions in thermodynamics , heat and temperature are closely related concepts with precise definitions . heat , $ \text q $ , is thermal energy transferred from a hotter system to a cooler system that are in contact . temperature is a measure of the average kinetic energy of the atoms or molecules in the system . the zeroth law of thermodynamics says that no heat is transferred between two objects in thermal equilibrium ; therefore , they are the same temperature . we can calculate the heat released or absorbed using the specific heat capacity $ \text c $ , the mass of the substance $ \text m $ , and the change in temperature $ \delta \text t $ in the following equation : $ \text q = \text { m } \times \text c \times \delta \text t $
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on an atomic level , the molecules in each object are constantly in motion and colliding with each other . every time molecules collide , kinetic energy can be transferred . when the two systems are in contact , heat will be transferred through molecular collisions from the hotter system to the cooler system .
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and also does it mean that only energy transferring between system is thermal and energy transferring between molecules are kinetic ?
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key points heat , $ \text q $ , is thermal energy transferred from a hotter system to a cooler system that are in contact . temperature is a measure of the average kinetic energy of the atoms or molecules in the system . the zeroth law of thermodynamics says that no heat is transferred between two objects in thermal equilibrium ; therefore , they are the same temperature . we can calculate the heat released or absorbed using the specific heat capacity $ \text c $ , the mass of the substance $ \text m $ , and the change in temperature $ \delta \text t $ in the equation : $ \text q = \text { m } \times \text c \times \delta \text t $ heat in thermodynamics what contains more heat , a cup of coffee or a glass of iced tea ? in chemistry class , that would be a trick question ( sorry ! ) . in thermodynamics , heat has a very specific meaning that is different from how we might use the word in everyday speech . scientists define heat as thermal energy transferred between two systems at different temperatures that come in contact . heat is written with the symbol q or q , and it has units of joules ( $ \text j $ ) . heat is sometimes called a process quantity , because it is defined in the context of a process by which energy can be transferred . we do n't talk about a cup of coffee containing heat , but we can talk about the heat transferred from the cup of hot coffee to your hand . heat is also an extensive property , so the change in temperature resulting from heat transferred to a system depends on how many molecules are in the system . relationship between heat and temperature heat and temperature are two different but closely related concepts . note that they have different units : temperature typically has units of degrees celsius ( $ ^\circ\text c $ ) or kelvin ( $ \text k $ ) , and heat has units of energy , joules ( $ \text j $ ) . temperature is a measure of the average kinetic energy of the atoms or molecules in the system . the water molecules in a cup of hot coffee have a higher average kinetic energy than the water molecules in a cup of iced tea , which also means they are moving at a higher velocity . temperature is also an intensive property , which means that the temperature does n't change no matter how much of a substance you have ( as long as it is all at the same temperature ! ) . this is why chemists can use the melting point to help identify a pure substance $ - $ the temperature at which it melts is a property of the substance with no dependence on the mass of a sample . on an atomic level , the molecules in each object are constantly in motion and colliding with each other . every time molecules collide , kinetic energy can be transferred . when the two systems are in contact , heat will be transferred through molecular collisions from the hotter system to the cooler system . the thermal energy will flow in that direction until the two objects are at the same temperature . when the two systems in contact are at the same temperature , we say they are in thermal equilibrium . zeroth law of thermodynamics : defining thermal equilibrium the zeroth law of thermodynamics defines thermal equilibrium within an isolated system . the zeroth law says when two objects at thermal equilibrium are in contact , there is no net heat transfer between the objects ; therefore , they are the same temperature . another way to state the zeroth law is to say that if two objects are both separately in thermal equilibrium with a third object , then they are in thermal equilibrium with each other . the zeroth law allows us to measure the temperature of objects . any time we use a thermometer , we are using the zeroth law of thermodynamics . let 's say we are measuring the temperature of a water bath . in order to make sure the reading is accurate , we usually want to wait for the temperature reading to stay constant . we are waiting for the thermometer and the water to reach thermal equilibrium ! at thermal equilibrium , the temperature of the thermometer bulb and the water bath will be the same , and there should be no net heat transfer from one object to the other ( assuming no other loss of heat to the surroundings ) . heat capacity : converting between heat and change in temperature how can we measure heat ? here are some things we know about heat so far : when a system absorbs or loses heat , the average kinetic energy of the molecules will change . thus , heat transfer results in a change in the system 's temperature as long as the system is not undergoing a phase change . the change in temperature resulting from heat transferred to or from a system depends on how many molecules are in the system . we can use a thermometer to measure the change in a system 's temperature . how can we use the change in temperature to calculate the heat transferred ? in order to figure out how the heat transferred to a system will change the temperature of the system , we need to know at least $ 2 $ things : the number of molecules in the system the heat capacity of the system the heat capacity tells us how much energy is needed to change the temperature of a given substance assuming that no phase changes are occurring . there are two main ways that heat capacity is reported . the specific heat capacity ( also called specific heat ) , represented by the symbol $ \text c $ or $ \text c $ , is how much energy is needed to increase the temperature of one gram of a substance by $ 1~^ { \circ } \text c $ or $ 1\ , \text k $ . specific heat capacity usually has units of $ \dfrac { \text j } { \text { grams } \cdot\text k } $ . the molar heat capacity , $ \text c_\text m $ or $ \text c_ { \text { mol } } $ , measures the amount of thermal energy it takes to raise the temperature of one mole of a substance by $ 1~^ { \circ } \text c $ or $ 1\ , \text k $ , and it usually has units of $ \dfrac { \text j } { \text { mol } \cdot\text k } $ . for example , the heat capacity of lead might be given as the specific heat capacity , $ 0.129\ , \dfrac { \text j } { \text { g } \cdot\text k } $ , or the molar heat capacity , $ 26.65\ , \dfrac { \text j } { \text { mol } \cdot\text k } $ . calculating $ \text q $ using the heat capacity we can use the heat capacity to determine the heat released or absorbed by a material using the following formula : $ \text q = \text { m } \times \text c \times \delta \text t $ where $ \text { m } $ is the mass of the substance ( in grams ) , $ \text { c } $ is the specific heat capacity , and $ \delta \text t $ is the change in temperature during the heat transfer . note that both mass and specific heat capacity can only have positive values , so the sign of $ \text q $ will depend on the sign of $ \delta \text t $ . we can calculate $ \delta \text t $ using the following equation : $ \delta \text t=\text t_ { \text { final } } -\text t_ { \text { initial } } $ where $ \text t_ { \text { final } } $ and $ \text t_ { \text { initial } } $ can have units of either $ ~^ { \circ } \text c $ or $ \text k $ . based on this equation , if $ \text q $ is positive ( energy of the system increases ) , then our system increases in temperature and $ \text t_ { \text { final } } & gt ; \text t_ { \text { initial } } $ . if $ \text q $ is negative ( energy of the system decreases ) , then our system 's temperature decreases and $ \text t_ { \text { final } } & lt ; \text t_ { \text { initial } } $ . example problem : cooling a cup of tea let 's say that we have $ 250\ , \text { ml } $ of hot tea which we would like to cool down before we try to drink it . the tea is currently at $ 370\ , \text k $ , and we 'd like to cool it down to $ 350\ , \text k $ . how much thermal energy has to be transferred from the tea to the surroundings to cool the tea ? we are going to assume that the tea is mostly water , so we can use the density and heat capacity of water in our calculations . the specific heat capacity of water is $ 4.18\ , \dfrac { \text j } { \text g \cdot \text k } $ , and the density of water is $ 1.00\ , \dfrac { \text g } { \text { ml } } $ . we can calculate the energy transferred in the process of cooling the tea using the following steps : 1 . calculate the mass of the substance we can calculate the mass of the tea/water using the volume and density of water : $ \text m=250\ , \cancel { \text { ml } } \times 1.00\ , \dfrac { \text g } { \cancel { \text { ml } } } =250\ , \text g $ 2 . calculate the change in temperature , $ \delta \text t $ we can calculate the change in temperature , $ \delta \text t $ , from the initial and final temperatures : $ \begin { align } \delta \text t & amp ; =\text t_ { \text { final } } -\text t_ { \text { initial } } \ \ & amp ; =350\ , \text k-370\ , \text k\ \ & amp ; =-20\ , \text k\end { align } $ since the temperature of the tea is decreasing and $ \delta \text t $ is negative , we would expect $ \text q $ to also be negative since our system is losing thermal energy . 3 . solve for $ \text q $ now we can solve for the heat transferred from the hot tea using the equation for heat : $ \begin { align } \text q & amp ; = \text { m } \times \text c \times \delta \text t\ & amp ; =250\ , \cancel { \text g } \times4.18\ , \dfrac { \text j } { \cancel { \text g } \cdot \cancel { \text k } } \times -20\ , \cancel { \text k } \ & amp ; =-21000\ , \text j\end { align } $ thus , we calculated that the tea will transfer $ 21000\ , \text j $ of energy to the surroundings when it cools down from $ 370\ , \text k $ to $ 350\ , \text k $ . conclusions in thermodynamics , heat and temperature are closely related concepts with precise definitions . heat , $ \text q $ , is thermal energy transferred from a hotter system to a cooler system that are in contact . temperature is a measure of the average kinetic energy of the atoms or molecules in the system . the zeroth law of thermodynamics says that no heat is transferred between two objects in thermal equilibrium ; therefore , they are the same temperature . we can calculate the heat released or absorbed using the specific heat capacity $ \text c $ , the mass of the substance $ \text m $ , and the change in temperature $ \delta \text t $ in the following equation : $ \text q = \text { m } \times \text c \times \delta \text t $
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heat is also an extensive property , so the change in temperature resulting from heat transferred to a system depends on how many molecules are in the system . relationship between heat and temperature heat and temperature are two different but closely related concepts . note that they have different units : temperature typically has units of degrees celsius ( $ ^\circ\text c $ ) or kelvin ( $ \text k $ ) , and heat has units of energy , joules ( $ \text j $ ) .
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so if the substance is not pure it would not have the same amount of temperature in different type of molecules ?
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key points heat , $ \text q $ , is thermal energy transferred from a hotter system to a cooler system that are in contact . temperature is a measure of the average kinetic energy of the atoms or molecules in the system . the zeroth law of thermodynamics says that no heat is transferred between two objects in thermal equilibrium ; therefore , they are the same temperature . we can calculate the heat released or absorbed using the specific heat capacity $ \text c $ , the mass of the substance $ \text m $ , and the change in temperature $ \delta \text t $ in the equation : $ \text q = \text { m } \times \text c \times \delta \text t $ heat in thermodynamics what contains more heat , a cup of coffee or a glass of iced tea ? in chemistry class , that would be a trick question ( sorry ! ) . in thermodynamics , heat has a very specific meaning that is different from how we might use the word in everyday speech . scientists define heat as thermal energy transferred between two systems at different temperatures that come in contact . heat is written with the symbol q or q , and it has units of joules ( $ \text j $ ) . heat is sometimes called a process quantity , because it is defined in the context of a process by which energy can be transferred . we do n't talk about a cup of coffee containing heat , but we can talk about the heat transferred from the cup of hot coffee to your hand . heat is also an extensive property , so the change in temperature resulting from heat transferred to a system depends on how many molecules are in the system . relationship between heat and temperature heat and temperature are two different but closely related concepts . note that they have different units : temperature typically has units of degrees celsius ( $ ^\circ\text c $ ) or kelvin ( $ \text k $ ) , and heat has units of energy , joules ( $ \text j $ ) . temperature is a measure of the average kinetic energy of the atoms or molecules in the system . the water molecules in a cup of hot coffee have a higher average kinetic energy than the water molecules in a cup of iced tea , which also means they are moving at a higher velocity . temperature is also an intensive property , which means that the temperature does n't change no matter how much of a substance you have ( as long as it is all at the same temperature ! ) . this is why chemists can use the melting point to help identify a pure substance $ - $ the temperature at which it melts is a property of the substance with no dependence on the mass of a sample . on an atomic level , the molecules in each object are constantly in motion and colliding with each other . every time molecules collide , kinetic energy can be transferred . when the two systems are in contact , heat will be transferred through molecular collisions from the hotter system to the cooler system . the thermal energy will flow in that direction until the two objects are at the same temperature . when the two systems in contact are at the same temperature , we say they are in thermal equilibrium . zeroth law of thermodynamics : defining thermal equilibrium the zeroth law of thermodynamics defines thermal equilibrium within an isolated system . the zeroth law says when two objects at thermal equilibrium are in contact , there is no net heat transfer between the objects ; therefore , they are the same temperature . another way to state the zeroth law is to say that if two objects are both separately in thermal equilibrium with a third object , then they are in thermal equilibrium with each other . the zeroth law allows us to measure the temperature of objects . any time we use a thermometer , we are using the zeroth law of thermodynamics . let 's say we are measuring the temperature of a water bath . in order to make sure the reading is accurate , we usually want to wait for the temperature reading to stay constant . we are waiting for the thermometer and the water to reach thermal equilibrium ! at thermal equilibrium , the temperature of the thermometer bulb and the water bath will be the same , and there should be no net heat transfer from one object to the other ( assuming no other loss of heat to the surroundings ) . heat capacity : converting between heat and change in temperature how can we measure heat ? here are some things we know about heat so far : when a system absorbs or loses heat , the average kinetic energy of the molecules will change . thus , heat transfer results in a change in the system 's temperature as long as the system is not undergoing a phase change . the change in temperature resulting from heat transferred to or from a system depends on how many molecules are in the system . we can use a thermometer to measure the change in a system 's temperature . how can we use the change in temperature to calculate the heat transferred ? in order to figure out how the heat transferred to a system will change the temperature of the system , we need to know at least $ 2 $ things : the number of molecules in the system the heat capacity of the system the heat capacity tells us how much energy is needed to change the temperature of a given substance assuming that no phase changes are occurring . there are two main ways that heat capacity is reported . the specific heat capacity ( also called specific heat ) , represented by the symbol $ \text c $ or $ \text c $ , is how much energy is needed to increase the temperature of one gram of a substance by $ 1~^ { \circ } \text c $ or $ 1\ , \text k $ . specific heat capacity usually has units of $ \dfrac { \text j } { \text { grams } \cdot\text k } $ . the molar heat capacity , $ \text c_\text m $ or $ \text c_ { \text { mol } } $ , measures the amount of thermal energy it takes to raise the temperature of one mole of a substance by $ 1~^ { \circ } \text c $ or $ 1\ , \text k $ , and it usually has units of $ \dfrac { \text j } { \text { mol } \cdot\text k } $ . for example , the heat capacity of lead might be given as the specific heat capacity , $ 0.129\ , \dfrac { \text j } { \text { g } \cdot\text k } $ , or the molar heat capacity , $ 26.65\ , \dfrac { \text j } { \text { mol } \cdot\text k } $ . calculating $ \text q $ using the heat capacity we can use the heat capacity to determine the heat released or absorbed by a material using the following formula : $ \text q = \text { m } \times \text c \times \delta \text t $ where $ \text { m } $ is the mass of the substance ( in grams ) , $ \text { c } $ is the specific heat capacity , and $ \delta \text t $ is the change in temperature during the heat transfer . note that both mass and specific heat capacity can only have positive values , so the sign of $ \text q $ will depend on the sign of $ \delta \text t $ . we can calculate $ \delta \text t $ using the following equation : $ \delta \text t=\text t_ { \text { final } } -\text t_ { \text { initial } } $ where $ \text t_ { \text { final } } $ and $ \text t_ { \text { initial } } $ can have units of either $ ~^ { \circ } \text c $ or $ \text k $ . based on this equation , if $ \text q $ is positive ( energy of the system increases ) , then our system increases in temperature and $ \text t_ { \text { final } } & gt ; \text t_ { \text { initial } } $ . if $ \text q $ is negative ( energy of the system decreases ) , then our system 's temperature decreases and $ \text t_ { \text { final } } & lt ; \text t_ { \text { initial } } $ . example problem : cooling a cup of tea let 's say that we have $ 250\ , \text { ml } $ of hot tea which we would like to cool down before we try to drink it . the tea is currently at $ 370\ , \text k $ , and we 'd like to cool it down to $ 350\ , \text k $ . how much thermal energy has to be transferred from the tea to the surroundings to cool the tea ? we are going to assume that the tea is mostly water , so we can use the density and heat capacity of water in our calculations . the specific heat capacity of water is $ 4.18\ , \dfrac { \text j } { \text g \cdot \text k } $ , and the density of water is $ 1.00\ , \dfrac { \text g } { \text { ml } } $ . we can calculate the energy transferred in the process of cooling the tea using the following steps : 1 . calculate the mass of the substance we can calculate the mass of the tea/water using the volume and density of water : $ \text m=250\ , \cancel { \text { ml } } \times 1.00\ , \dfrac { \text g } { \cancel { \text { ml } } } =250\ , \text g $ 2 . calculate the change in temperature , $ \delta \text t $ we can calculate the change in temperature , $ \delta \text t $ , from the initial and final temperatures : $ \begin { align } \delta \text t & amp ; =\text t_ { \text { final } } -\text t_ { \text { initial } } \ \ & amp ; =350\ , \text k-370\ , \text k\ \ & amp ; =-20\ , \text k\end { align } $ since the temperature of the tea is decreasing and $ \delta \text t $ is negative , we would expect $ \text q $ to also be negative since our system is losing thermal energy . 3 . solve for $ \text q $ now we can solve for the heat transferred from the hot tea using the equation for heat : $ \begin { align } \text q & amp ; = \text { m } \times \text c \times \delta \text t\ & amp ; =250\ , \cancel { \text g } \times4.18\ , \dfrac { \text j } { \cancel { \text g } \cdot \cancel { \text k } } \times -20\ , \cancel { \text k } \ & amp ; =-21000\ , \text j\end { align } $ thus , we calculated that the tea will transfer $ 21000\ , \text j $ of energy to the surroundings when it cools down from $ 370\ , \text k $ to $ 350\ , \text k $ . conclusions in thermodynamics , heat and temperature are closely related concepts with precise definitions . heat , $ \text q $ , is thermal energy transferred from a hotter system to a cooler system that are in contact . temperature is a measure of the average kinetic energy of the atoms or molecules in the system . the zeroth law of thermodynamics says that no heat is transferred between two objects in thermal equilibrium ; therefore , they are the same temperature . we can calculate the heat released or absorbed using the specific heat capacity $ \text c $ , the mass of the substance $ \text m $ , and the change in temperature $ \delta \text t $ in the following equation : $ \text q = \text { m } \times \text c \times \delta \text t $
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note that both mass and specific heat capacity can only have positive values , so the sign of $ \text q $ will depend on the sign of $ \delta \text t $ . we can calculate $ \delta \text t $ using the following equation : $ \delta \text t=\text t_ { \text { final } } -\text t_ { \text { initial } } $ where $ \text t_ { \text { final } } $ and $ \text t_ { \text { initial } } $ can have units of either $ ~^ { \circ } \text c $ or $ \text k $ . based on this equation , if $ \text q $ is positive ( energy of the system increases ) , then our system increases in temperature and $ \text t_ { \text { final } } & gt ; \text t_ { \text { initial } } $ .
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is it possible for me to request a copy of these notes either via email or a link to be able to download for reference when studying ?
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key points heat , $ \text q $ , is thermal energy transferred from a hotter system to a cooler system that are in contact . temperature is a measure of the average kinetic energy of the atoms or molecules in the system . the zeroth law of thermodynamics says that no heat is transferred between two objects in thermal equilibrium ; therefore , they are the same temperature . we can calculate the heat released or absorbed using the specific heat capacity $ \text c $ , the mass of the substance $ \text m $ , and the change in temperature $ \delta \text t $ in the equation : $ \text q = \text { m } \times \text c \times \delta \text t $ heat in thermodynamics what contains more heat , a cup of coffee or a glass of iced tea ? in chemistry class , that would be a trick question ( sorry ! ) . in thermodynamics , heat has a very specific meaning that is different from how we might use the word in everyday speech . scientists define heat as thermal energy transferred between two systems at different temperatures that come in contact . heat is written with the symbol q or q , and it has units of joules ( $ \text j $ ) . heat is sometimes called a process quantity , because it is defined in the context of a process by which energy can be transferred . we do n't talk about a cup of coffee containing heat , but we can talk about the heat transferred from the cup of hot coffee to your hand . heat is also an extensive property , so the change in temperature resulting from heat transferred to a system depends on how many molecules are in the system . relationship between heat and temperature heat and temperature are two different but closely related concepts . note that they have different units : temperature typically has units of degrees celsius ( $ ^\circ\text c $ ) or kelvin ( $ \text k $ ) , and heat has units of energy , joules ( $ \text j $ ) . temperature is a measure of the average kinetic energy of the atoms or molecules in the system . the water molecules in a cup of hot coffee have a higher average kinetic energy than the water molecules in a cup of iced tea , which also means they are moving at a higher velocity . temperature is also an intensive property , which means that the temperature does n't change no matter how much of a substance you have ( as long as it is all at the same temperature ! ) . this is why chemists can use the melting point to help identify a pure substance $ - $ the temperature at which it melts is a property of the substance with no dependence on the mass of a sample . on an atomic level , the molecules in each object are constantly in motion and colliding with each other . every time molecules collide , kinetic energy can be transferred . when the two systems are in contact , heat will be transferred through molecular collisions from the hotter system to the cooler system . the thermal energy will flow in that direction until the two objects are at the same temperature . when the two systems in contact are at the same temperature , we say they are in thermal equilibrium . zeroth law of thermodynamics : defining thermal equilibrium the zeroth law of thermodynamics defines thermal equilibrium within an isolated system . the zeroth law says when two objects at thermal equilibrium are in contact , there is no net heat transfer between the objects ; therefore , they are the same temperature . another way to state the zeroth law is to say that if two objects are both separately in thermal equilibrium with a third object , then they are in thermal equilibrium with each other . the zeroth law allows us to measure the temperature of objects . any time we use a thermometer , we are using the zeroth law of thermodynamics . let 's say we are measuring the temperature of a water bath . in order to make sure the reading is accurate , we usually want to wait for the temperature reading to stay constant . we are waiting for the thermometer and the water to reach thermal equilibrium ! at thermal equilibrium , the temperature of the thermometer bulb and the water bath will be the same , and there should be no net heat transfer from one object to the other ( assuming no other loss of heat to the surroundings ) . heat capacity : converting between heat and change in temperature how can we measure heat ? here are some things we know about heat so far : when a system absorbs or loses heat , the average kinetic energy of the molecules will change . thus , heat transfer results in a change in the system 's temperature as long as the system is not undergoing a phase change . the change in temperature resulting from heat transferred to or from a system depends on how many molecules are in the system . we can use a thermometer to measure the change in a system 's temperature . how can we use the change in temperature to calculate the heat transferred ? in order to figure out how the heat transferred to a system will change the temperature of the system , we need to know at least $ 2 $ things : the number of molecules in the system the heat capacity of the system the heat capacity tells us how much energy is needed to change the temperature of a given substance assuming that no phase changes are occurring . there are two main ways that heat capacity is reported . the specific heat capacity ( also called specific heat ) , represented by the symbol $ \text c $ or $ \text c $ , is how much energy is needed to increase the temperature of one gram of a substance by $ 1~^ { \circ } \text c $ or $ 1\ , \text k $ . specific heat capacity usually has units of $ \dfrac { \text j } { \text { grams } \cdot\text k } $ . the molar heat capacity , $ \text c_\text m $ or $ \text c_ { \text { mol } } $ , measures the amount of thermal energy it takes to raise the temperature of one mole of a substance by $ 1~^ { \circ } \text c $ or $ 1\ , \text k $ , and it usually has units of $ \dfrac { \text j } { \text { mol } \cdot\text k } $ . for example , the heat capacity of lead might be given as the specific heat capacity , $ 0.129\ , \dfrac { \text j } { \text { g } \cdot\text k } $ , or the molar heat capacity , $ 26.65\ , \dfrac { \text j } { \text { mol } \cdot\text k } $ . calculating $ \text q $ using the heat capacity we can use the heat capacity to determine the heat released or absorbed by a material using the following formula : $ \text q = \text { m } \times \text c \times \delta \text t $ where $ \text { m } $ is the mass of the substance ( in grams ) , $ \text { c } $ is the specific heat capacity , and $ \delta \text t $ is the change in temperature during the heat transfer . note that both mass and specific heat capacity can only have positive values , so the sign of $ \text q $ will depend on the sign of $ \delta \text t $ . we can calculate $ \delta \text t $ using the following equation : $ \delta \text t=\text t_ { \text { final } } -\text t_ { \text { initial } } $ where $ \text t_ { \text { final } } $ and $ \text t_ { \text { initial } } $ can have units of either $ ~^ { \circ } \text c $ or $ \text k $ . based on this equation , if $ \text q $ is positive ( energy of the system increases ) , then our system increases in temperature and $ \text t_ { \text { final } } & gt ; \text t_ { \text { initial } } $ . if $ \text q $ is negative ( energy of the system decreases ) , then our system 's temperature decreases and $ \text t_ { \text { final } } & lt ; \text t_ { \text { initial } } $ . example problem : cooling a cup of tea let 's say that we have $ 250\ , \text { ml } $ of hot tea which we would like to cool down before we try to drink it . the tea is currently at $ 370\ , \text k $ , and we 'd like to cool it down to $ 350\ , \text k $ . how much thermal energy has to be transferred from the tea to the surroundings to cool the tea ? we are going to assume that the tea is mostly water , so we can use the density and heat capacity of water in our calculations . the specific heat capacity of water is $ 4.18\ , \dfrac { \text j } { \text g \cdot \text k } $ , and the density of water is $ 1.00\ , \dfrac { \text g } { \text { ml } } $ . we can calculate the energy transferred in the process of cooling the tea using the following steps : 1 . calculate the mass of the substance we can calculate the mass of the tea/water using the volume and density of water : $ \text m=250\ , \cancel { \text { ml } } \times 1.00\ , \dfrac { \text g } { \cancel { \text { ml } } } =250\ , \text g $ 2 . calculate the change in temperature , $ \delta \text t $ we can calculate the change in temperature , $ \delta \text t $ , from the initial and final temperatures : $ \begin { align } \delta \text t & amp ; =\text t_ { \text { final } } -\text t_ { \text { initial } } \ \ & amp ; =350\ , \text k-370\ , \text k\ \ & amp ; =-20\ , \text k\end { align } $ since the temperature of the tea is decreasing and $ \delta \text t $ is negative , we would expect $ \text q $ to also be negative since our system is losing thermal energy . 3 . solve for $ \text q $ now we can solve for the heat transferred from the hot tea using the equation for heat : $ \begin { align } \text q & amp ; = \text { m } \times \text c \times \delta \text t\ & amp ; =250\ , \cancel { \text g } \times4.18\ , \dfrac { \text j } { \cancel { \text g } \cdot \cancel { \text k } } \times -20\ , \cancel { \text k } \ & amp ; =-21000\ , \text j\end { align } $ thus , we calculated that the tea will transfer $ 21000\ , \text j $ of energy to the surroundings when it cools down from $ 370\ , \text k $ to $ 350\ , \text k $ . conclusions in thermodynamics , heat and temperature are closely related concepts with precise definitions . heat , $ \text q $ , is thermal energy transferred from a hotter system to a cooler system that are in contact . temperature is a measure of the average kinetic energy of the atoms or molecules in the system . the zeroth law of thermodynamics says that no heat is transferred between two objects in thermal equilibrium ; therefore , they are the same temperature . we can calculate the heat released or absorbed using the specific heat capacity $ \text c $ , the mass of the substance $ \text m $ , and the change in temperature $ \delta \text t $ in the following equation : $ \text q = \text { m } \times \text c \times \delta \text t $
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if $ \text q $ is negative ( energy of the system decreases ) , then our system 's temperature decreases and $ \text t_ { \text { final } } & lt ; \text t_ { \text { initial } } $ . example problem : cooling a cup of tea let 's say that we have $ 250\ , \text { ml } $ of hot tea which we would like to cool down before we try to drink it . the tea is currently at $ 370\ , \text k $ , and we 'd like to cool it down to $ 350\ , \text k $ .
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for the example problem with `` cooling a cup of tea '' , now that we figured out how much heat needs to be transferred to the surroundings , is there a way to figure out how much time that would take using just the information given , or would more information be needed to figure that out ?
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key points heat , $ \text q $ , is thermal energy transferred from a hotter system to a cooler system that are in contact . temperature is a measure of the average kinetic energy of the atoms or molecules in the system . the zeroth law of thermodynamics says that no heat is transferred between two objects in thermal equilibrium ; therefore , they are the same temperature . we can calculate the heat released or absorbed using the specific heat capacity $ \text c $ , the mass of the substance $ \text m $ , and the change in temperature $ \delta \text t $ in the equation : $ \text q = \text { m } \times \text c \times \delta \text t $ heat in thermodynamics what contains more heat , a cup of coffee or a glass of iced tea ? in chemistry class , that would be a trick question ( sorry ! ) . in thermodynamics , heat has a very specific meaning that is different from how we might use the word in everyday speech . scientists define heat as thermal energy transferred between two systems at different temperatures that come in contact . heat is written with the symbol q or q , and it has units of joules ( $ \text j $ ) . heat is sometimes called a process quantity , because it is defined in the context of a process by which energy can be transferred . we do n't talk about a cup of coffee containing heat , but we can talk about the heat transferred from the cup of hot coffee to your hand . heat is also an extensive property , so the change in temperature resulting from heat transferred to a system depends on how many molecules are in the system . relationship between heat and temperature heat and temperature are two different but closely related concepts . note that they have different units : temperature typically has units of degrees celsius ( $ ^\circ\text c $ ) or kelvin ( $ \text k $ ) , and heat has units of energy , joules ( $ \text j $ ) . temperature is a measure of the average kinetic energy of the atoms or molecules in the system . the water molecules in a cup of hot coffee have a higher average kinetic energy than the water molecules in a cup of iced tea , which also means they are moving at a higher velocity . temperature is also an intensive property , which means that the temperature does n't change no matter how much of a substance you have ( as long as it is all at the same temperature ! ) . this is why chemists can use the melting point to help identify a pure substance $ - $ the temperature at which it melts is a property of the substance with no dependence on the mass of a sample . on an atomic level , the molecules in each object are constantly in motion and colliding with each other . every time molecules collide , kinetic energy can be transferred . when the two systems are in contact , heat will be transferred through molecular collisions from the hotter system to the cooler system . the thermal energy will flow in that direction until the two objects are at the same temperature . when the two systems in contact are at the same temperature , we say they are in thermal equilibrium . zeroth law of thermodynamics : defining thermal equilibrium the zeroth law of thermodynamics defines thermal equilibrium within an isolated system . the zeroth law says when two objects at thermal equilibrium are in contact , there is no net heat transfer between the objects ; therefore , they are the same temperature . another way to state the zeroth law is to say that if two objects are both separately in thermal equilibrium with a third object , then they are in thermal equilibrium with each other . the zeroth law allows us to measure the temperature of objects . any time we use a thermometer , we are using the zeroth law of thermodynamics . let 's say we are measuring the temperature of a water bath . in order to make sure the reading is accurate , we usually want to wait for the temperature reading to stay constant . we are waiting for the thermometer and the water to reach thermal equilibrium ! at thermal equilibrium , the temperature of the thermometer bulb and the water bath will be the same , and there should be no net heat transfer from one object to the other ( assuming no other loss of heat to the surroundings ) . heat capacity : converting between heat and change in temperature how can we measure heat ? here are some things we know about heat so far : when a system absorbs or loses heat , the average kinetic energy of the molecules will change . thus , heat transfer results in a change in the system 's temperature as long as the system is not undergoing a phase change . the change in temperature resulting from heat transferred to or from a system depends on how many molecules are in the system . we can use a thermometer to measure the change in a system 's temperature . how can we use the change in temperature to calculate the heat transferred ? in order to figure out how the heat transferred to a system will change the temperature of the system , we need to know at least $ 2 $ things : the number of molecules in the system the heat capacity of the system the heat capacity tells us how much energy is needed to change the temperature of a given substance assuming that no phase changes are occurring . there are two main ways that heat capacity is reported . the specific heat capacity ( also called specific heat ) , represented by the symbol $ \text c $ or $ \text c $ , is how much energy is needed to increase the temperature of one gram of a substance by $ 1~^ { \circ } \text c $ or $ 1\ , \text k $ . specific heat capacity usually has units of $ \dfrac { \text j } { \text { grams } \cdot\text k } $ . the molar heat capacity , $ \text c_\text m $ or $ \text c_ { \text { mol } } $ , measures the amount of thermal energy it takes to raise the temperature of one mole of a substance by $ 1~^ { \circ } \text c $ or $ 1\ , \text k $ , and it usually has units of $ \dfrac { \text j } { \text { mol } \cdot\text k } $ . for example , the heat capacity of lead might be given as the specific heat capacity , $ 0.129\ , \dfrac { \text j } { \text { g } \cdot\text k } $ , or the molar heat capacity , $ 26.65\ , \dfrac { \text j } { \text { mol } \cdot\text k } $ . calculating $ \text q $ using the heat capacity we can use the heat capacity to determine the heat released or absorbed by a material using the following formula : $ \text q = \text { m } \times \text c \times \delta \text t $ where $ \text { m } $ is the mass of the substance ( in grams ) , $ \text { c } $ is the specific heat capacity , and $ \delta \text t $ is the change in temperature during the heat transfer . note that both mass and specific heat capacity can only have positive values , so the sign of $ \text q $ will depend on the sign of $ \delta \text t $ . we can calculate $ \delta \text t $ using the following equation : $ \delta \text t=\text t_ { \text { final } } -\text t_ { \text { initial } } $ where $ \text t_ { \text { final } } $ and $ \text t_ { \text { initial } } $ can have units of either $ ~^ { \circ } \text c $ or $ \text k $ . based on this equation , if $ \text q $ is positive ( energy of the system increases ) , then our system increases in temperature and $ \text t_ { \text { final } } & gt ; \text t_ { \text { initial } } $ . if $ \text q $ is negative ( energy of the system decreases ) , then our system 's temperature decreases and $ \text t_ { \text { final } } & lt ; \text t_ { \text { initial } } $ . example problem : cooling a cup of tea let 's say that we have $ 250\ , \text { ml } $ of hot tea which we would like to cool down before we try to drink it . the tea is currently at $ 370\ , \text k $ , and we 'd like to cool it down to $ 350\ , \text k $ . how much thermal energy has to be transferred from the tea to the surroundings to cool the tea ? we are going to assume that the tea is mostly water , so we can use the density and heat capacity of water in our calculations . the specific heat capacity of water is $ 4.18\ , \dfrac { \text j } { \text g \cdot \text k } $ , and the density of water is $ 1.00\ , \dfrac { \text g } { \text { ml } } $ . we can calculate the energy transferred in the process of cooling the tea using the following steps : 1 . calculate the mass of the substance we can calculate the mass of the tea/water using the volume and density of water : $ \text m=250\ , \cancel { \text { ml } } \times 1.00\ , \dfrac { \text g } { \cancel { \text { ml } } } =250\ , \text g $ 2 . calculate the change in temperature , $ \delta \text t $ we can calculate the change in temperature , $ \delta \text t $ , from the initial and final temperatures : $ \begin { align } \delta \text t & amp ; =\text t_ { \text { final } } -\text t_ { \text { initial } } \ \ & amp ; =350\ , \text k-370\ , \text k\ \ & amp ; =-20\ , \text k\end { align } $ since the temperature of the tea is decreasing and $ \delta \text t $ is negative , we would expect $ \text q $ to also be negative since our system is losing thermal energy . 3 . solve for $ \text q $ now we can solve for the heat transferred from the hot tea using the equation for heat : $ \begin { align } \text q & amp ; = \text { m } \times \text c \times \delta \text t\ & amp ; =250\ , \cancel { \text g } \times4.18\ , \dfrac { \text j } { \cancel { \text g } \cdot \cancel { \text k } } \times -20\ , \cancel { \text k } \ & amp ; =-21000\ , \text j\end { align } $ thus , we calculated that the tea will transfer $ 21000\ , \text j $ of energy to the surroundings when it cools down from $ 370\ , \text k $ to $ 350\ , \text k $ . conclusions in thermodynamics , heat and temperature are closely related concepts with precise definitions . heat , $ \text q $ , is thermal energy transferred from a hotter system to a cooler system that are in contact . temperature is a measure of the average kinetic energy of the atoms or molecules in the system . the zeroth law of thermodynamics says that no heat is transferred between two objects in thermal equilibrium ; therefore , they are the same temperature . we can calculate the heat released or absorbed using the specific heat capacity $ \text c $ , the mass of the substance $ \text m $ , and the change in temperature $ \delta \text t $ in the following equation : $ \text q = \text { m } \times \text c \times \delta \text t $
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the water molecules in a cup of hot coffee have a higher average kinetic energy than the water molecules in a cup of iced tea , which also means they are moving at a higher velocity . temperature is also an intensive property , which means that the temperature does n't change no matter how much of a substance you have ( as long as it is all at the same temperature ! ) . this is why chemists can use the melting point to help identify a pure substance $ - $ the temperature at which it melts is a property of the substance with no dependence on the mass of a sample .
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so temperature is similar to internal energy in a way ?
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key points heat , $ \text q $ , is thermal energy transferred from a hotter system to a cooler system that are in contact . temperature is a measure of the average kinetic energy of the atoms or molecules in the system . the zeroth law of thermodynamics says that no heat is transferred between two objects in thermal equilibrium ; therefore , they are the same temperature . we can calculate the heat released or absorbed using the specific heat capacity $ \text c $ , the mass of the substance $ \text m $ , and the change in temperature $ \delta \text t $ in the equation : $ \text q = \text { m } \times \text c \times \delta \text t $ heat in thermodynamics what contains more heat , a cup of coffee or a glass of iced tea ? in chemistry class , that would be a trick question ( sorry ! ) . in thermodynamics , heat has a very specific meaning that is different from how we might use the word in everyday speech . scientists define heat as thermal energy transferred between two systems at different temperatures that come in contact . heat is written with the symbol q or q , and it has units of joules ( $ \text j $ ) . heat is sometimes called a process quantity , because it is defined in the context of a process by which energy can be transferred . we do n't talk about a cup of coffee containing heat , but we can talk about the heat transferred from the cup of hot coffee to your hand . heat is also an extensive property , so the change in temperature resulting from heat transferred to a system depends on how many molecules are in the system . relationship between heat and temperature heat and temperature are two different but closely related concepts . note that they have different units : temperature typically has units of degrees celsius ( $ ^\circ\text c $ ) or kelvin ( $ \text k $ ) , and heat has units of energy , joules ( $ \text j $ ) . temperature is a measure of the average kinetic energy of the atoms or molecules in the system . the water molecules in a cup of hot coffee have a higher average kinetic energy than the water molecules in a cup of iced tea , which also means they are moving at a higher velocity . temperature is also an intensive property , which means that the temperature does n't change no matter how much of a substance you have ( as long as it is all at the same temperature ! ) . this is why chemists can use the melting point to help identify a pure substance $ - $ the temperature at which it melts is a property of the substance with no dependence on the mass of a sample . on an atomic level , the molecules in each object are constantly in motion and colliding with each other . every time molecules collide , kinetic energy can be transferred . when the two systems are in contact , heat will be transferred through molecular collisions from the hotter system to the cooler system . the thermal energy will flow in that direction until the two objects are at the same temperature . when the two systems in contact are at the same temperature , we say they are in thermal equilibrium . zeroth law of thermodynamics : defining thermal equilibrium the zeroth law of thermodynamics defines thermal equilibrium within an isolated system . the zeroth law says when two objects at thermal equilibrium are in contact , there is no net heat transfer between the objects ; therefore , they are the same temperature . another way to state the zeroth law is to say that if two objects are both separately in thermal equilibrium with a third object , then they are in thermal equilibrium with each other . the zeroth law allows us to measure the temperature of objects . any time we use a thermometer , we are using the zeroth law of thermodynamics . let 's say we are measuring the temperature of a water bath . in order to make sure the reading is accurate , we usually want to wait for the temperature reading to stay constant . we are waiting for the thermometer and the water to reach thermal equilibrium ! at thermal equilibrium , the temperature of the thermometer bulb and the water bath will be the same , and there should be no net heat transfer from one object to the other ( assuming no other loss of heat to the surroundings ) . heat capacity : converting between heat and change in temperature how can we measure heat ? here are some things we know about heat so far : when a system absorbs or loses heat , the average kinetic energy of the molecules will change . thus , heat transfer results in a change in the system 's temperature as long as the system is not undergoing a phase change . the change in temperature resulting from heat transferred to or from a system depends on how many molecules are in the system . we can use a thermometer to measure the change in a system 's temperature . how can we use the change in temperature to calculate the heat transferred ? in order to figure out how the heat transferred to a system will change the temperature of the system , we need to know at least $ 2 $ things : the number of molecules in the system the heat capacity of the system the heat capacity tells us how much energy is needed to change the temperature of a given substance assuming that no phase changes are occurring . there are two main ways that heat capacity is reported . the specific heat capacity ( also called specific heat ) , represented by the symbol $ \text c $ or $ \text c $ , is how much energy is needed to increase the temperature of one gram of a substance by $ 1~^ { \circ } \text c $ or $ 1\ , \text k $ . specific heat capacity usually has units of $ \dfrac { \text j } { \text { grams } \cdot\text k } $ . the molar heat capacity , $ \text c_\text m $ or $ \text c_ { \text { mol } } $ , measures the amount of thermal energy it takes to raise the temperature of one mole of a substance by $ 1~^ { \circ } \text c $ or $ 1\ , \text k $ , and it usually has units of $ \dfrac { \text j } { \text { mol } \cdot\text k } $ . for example , the heat capacity of lead might be given as the specific heat capacity , $ 0.129\ , \dfrac { \text j } { \text { g } \cdot\text k } $ , or the molar heat capacity , $ 26.65\ , \dfrac { \text j } { \text { mol } \cdot\text k } $ . calculating $ \text q $ using the heat capacity we can use the heat capacity to determine the heat released or absorbed by a material using the following formula : $ \text q = \text { m } \times \text c \times \delta \text t $ where $ \text { m } $ is the mass of the substance ( in grams ) , $ \text { c } $ is the specific heat capacity , and $ \delta \text t $ is the change in temperature during the heat transfer . note that both mass and specific heat capacity can only have positive values , so the sign of $ \text q $ will depend on the sign of $ \delta \text t $ . we can calculate $ \delta \text t $ using the following equation : $ \delta \text t=\text t_ { \text { final } } -\text t_ { \text { initial } } $ where $ \text t_ { \text { final } } $ and $ \text t_ { \text { initial } } $ can have units of either $ ~^ { \circ } \text c $ or $ \text k $ . based on this equation , if $ \text q $ is positive ( energy of the system increases ) , then our system increases in temperature and $ \text t_ { \text { final } } & gt ; \text t_ { \text { initial } } $ . if $ \text q $ is negative ( energy of the system decreases ) , then our system 's temperature decreases and $ \text t_ { \text { final } } & lt ; \text t_ { \text { initial } } $ . example problem : cooling a cup of tea let 's say that we have $ 250\ , \text { ml } $ of hot tea which we would like to cool down before we try to drink it . the tea is currently at $ 370\ , \text k $ , and we 'd like to cool it down to $ 350\ , \text k $ . how much thermal energy has to be transferred from the tea to the surroundings to cool the tea ? we are going to assume that the tea is mostly water , so we can use the density and heat capacity of water in our calculations . the specific heat capacity of water is $ 4.18\ , \dfrac { \text j } { \text g \cdot \text k } $ , and the density of water is $ 1.00\ , \dfrac { \text g } { \text { ml } } $ . we can calculate the energy transferred in the process of cooling the tea using the following steps : 1 . calculate the mass of the substance we can calculate the mass of the tea/water using the volume and density of water : $ \text m=250\ , \cancel { \text { ml } } \times 1.00\ , \dfrac { \text g } { \cancel { \text { ml } } } =250\ , \text g $ 2 . calculate the change in temperature , $ \delta \text t $ we can calculate the change in temperature , $ \delta \text t $ , from the initial and final temperatures : $ \begin { align } \delta \text t & amp ; =\text t_ { \text { final } } -\text t_ { \text { initial } } \ \ & amp ; =350\ , \text k-370\ , \text k\ \ & amp ; =-20\ , \text k\end { align } $ since the temperature of the tea is decreasing and $ \delta \text t $ is negative , we would expect $ \text q $ to also be negative since our system is losing thermal energy . 3 . solve for $ \text q $ now we can solve for the heat transferred from the hot tea using the equation for heat : $ \begin { align } \text q & amp ; = \text { m } \times \text c \times \delta \text t\ & amp ; =250\ , \cancel { \text g } \times4.18\ , \dfrac { \text j } { \cancel { \text g } \cdot \cancel { \text k } } \times -20\ , \cancel { \text k } \ & amp ; =-21000\ , \text j\end { align } $ thus , we calculated that the tea will transfer $ 21000\ , \text j $ of energy to the surroundings when it cools down from $ 370\ , \text k $ to $ 350\ , \text k $ . conclusions in thermodynamics , heat and temperature are closely related concepts with precise definitions . heat , $ \text q $ , is thermal energy transferred from a hotter system to a cooler system that are in contact . temperature is a measure of the average kinetic energy of the atoms or molecules in the system . the zeroth law of thermodynamics says that no heat is transferred between two objects in thermal equilibrium ; therefore , they are the same temperature . we can calculate the heat released or absorbed using the specific heat capacity $ \text c $ , the mass of the substance $ \text m $ , and the change in temperature $ \delta \text t $ in the following equation : $ \text q = \text { m } \times \text c \times \delta \text t $
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heat is also an extensive property , so the change in temperature resulting from heat transferred to a system depends on how many molecules are in the system . relationship between heat and temperature heat and temperature are two different but closely related concepts . note that they have different units : temperature typically has units of degrees celsius ( $ ^\circ\text c $ ) or kelvin ( $ \text k $ ) , and heat has units of energy , joules ( $ \text j $ ) .
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in the first paragraph oon the `` relationship between heat and temperature '' how do chemists know if a sample substance is a pure one ?
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