id int64 1 14M | domain stringclasses 6
values | topic stringclasses 23
values | subtopic stringclasses 37
values | difficulty int64 1 8 | unit_type stringclasses 3
values | title stringlengths 14 86 | content stringlengths 203 553 | key_equations stringclasses 23
values | prerequisites stringclasses 29
values | learning_objective stringclasses 37
values |
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5,501 | physics | mechanics | newton_second_law | 3 | worked_example | Newton's second law: mass 13.42 kg, acceleration 2.246 m/s² | A net force acting on a mass of 13.42 kg produces an acceleration of 2.246 m/s². By Newton's second law, F_net = m a = 13.42 × 2.246 = 30.15 N. Direction of F_net is the same as the direction of the acceleration. This relation defines the inertial mass and is the foundation of classical dynamics. | F_net = m a | kinematics_1d | Compute net force from mass and acceleration using Newton's second law. |
5,502 | physics | mechanics | newton_second_law | 3 | worked_example | Newton's second law: mass 1.752 kg, acceleration 9.809 m/s² | A net force acting on a mass of 1.752 kg produces an acceleration of 9.809 m/s². By Newton's second law, F_net = m a = 1.752 × 9.809 = 17.18 N. Direction of F_net is the same as the direction of the acceleration. This relation defines the inertial mass and is the foundation of classical dynamics. | F_net = m a | kinematics_1d | Compute net force from mass and acceleration using Newton's second law. |
5,503 | physics | mechanics | newton_second_law | 3 | worked_example | Newton's second law: mass 38.24 kg, acceleration 1.273 m/s² | A net force acting on a mass of 38.24 kg produces an acceleration of 1.273 m/s². By Newton's second law, F_net = m a = 38.24 × 1.273 = 48.67 N. Direction of F_net is the same as the direction of the acceleration. This relation defines the inertial mass and is the foundation of classical dynamics. | F_net = m a | kinematics_1d | Compute net force from mass and acceleration using Newton's second law. |
5,504 | physics | mechanics | newton_second_law | 3 | worked_example | Newton's second law: mass 23.51 kg, acceleration 13.66 m/s² | A net force acting on a mass of 23.51 kg produces an acceleration of 13.66 m/s². By Newton's second law, F_net = m a = 23.51 × 13.66 = 321.1 N. Direction of F_net is the same as the direction of the acceleration. This relation defines the inertial mass and is the foundation of classical dynamics. | F_net = m a | kinematics_1d | Compute net force from mass and acceleration using Newton's second law. |
5,505 | physics | mechanics | newton_second_law | 3 | worked_example | Newton's second law: mass 7.807 kg, acceleration 9.426 m/s² | A net force acting on a mass of 7.807 kg produces an acceleration of 9.426 m/s². By Newton's second law, F_net = m a = 7.807 × 9.426 = 73.59 N. Direction of F_net is the same as the direction of the acceleration. This relation defines the inertial mass and is the foundation of classical dynamics. | F_net = m a | kinematics_1d | Compute net force from mass and acceleration using Newton's second law. |
5,506 | physics | mechanics | newton_second_law | 3 | worked_example | Newton's second law: mass 27.99 kg, acceleration 8.948 m/s² | A net force acting on a mass of 27.99 kg produces an acceleration of 8.948 m/s². By Newton's second law, F_net = m a = 27.99 × 8.948 = 250.4 N. Direction of F_net is the same as the direction of the acceleration. This relation defines the inertial mass and is the foundation of classical dynamics. | F_net = m a | kinematics_1d | Compute net force from mass and acceleration using Newton's second law. |
5,507 | physics | mechanics | newton_second_law | 3 | worked_example | Newton's second law: mass 38.04 kg, acceleration 1.705 m/s² | A net force acting on a mass of 38.04 kg produces an acceleration of 1.705 m/s². By Newton's second law, F_net = m a = 38.04 × 1.705 = 64.88 N. Direction of F_net is the same as the direction of the acceleration. This relation defines the inertial mass and is the foundation of classical dynamics. | F_net = m a | kinematics_1d | Compute net force from mass and acceleration using Newton's second law. |
5,508 | physics | mechanics | newton_second_law | 3 | worked_example | Newton's second law: mass 14.04 kg, acceleration 4.212 m/s² | A net force acting on a mass of 14.04 kg produces an acceleration of 4.212 m/s². By Newton's second law, F_net = m a = 14.04 × 4.212 = 59.14 N. Direction of F_net is the same as the direction of the acceleration. This relation defines the inertial mass and is the foundation of classical dynamics. | F_net = m a | kinematics_1d | Compute net force from mass and acceleration using Newton's second law. |
5,509 | physics | mechanics | newton_second_law | 3 | worked_example | Newton's second law: mass 35.46 kg, acceleration 6.855 m/s² | A net force acting on a mass of 35.46 kg produces an acceleration of 6.855 m/s². By Newton's second law, F_net = m a = 35.46 × 6.855 = 243.1 N. Direction of F_net is the same as the direction of the acceleration. This relation defines the inertial mass and is the foundation of classical dynamics. | F_net = m a | kinematics_1d | Compute net force from mass and acceleration using Newton's second law. |
5,510 | physics | mechanics | newton_second_law | 3 | worked_example | Newton's second law: mass 8.564 kg, acceleration 2.015 m/s² | A net force acting on a mass of 8.564 kg produces an acceleration of 2.015 m/s². By Newton's second law, F_net = m a = 8.564 × 2.015 = 17.26 N. Direction of F_net is the same as the direction of the acceleration. This relation defines the inertial mass and is the foundation of classical dynamics. | F_net = m a | kinematics_1d | Compute net force from mass and acceleration using Newton's second law. |
5,511 | physics | mechanics | newton_second_law | 3 | worked_example | Newton's second law: mass 47.83 kg, acceleration 1.946 m/s² | A net force acting on a mass of 47.83 kg produces an acceleration of 1.946 m/s². By Newton's second law, F_net = m a = 47.83 × 1.946 = 93.09 N. Direction of F_net is the same as the direction of the acceleration. This relation defines the inertial mass and is the foundation of classical dynamics. | F_net = m a | kinematics_1d | Compute net force from mass and acceleration using Newton's second law. |
5,512 | physics | mechanics | newton_second_law | 3 | worked_example | Newton's second law: mass 43.45 kg, acceleration 8.333 m/s² | A net force acting on a mass of 43.45 kg produces an acceleration of 8.333 m/s². By Newton's second law, F_net = m a = 43.45 × 8.333 = 362.1 N. Direction of F_net is the same as the direction of the acceleration. This relation defines the inertial mass and is the foundation of classical dynamics. | F_net = m a | kinematics_1d | Compute net force from mass and acceleration using Newton's second law. |
5,513 | physics | mechanics | newton_second_law | 3 | worked_example | Newton's second law: mass 18.76 kg, acceleration 4.798 m/s² | A net force acting on a mass of 18.76 kg produces an acceleration of 4.798 m/s². By Newton's second law, F_net = m a = 18.76 × 4.798 = 89.99 N. Direction of F_net is the same as the direction of the acceleration. This relation defines the inertial mass and is the foundation of classical dynamics. | F_net = m a | kinematics_1d | Compute net force from mass and acceleration using Newton's second law. |
5,514 | physics | mechanics | newton_second_law | 3 | worked_example | Newton's second law: mass 36.16 kg, acceleration 8.297 m/s² | A net force acting on a mass of 36.16 kg produces an acceleration of 8.297 m/s². By Newton's second law, F_net = m a = 36.16 × 8.297 = 300.1 N. Direction of F_net is the same as the direction of the acceleration. This relation defines the inertial mass and is the foundation of classical dynamics. | F_net = m a | kinematics_1d | Compute net force from mass and acceleration using Newton's second law. |
5,515 | physics | mechanics | newton_second_law | 3 | worked_example | Newton's second law: mass 31.63 kg, acceleration 9.73 m/s² | A net force acting on a mass of 31.63 kg produces an acceleration of 9.73 m/s². By Newton's second law, F_net = m a = 31.63 × 9.73 = 307.8 N. Direction of F_net is the same as the direction of the acceleration. This relation defines the inertial mass and is the foundation of classical dynamics. | F_net = m a | kinematics_1d | Compute net force from mass and acceleration using Newton's second law. |
5,516 | physics | mechanics | newton_second_law | 3 | worked_example | Newton's second law: mass 20.08 kg, acceleration 4.966 m/s² | A net force acting on a mass of 20.08 kg produces an acceleration of 4.966 m/s². By Newton's second law, F_net = m a = 20.08 × 4.966 = 99.72 N. Direction of F_net is the same as the direction of the acceleration. This relation defines the inertial mass and is the foundation of classical dynamics. | F_net = m a | kinematics_1d | Compute net force from mass and acceleration using Newton's second law. |
5,517 | physics | mechanics | newton_second_law | 3 | worked_example | Newton's second law: mass 41.89 kg, acceleration 0.6146 m/s² | A net force acting on a mass of 41.89 kg produces an acceleration of 0.6146 m/s². By Newton's second law, F_net = m a = 41.89 × 0.6146 = 25.74 N. Direction of F_net is the same as the direction of the acceleration. This relation defines the inertial mass and is the foundation of classical dynamics. | F_net = m a | kinematics_1d | Compute net force from mass and acceleration using Newton's second law. |
5,518 | physics | mechanics | newton_second_law | 3 | worked_example | Newton's second law: mass 14.5 kg, acceleration 8.528 m/s² | A net force acting on a mass of 14.5 kg produces an acceleration of 8.528 m/s². By Newton's second law, F_net = m a = 14.5 × 8.528 = 123.7 N. Direction of F_net is the same as the direction of the acceleration. This relation defines the inertial mass and is the foundation of classical dynamics. | F_net = m a | kinematics_1d | Compute net force from mass and acceleration using Newton's second law. |
5,519 | physics | mechanics | newton_second_law | 3 | worked_example | Newton's second law: mass 28.64 kg, acceleration 13.26 m/s² | A net force acting on a mass of 28.64 kg produces an acceleration of 13.26 m/s². By Newton's second law, F_net = m a = 28.64 × 13.26 = 379.8 N. Direction of F_net is the same as the direction of the acceleration. This relation defines the inertial mass and is the foundation of classical dynamics. | F_net = m a | kinematics_1d | Compute net force from mass and acceleration using Newton's second law. |
5,520 | physics | mechanics | newton_second_law | 3 | worked_example | Newton's second law: mass 3.265 kg, acceleration 10.79 m/s² | A net force acting on a mass of 3.265 kg produces an acceleration of 10.79 m/s². By Newton's second law, F_net = m a = 3.265 × 10.79 = 35.22 N. Direction of F_net is the same as the direction of the acceleration. This relation defines the inertial mass and is the foundation of classical dynamics. | F_net = m a | kinematics_1d | Compute net force from mass and acceleration using Newton's second law. |
5,521 | physics | mechanics | newton_second_law | 3 | worked_example | Newton's second law: mass 7.027 kg, acceleration 1.052 m/s² | A net force acting on a mass of 7.027 kg produces an acceleration of 1.052 m/s². By Newton's second law, F_net = m a = 7.027 × 1.052 = 7.391 N. Direction of F_net is the same as the direction of the acceleration. This relation defines the inertial mass and is the foundation of classical dynamics. | F_net = m a | kinematics_1d | Compute net force from mass and acceleration using Newton's second law. |
5,522 | physics | mechanics | newton_second_law | 3 | worked_example | Newton's second law: mass 3.861 kg, acceleration 2.358 m/s² | A net force acting on a mass of 3.861 kg produces an acceleration of 2.358 m/s². By Newton's second law, F_net = m a = 3.861 × 2.358 = 9.105 N. Direction of F_net is the same as the direction of the acceleration. This relation defines the inertial mass and is the foundation of classical dynamics. | F_net = m a | kinematics_1d | Compute net force from mass and acceleration using Newton's second law. |
5,523 | physics | mechanics | newton_second_law | 3 | worked_example | Newton's second law: mass 26.38 kg, acceleration 14.11 m/s² | A net force acting on a mass of 26.38 kg produces an acceleration of 14.11 m/s². By Newton's second law, F_net = m a = 26.38 × 14.11 = 372.1 N. Direction of F_net is the same as the direction of the acceleration. This relation defines the inertial mass and is the foundation of classical dynamics. | F_net = m a | kinematics_1d | Compute net force from mass and acceleration using Newton's second law. |
5,524 | physics | mechanics | newton_second_law | 3 | worked_example | Newton's second law: mass 12.13 kg, acceleration 11.34 m/s² | A net force acting on a mass of 12.13 kg produces an acceleration of 11.34 m/s². By Newton's second law, F_net = m a = 12.13 × 11.34 = 137.6 N. Direction of F_net is the same as the direction of the acceleration. This relation defines the inertial mass and is the foundation of classical dynamics. | F_net = m a | kinematics_1d | Compute net force from mass and acceleration using Newton's second law. |
5,525 | physics | mechanics | newton_second_law | 3 | worked_example | Newton's second law: mass 48.63 kg, acceleration 5.139 m/s² | A net force acting on a mass of 48.63 kg produces an acceleration of 5.139 m/s². By Newton's second law, F_net = m a = 48.63 × 5.139 = 249.9 N. Direction of F_net is the same as the direction of the acceleration. This relation defines the inertial mass and is the foundation of classical dynamics. | F_net = m a | kinematics_1d | Compute net force from mass and acceleration using Newton's second law. |
5,526 | physics | mechanics | newton_second_law | 3 | worked_example | Newton's second law: mass 20.93 kg, acceleration 11.29 m/s² | A net force acting on a mass of 20.93 kg produces an acceleration of 11.29 m/s². By Newton's second law, F_net = m a = 20.93 × 11.29 = 236.2 N. Direction of F_net is the same as the direction of the acceleration. This relation defines the inertial mass and is the foundation of classical dynamics. | F_net = m a | kinematics_1d | Compute net force from mass and acceleration using Newton's second law. |
5,527 | physics | mechanics | newton_second_law | 3 | worked_example | Newton's second law: mass 24.88 kg, acceleration 8.222 m/s² | A net force acting on a mass of 24.88 kg produces an acceleration of 8.222 m/s². By Newton's second law, F_net = m a = 24.88 × 8.222 = 204.5 N. Direction of F_net is the same as the direction of the acceleration. This relation defines the inertial mass and is the foundation of classical dynamics. | F_net = m a | kinematics_1d | Compute net force from mass and acceleration using Newton's second law. |
5,528 | physics | mechanics | newton_second_law | 3 | worked_example | Newton's second law: mass 26.47 kg, acceleration 9.663 m/s² | A net force acting on a mass of 26.47 kg produces an acceleration of 9.663 m/s². By Newton's second law, F_net = m a = 26.47 × 9.663 = 255.8 N. Direction of F_net is the same as the direction of the acceleration. This relation defines the inertial mass and is the foundation of classical dynamics. | F_net = m a | kinematics_1d | Compute net force from mass and acceleration using Newton's second law. |
5,529 | physics | mechanics | newton_second_law | 3 | worked_example | Newton's second law: mass 20.1 kg, acceleration 2.409 m/s² | A net force acting on a mass of 20.1 kg produces an acceleration of 2.409 m/s². By Newton's second law, F_net = m a = 20.1 × 2.409 = 48.43 N. Direction of F_net is the same as the direction of the acceleration. This relation defines the inertial mass and is the foundation of classical dynamics. | F_net = m a | kinematics_1d | Compute net force from mass and acceleration using Newton's second law. |
5,530 | physics | mechanics | newton_second_law | 3 | worked_example | Newton's second law: mass 41.23 kg, acceleration 12.23 m/s² | A net force acting on a mass of 41.23 kg produces an acceleration of 12.23 m/s². By Newton's second law, F_net = m a = 41.23 × 12.23 = 504.4 N. Direction of F_net is the same as the direction of the acceleration. This relation defines the inertial mass and is the foundation of classical dynamics. | F_net = m a | kinematics_1d | Compute net force from mass and acceleration using Newton's second law. |
5,531 | physics | mechanics | newton_second_law | 3 | worked_example | Newton's second law: mass 22.7 kg, acceleration 1.817 m/s² | A net force acting on a mass of 22.7 kg produces an acceleration of 1.817 m/s². By Newton's second law, F_net = m a = 22.7 × 1.817 = 41.25 N. Direction of F_net is the same as the direction of the acceleration. This relation defines the inertial mass and is the foundation of classical dynamics. | F_net = m a | kinematics_1d | Compute net force from mass and acceleration using Newton's second law. |
5,532 | physics | mechanics | newton_second_law | 3 | worked_example | Newton's second law: mass 45.62 kg, acceleration 3.985 m/s² | A net force acting on a mass of 45.62 kg produces an acceleration of 3.985 m/s². By Newton's second law, F_net = m a = 45.62 × 3.985 = 181.8 N. Direction of F_net is the same as the direction of the acceleration. This relation defines the inertial mass and is the foundation of classical dynamics. | F_net = m a | kinematics_1d | Compute net force from mass and acceleration using Newton's second law. |
5,533 | physics | mechanics | newton_second_law | 3 | worked_example | Newton's second law: mass 12.34 kg, acceleration 9.707 m/s² | A net force acting on a mass of 12.34 kg produces an acceleration of 9.707 m/s². By Newton's second law, F_net = m a = 12.34 × 9.707 = 119.8 N. Direction of F_net is the same as the direction of the acceleration. This relation defines the inertial mass and is the foundation of classical dynamics. | F_net = m a | kinematics_1d | Compute net force from mass and acceleration using Newton's second law. |
5,534 | physics | mechanics | newton_second_law | 3 | worked_example | Newton's second law: mass 28.11 kg, acceleration 0.882 m/s² | A net force acting on a mass of 28.11 kg produces an acceleration of 0.882 m/s². By Newton's second law, F_net = m a = 28.11 × 0.882 = 24.79 N. Direction of F_net is the same as the direction of the acceleration. This relation defines the inertial mass and is the foundation of classical dynamics. | F_net = m a | kinematics_1d | Compute net force from mass and acceleration using Newton's second law. |
5,535 | physics | mechanics | newton_second_law | 3 | worked_example | Newton's second law: mass 42.22 kg, acceleration 3.46 m/s² | A net force acting on a mass of 42.22 kg produces an acceleration of 3.46 m/s². By Newton's second law, F_net = m a = 42.22 × 3.46 = 146.1 N. Direction of F_net is the same as the direction of the acceleration. This relation defines the inertial mass and is the foundation of classical dynamics. | F_net = m a | kinematics_1d | Compute net force from mass and acceleration using Newton's second law. |
5,536 | physics | mechanics | newton_second_law | 3 | worked_example | Newton's second law: mass 2.599 kg, acceleration 3.363 m/s² | A net force acting on a mass of 2.599 kg produces an acceleration of 3.363 m/s². By Newton's second law, F_net = m a = 2.599 × 3.363 = 8.739 N. Direction of F_net is the same as the direction of the acceleration. This relation defines the inertial mass and is the foundation of classical dynamics. | F_net = m a | kinematics_1d | Compute net force from mass and acceleration using Newton's second law. |
5,537 | physics | mechanics | newton_second_law | 3 | worked_example | Newton's second law: mass 12.02 kg, acceleration 2.374 m/s² | A net force acting on a mass of 12.02 kg produces an acceleration of 2.374 m/s². By Newton's second law, F_net = m a = 12.02 × 2.374 = 28.53 N. Direction of F_net is the same as the direction of the acceleration. This relation defines the inertial mass and is the foundation of classical dynamics. | F_net = m a | kinematics_1d | Compute net force from mass and acceleration using Newton's second law. |
5,538 | physics | mechanics | newton_second_law | 3 | worked_example | Newton's second law: mass 43.59 kg, acceleration 7.559 m/s² | A net force acting on a mass of 43.59 kg produces an acceleration of 7.559 m/s². By Newton's second law, F_net = m a = 43.59 × 7.559 = 329.5 N. Direction of F_net is the same as the direction of the acceleration. This relation defines the inertial mass and is the foundation of classical dynamics. | F_net = m a | kinematics_1d | Compute net force from mass and acceleration using Newton's second law. |
5,539 | physics | mechanics | newton_second_law | 3 | worked_example | Newton's second law: mass 16.49 kg, acceleration 11.36 m/s² | A net force acting on a mass of 16.49 kg produces an acceleration of 11.36 m/s². By Newton's second law, F_net = m a = 16.49 × 11.36 = 187.3 N. Direction of F_net is the same as the direction of the acceleration. This relation defines the inertial mass and is the foundation of classical dynamics. | F_net = m a | kinematics_1d | Compute net force from mass and acceleration using Newton's second law. |
5,540 | physics | mechanics | newton_second_law | 3 | worked_example | Newton's second law: mass 33.68 kg, acceleration 0.3869 m/s² | A net force acting on a mass of 33.68 kg produces an acceleration of 0.3869 m/s². By Newton's second law, F_net = m a = 33.68 × 0.3869 = 13.03 N. Direction of F_net is the same as the direction of the acceleration. This relation defines the inertial mass and is the foundation of classical dynamics. | F_net = m a | kinematics_1d | Compute net force from mass and acceleration using Newton's second law. |
5,541 | physics | mechanics | newton_second_law | 3 | worked_example | Newton's second law: mass 38.39 kg, acceleration 7.434 m/s² | A net force acting on a mass of 38.39 kg produces an acceleration of 7.434 m/s². By Newton's second law, F_net = m a = 38.39 × 7.434 = 285.4 N. Direction of F_net is the same as the direction of the acceleration. This relation defines the inertial mass and is the foundation of classical dynamics. | F_net = m a | kinematics_1d | Compute net force from mass and acceleration using Newton's second law. |
5,542 | physics | mechanics | newton_second_law | 3 | worked_example | Newton's second law: mass 33.65 kg, acceleration 10.49 m/s² | A net force acting on a mass of 33.65 kg produces an acceleration of 10.49 m/s². By Newton's second law, F_net = m a = 33.65 × 10.49 = 352.9 N. Direction of F_net is the same as the direction of the acceleration. This relation defines the inertial mass and is the foundation of classical dynamics. | F_net = m a | kinematics_1d | Compute net force from mass and acceleration using Newton's second law. |
5,543 | physics | mechanics | newton_second_law | 3 | worked_example | Newton's second law: mass 27.73 kg, acceleration 0.6979 m/s² | A net force acting on a mass of 27.73 kg produces an acceleration of 0.6979 m/s². By Newton's second law, F_net = m a = 27.73 × 0.6979 = 19.35 N. Direction of F_net is the same as the direction of the acceleration. This relation defines the inertial mass and is the foundation of classical dynamics. | F_net = m a | kinematics_1d | Compute net force from mass and acceleration using Newton's second law. |
5,544 | physics | mechanics | newton_second_law | 3 | worked_example | Newton's second law: mass 29.69 kg, acceleration 3.212 m/s² | A net force acting on a mass of 29.69 kg produces an acceleration of 3.212 m/s². By Newton's second law, F_net = m a = 29.69 × 3.212 = 95.35 N. Direction of F_net is the same as the direction of the acceleration. This relation defines the inertial mass and is the foundation of classical dynamics. | F_net = m a | kinematics_1d | Compute net force from mass and acceleration using Newton's second law. |
5,545 | physics | mechanics | newton_second_law | 3 | worked_example | Newton's second law: mass 29.35 kg, acceleration 0.2066 m/s² | A net force acting on a mass of 29.35 kg produces an acceleration of 0.2066 m/s². By Newton's second law, F_net = m a = 29.35 × 0.2066 = 6.062 N. Direction of F_net is the same as the direction of the acceleration. This relation defines the inertial mass and is the foundation of classical dynamics. | F_net = m a | kinematics_1d | Compute net force from mass and acceleration using Newton's second law. |
5,546 | physics | mechanics | newton_second_law | 3 | worked_example | Newton's second law: mass 35.24 kg, acceleration 8.766 m/s² | A net force acting on a mass of 35.24 kg produces an acceleration of 8.766 m/s². By Newton's second law, F_net = m a = 35.24 × 8.766 = 308.9 N. Direction of F_net is the same as the direction of the acceleration. This relation defines the inertial mass and is the foundation of classical dynamics. | F_net = m a | kinematics_1d | Compute net force from mass and acceleration using Newton's second law. |
5,547 | physics | mechanics | newton_second_law | 3 | worked_example | Newton's second law: mass 18.72 kg, acceleration 13.49 m/s² | A net force acting on a mass of 18.72 kg produces an acceleration of 13.49 m/s². By Newton's second law, F_net = m a = 18.72 × 13.49 = 252.5 N. Direction of F_net is the same as the direction of the acceleration. This relation defines the inertial mass and is the foundation of classical dynamics. | F_net = m a | kinematics_1d | Compute net force from mass and acceleration using Newton's second law. |
5,548 | physics | mechanics | newton_second_law | 3 | worked_example | Newton's second law: mass 8.627 kg, acceleration 3.27 m/s² | A net force acting on a mass of 8.627 kg produces an acceleration of 3.27 m/s². By Newton's second law, F_net = m a = 8.627 × 3.27 = 28.21 N. Direction of F_net is the same as the direction of the acceleration. This relation defines the inertial mass and is the foundation of classical dynamics. | F_net = m a | kinematics_1d | Compute net force from mass and acceleration using Newton's second law. |
5,549 | physics | mechanics | newton_second_law | 3 | worked_example | Newton's second law: mass 27.04 kg, acceleration 13.74 m/s² | A net force acting on a mass of 27.04 kg produces an acceleration of 13.74 m/s². By Newton's second law, F_net = m a = 27.04 × 13.74 = 371.5 N. Direction of F_net is the same as the direction of the acceleration. This relation defines the inertial mass and is the foundation of classical dynamics. | F_net = m a | kinematics_1d | Compute net force from mass and acceleration using Newton's second law. |
5,550 | physics | mechanics | newton_second_law | 3 | worked_example | Newton's second law: mass 37.34 kg, acceleration 4.293 m/s² | A net force acting on a mass of 37.34 kg produces an acceleration of 4.293 m/s². By Newton's second law, F_net = m a = 37.34 × 4.293 = 160.3 N. Direction of F_net is the same as the direction of the acceleration. This relation defines the inertial mass and is the foundation of classical dynamics. | F_net = m a | kinematics_1d | Compute net force from mass and acceleration using Newton's second law. |
5,551 | physics | mechanics | newton_second_law | 3 | worked_example | Newton's second law: mass 28.04 kg, acceleration 7.712 m/s² | A net force acting on a mass of 28.04 kg produces an acceleration of 7.712 m/s². By Newton's second law, F_net = m a = 28.04 × 7.712 = 216.2 N. Direction of F_net is the same as the direction of the acceleration. This relation defines the inertial mass and is the foundation of classical dynamics. | F_net = m a | kinematics_1d | Compute net force from mass and acceleration using Newton's second law. |
5,552 | physics | mechanics | newton_second_law | 3 | worked_example | Newton's second law: mass 8.413 kg, acceleration 3.104 m/s² | A net force acting on a mass of 8.413 kg produces an acceleration of 3.104 m/s². By Newton's second law, F_net = m a = 8.413 × 3.104 = 26.11 N. Direction of F_net is the same as the direction of the acceleration. This relation defines the inertial mass and is the foundation of classical dynamics. | F_net = m a | kinematics_1d | Compute net force from mass and acceleration using Newton's second law. |
5,553 | physics | mechanics | newton_second_law | 3 | worked_example | Newton's second law: mass 30.58 kg, acceleration 5.084 m/s² | A net force acting on a mass of 30.58 kg produces an acceleration of 5.084 m/s². By Newton's second law, F_net = m a = 30.58 × 5.084 = 155.4 N. Direction of F_net is the same as the direction of the acceleration. This relation defines the inertial mass and is the foundation of classical dynamics. | F_net = m a | kinematics_1d | Compute net force from mass and acceleration using Newton's second law. |
5,554 | physics | mechanics | newton_second_law | 3 | worked_example | Newton's second law: mass 16.69 kg, acceleration 3.979 m/s² | A net force acting on a mass of 16.69 kg produces an acceleration of 3.979 m/s². By Newton's second law, F_net = m a = 16.69 × 3.979 = 66.41 N. Direction of F_net is the same as the direction of the acceleration. This relation defines the inertial mass and is the foundation of classical dynamics. | F_net = m a | kinematics_1d | Compute net force from mass and acceleration using Newton's second law. |
5,555 | physics | mechanics | newton_second_law | 3 | worked_example | Newton's second law: mass 6.399 kg, acceleration 4.352 m/s² | A net force acting on a mass of 6.399 kg produces an acceleration of 4.352 m/s². By Newton's second law, F_net = m a = 6.399 × 4.352 = 27.85 N. Direction of F_net is the same as the direction of the acceleration. This relation defines the inertial mass and is the foundation of classical dynamics. | F_net = m a | kinematics_1d | Compute net force from mass and acceleration using Newton's second law. |
5,556 | physics | mechanics | newton_second_law | 3 | worked_example | Newton's second law: mass 1.926 kg, acceleration 9.09 m/s² | A net force acting on a mass of 1.926 kg produces an acceleration of 9.09 m/s². By Newton's second law, F_net = m a = 1.926 × 9.09 = 17.51 N. Direction of F_net is the same as the direction of the acceleration. This relation defines the inertial mass and is the foundation of classical dynamics. | F_net = m a | kinematics_1d | Compute net force from mass and acceleration using Newton's second law. |
5,557 | physics | mechanics | newton_second_law | 3 | worked_example | Newton's second law: mass 32.74 kg, acceleration 7.726 m/s² | A net force acting on a mass of 32.74 kg produces an acceleration of 7.726 m/s². By Newton's second law, F_net = m a = 32.74 × 7.726 = 253 N. Direction of F_net is the same as the direction of the acceleration. This relation defines the inertial mass and is the foundation of classical dynamics. | F_net = m a | kinematics_1d | Compute net force from mass and acceleration using Newton's second law. |
5,558 | physics | mechanics | newton_second_law | 3 | worked_example | Newton's second law: mass 20.58 kg, acceleration 2.854 m/s² | A net force acting on a mass of 20.58 kg produces an acceleration of 2.854 m/s². By Newton's second law, F_net = m a = 20.58 × 2.854 = 58.74 N. Direction of F_net is the same as the direction of the acceleration. This relation defines the inertial mass and is the foundation of classical dynamics. | F_net = m a | kinematics_1d | Compute net force from mass and acceleration using Newton's second law. |
5,559 | physics | mechanics | newton_second_law | 3 | worked_example | Newton's second law: mass 2.562 kg, acceleration 12.09 m/s² | A net force acting on a mass of 2.562 kg produces an acceleration of 12.09 m/s². By Newton's second law, F_net = m a = 2.562 × 12.09 = 30.97 N. Direction of F_net is the same as the direction of the acceleration. This relation defines the inertial mass and is the foundation of classical dynamics. | F_net = m a | kinematics_1d | Compute net force from mass and acceleration using Newton's second law. |
5,560 | physics | mechanics | newton_second_law | 3 | worked_example | Newton's second law: mass 37.27 kg, acceleration 13.13 m/s² | A net force acting on a mass of 37.27 kg produces an acceleration of 13.13 m/s². By Newton's second law, F_net = m a = 37.27 × 13.13 = 489.2 N. Direction of F_net is the same as the direction of the acceleration. This relation defines the inertial mass and is the foundation of classical dynamics. | F_net = m a | kinematics_1d | Compute net force from mass and acceleration using Newton's second law. |
5,561 | physics | mechanics | newton_second_law | 3 | worked_example | Newton's second law: mass 10.41 kg, acceleration 8.596 m/s² | A net force acting on a mass of 10.41 kg produces an acceleration of 8.596 m/s². By Newton's second law, F_net = m a = 10.41 × 8.596 = 89.46 N. Direction of F_net is the same as the direction of the acceleration. This relation defines the inertial mass and is the foundation of classical dynamics. | F_net = m a | kinematics_1d | Compute net force from mass and acceleration using Newton's second law. |
5,562 | physics | mechanics | newton_second_law | 3 | worked_example | Newton's second law: mass 37.49 kg, acceleration 12.51 m/s² | A net force acting on a mass of 37.49 kg produces an acceleration of 12.51 m/s². By Newton's second law, F_net = m a = 37.49 × 12.51 = 469 N. Direction of F_net is the same as the direction of the acceleration. This relation defines the inertial mass and is the foundation of classical dynamics. | F_net = m a | kinematics_1d | Compute net force from mass and acceleration using Newton's second law. |
5,563 | physics | mechanics | newton_second_law | 3 | worked_example | Newton's second law: mass 33.87 kg, acceleration 13.31 m/s² | A net force acting on a mass of 33.87 kg produces an acceleration of 13.31 m/s². By Newton's second law, F_net = m a = 33.87 × 13.31 = 450.9 N. Direction of F_net is the same as the direction of the acceleration. This relation defines the inertial mass and is the foundation of classical dynamics. | F_net = m a | kinematics_1d | Compute net force from mass and acceleration using Newton's second law. |
5,564 | physics | mechanics | newton_second_law | 3 | worked_example | Newton's second law: mass 40.98 kg, acceleration 5.478 m/s² | A net force acting on a mass of 40.98 kg produces an acceleration of 5.478 m/s². By Newton's second law, F_net = m a = 40.98 × 5.478 = 224.5 N. Direction of F_net is the same as the direction of the acceleration. This relation defines the inertial mass and is the foundation of classical dynamics. | F_net = m a | kinematics_1d | Compute net force from mass and acceleration using Newton's second law. |
5,565 | physics | mechanics | newton_second_law | 3 | worked_example | Newton's second law: mass 3.191 kg, acceleration 9.429 m/s² | A net force acting on a mass of 3.191 kg produces an acceleration of 9.429 m/s². By Newton's second law, F_net = m a = 3.191 × 9.429 = 30.09 N. Direction of F_net is the same as the direction of the acceleration. This relation defines the inertial mass and is the foundation of classical dynamics. | F_net = m a | kinematics_1d | Compute net force from mass and acceleration using Newton's second law. |
5,566 | physics | mechanics | newton_second_law | 3 | worked_example | Newton's second law: mass 28.56 kg, acceleration 13.28 m/s² | A net force acting on a mass of 28.56 kg produces an acceleration of 13.28 m/s². By Newton's second law, F_net = m a = 28.56 × 13.28 = 379.2 N. Direction of F_net is the same as the direction of the acceleration. This relation defines the inertial mass and is the foundation of classical dynamics. | F_net = m a | kinematics_1d | Compute net force from mass and acceleration using Newton's second law. |
5,567 | physics | mechanics | mechanical_energy | 4 | worked_example | Conservation of mechanical energy: drop from height 20.69 m | An object of mass 5.196 kg is released from rest at height 20.69 m above a reference level. Taking gravitational potential energy as m g h and kinetic energy as (1/2) m v², conservation of mechanical energy (neglecting non-conservative work) yields (1/2) m v² = m g h, so v = sqrt(2 g h) = 20.14 m/s at the reference lev... | K + U = constant (conservative systems); U_g = m g h; K = (1/2) m v^2 | newton_second_law; work-energy theorem | Apply conservation of mechanical energy to free-fall motion. |
5,568 | physics | mechanics | mechanical_energy | 4 | worked_example | Conservation of mechanical energy: drop from height 9.049 m | An object of mass 6.336 kg is released from rest at height 9.049 m above a reference level. Taking gravitational potential energy as m g h and kinetic energy as (1/2) m v², conservation of mechanical energy (neglecting non-conservative work) yields (1/2) m v² = m g h, so v = sqrt(2 g h) = 13.32 m/s at the reference lev... | K + U = constant (conservative systems); U_g = m g h; K = (1/2) m v^2 | newton_second_law; work-energy theorem | Apply conservation of mechanical energy to free-fall motion. |
5,569 | physics | mechanics | mechanical_energy | 4 | worked_example | Conservation of mechanical energy: drop from height 13.69 m | An object of mass 3.782 kg is released from rest at height 13.69 m above a reference level. Taking gravitational potential energy as m g h and kinetic energy as (1/2) m v², conservation of mechanical energy (neglecting non-conservative work) yields (1/2) m v² = m g h, so v = sqrt(2 g h) = 16.39 m/s at the reference lev... | K + U = constant (conservative systems); U_g = m g h; K = (1/2) m v^2 | newton_second_law; work-energy theorem | Apply conservation of mechanical energy to free-fall motion. |
5,570 | physics | mechanics | mechanical_energy | 4 | worked_example | Conservation of mechanical energy: drop from height 28.94 m | An object of mass 10.09 kg is released from rest at height 28.94 m above a reference level. Taking gravitational potential energy as m g h and kinetic energy as (1/2) m v², conservation of mechanical energy (neglecting non-conservative work) yields (1/2) m v² = m g h, so v = sqrt(2 g h) = 23.82 m/s at the reference lev... | K + U = constant (conservative systems); U_g = m g h; K = (1/2) m v^2 | newton_second_law; work-energy theorem | Apply conservation of mechanical energy to free-fall motion. |
5,571 | physics | mechanics | mechanical_energy | 4 | worked_example | Conservation of mechanical energy: drop from height 18.59 m | An object of mass 14.46 kg is released from rest at height 18.59 m above a reference level. Taking gravitational potential energy as m g h and kinetic energy as (1/2) m v², conservation of mechanical energy (neglecting non-conservative work) yields (1/2) m v² = m g h, so v = sqrt(2 g h) = 19.09 m/s at the reference lev... | K + U = constant (conservative systems); U_g = m g h; K = (1/2) m v^2 | newton_second_law; work-energy theorem | Apply conservation of mechanical energy to free-fall motion. |
5,572 | physics | mechanics | mechanical_energy | 4 | worked_example | Conservation of mechanical energy: drop from height 30.28 m | An object of mass 7.937 kg is released from rest at height 30.28 m above a reference level. Taking gravitational potential energy as m g h and kinetic energy as (1/2) m v², conservation of mechanical energy (neglecting non-conservative work) yields (1/2) m v² = m g h, so v = sqrt(2 g h) = 24.37 m/s at the reference lev... | K + U = constant (conservative systems); U_g = m g h; K = (1/2) m v^2 | newton_second_law; work-energy theorem | Apply conservation of mechanical energy to free-fall motion. |
5,573 | physics | mechanics | mechanical_energy | 4 | worked_example | Conservation of mechanical energy: drop from height 19.47 m | An object of mass 18.43 kg is released from rest at height 19.47 m above a reference level. Taking gravitational potential energy as m g h and kinetic energy as (1/2) m v², conservation of mechanical energy (neglecting non-conservative work) yields (1/2) m v² = m g h, so v = sqrt(2 g h) = 19.54 m/s at the reference lev... | K + U = constant (conservative systems); U_g = m g h; K = (1/2) m v^2 | newton_second_law; work-energy theorem | Apply conservation of mechanical energy to free-fall motion. |
5,574 | physics | mechanics | mechanical_energy | 4 | worked_example | Conservation of mechanical energy: drop from height 31.9 m | An object of mass 12.02 kg is released from rest at height 31.9 m above a reference level. Taking gravitational potential energy as m g h and kinetic energy as (1/2) m v², conservation of mechanical energy (neglecting non-conservative work) yields (1/2) m v² = m g h, so v = sqrt(2 g h) = 25.01 m/s at the reference leve... | K + U = constant (conservative systems); U_g = m g h; K = (1/2) m v^2 | newton_second_law; work-energy theorem | Apply conservation of mechanical energy to free-fall motion. |
5,575 | physics | mechanics | mechanical_energy | 4 | worked_example | Conservation of mechanical energy: drop from height 37.54 m | An object of mass 13.41 kg is released from rest at height 37.54 m above a reference level. Taking gravitational potential energy as m g h and kinetic energy as (1/2) m v², conservation of mechanical energy (neglecting non-conservative work) yields (1/2) m v² = m g h, so v = sqrt(2 g h) = 27.14 m/s at the reference lev... | K + U = constant (conservative systems); U_g = m g h; K = (1/2) m v^2 | newton_second_law; work-energy theorem | Apply conservation of mechanical energy to free-fall motion. |
5,576 | physics | mechanics | mechanical_energy | 4 | worked_example | Conservation of mechanical energy: drop from height 4.737 m | An object of mass 1.738 kg is released from rest at height 4.737 m above a reference level. Taking gravitational potential energy as m g h and kinetic energy as (1/2) m v², conservation of mechanical energy (neglecting non-conservative work) yields (1/2) m v² = m g h, so v = sqrt(2 g h) = 9.639 m/s at the reference lev... | K + U = constant (conservative systems); U_g = m g h; K = (1/2) m v^2 | newton_second_law; work-energy theorem | Apply conservation of mechanical energy to free-fall motion. |
5,577 | physics | mechanics | mechanical_energy | 4 | worked_example | Conservation of mechanical energy: drop from height 1.732 m | An object of mass 0.2054 kg is released from rest at height 1.732 m above a reference level. Taking gravitational potential energy as m g h and kinetic energy as (1/2) m v², conservation of mechanical energy (neglecting non-conservative work) yields (1/2) m v² = m g h, so v = sqrt(2 g h) = 5.829 m/s at the reference le... | K + U = constant (conservative systems); U_g = m g h; K = (1/2) m v^2 | newton_second_law; work-energy theorem | Apply conservation of mechanical energy to free-fall motion. |
5,578 | physics | mechanics | mechanical_energy | 4 | worked_example | Conservation of mechanical energy: drop from height 0.9429 m | An object of mass 1.191 kg is released from rest at height 0.9429 m above a reference level. Taking gravitational potential energy as m g h and kinetic energy as (1/2) m v², conservation of mechanical energy (neglecting non-conservative work) yields (1/2) m v² = m g h, so v = sqrt(2 g h) = 4.3 m/s at the reference leve... | K + U = constant (conservative systems); U_g = m g h; K = (1/2) m v^2 | newton_second_law; work-energy theorem | Apply conservation of mechanical energy to free-fall motion. |
5,579 | physics | mechanics | mechanical_energy | 4 | worked_example | Conservation of mechanical energy: drop from height 0.5004 m | An object of mass 5.63 kg is released from rest at height 0.5004 m above a reference level. Taking gravitational potential energy as m g h and kinetic energy as (1/2) m v², conservation of mechanical energy (neglecting non-conservative work) yields (1/2) m v² = m g h, so v = sqrt(2 g h) = 3.133 m/s at the reference lev... | K + U = constant (conservative systems); U_g = m g h; K = (1/2) m v^2 | newton_second_law; work-energy theorem | Apply conservation of mechanical energy to free-fall motion. |
5,580 | physics | mechanics | mechanical_energy | 4 | worked_example | Conservation of mechanical energy: drop from height 29.29 m | An object of mass 9.012 kg is released from rest at height 29.29 m above a reference level. Taking gravitational potential energy as m g h and kinetic energy as (1/2) m v², conservation of mechanical energy (neglecting non-conservative work) yields (1/2) m v² = m g h, so v = sqrt(2 g h) = 23.97 m/s at the reference lev... | K + U = constant (conservative systems); U_g = m g h; K = (1/2) m v^2 | newton_second_law; work-energy theorem | Apply conservation of mechanical energy to free-fall motion. |
5,581 | physics | mechanics | mechanical_energy | 4 | worked_example | Conservation of mechanical energy: drop from height 4.341 m | An object of mass 19.23 kg is released from rest at height 4.341 m above a reference level. Taking gravitational potential energy as m g h and kinetic energy as (1/2) m v², conservation of mechanical energy (neglecting non-conservative work) yields (1/2) m v² = m g h, so v = sqrt(2 g h) = 9.227 m/s at the reference lev... | K + U = constant (conservative systems); U_g = m g h; K = (1/2) m v^2 | newton_second_law; work-energy theorem | Apply conservation of mechanical energy to free-fall motion. |
5,582 | physics | mechanics | mechanical_energy | 4 | worked_example | Conservation of mechanical energy: drop from height 26.01 m | An object of mass 7.487 kg is released from rest at height 26.01 m above a reference level. Taking gravitational potential energy as m g h and kinetic energy as (1/2) m v², conservation of mechanical energy (neglecting non-conservative work) yields (1/2) m v² = m g h, so v = sqrt(2 g h) = 22.59 m/s at the reference lev... | K + U = constant (conservative systems); U_g = m g h; K = (1/2) m v^2 | newton_second_law; work-energy theorem | Apply conservation of mechanical energy to free-fall motion. |
5,583 | physics | mechanics | mechanical_energy | 4 | worked_example | Conservation of mechanical energy: drop from height 29.8 m | An object of mass 19.42 kg is released from rest at height 29.8 m above a reference level. Taking gravitational potential energy as m g h and kinetic energy as (1/2) m v², conservation of mechanical energy (neglecting non-conservative work) yields (1/2) m v² = m g h, so v = sqrt(2 g h) = 24.18 m/s at the reference leve... | K + U = constant (conservative systems); U_g = m g h; K = (1/2) m v^2 | newton_second_law; work-energy theorem | Apply conservation of mechanical energy to free-fall motion. |
5,584 | physics | mechanics | mechanical_energy | 4 | worked_example | Conservation of mechanical energy: drop from height 11.56 m | An object of mass 7.582 kg is released from rest at height 11.56 m above a reference level. Taking gravitational potential energy as m g h and kinetic energy as (1/2) m v², conservation of mechanical energy (neglecting non-conservative work) yields (1/2) m v² = m g h, so v = sqrt(2 g h) = 15.06 m/s at the reference lev... | K + U = constant (conservative systems); U_g = m g h; K = (1/2) m v^2 | newton_second_law; work-energy theorem | Apply conservation of mechanical energy to free-fall motion. |
5,585 | physics | mechanics | mechanical_energy | 4 | worked_example | Conservation of mechanical energy: drop from height 12.25 m | An object of mass 13.14 kg is released from rest at height 12.25 m above a reference level. Taking gravitational potential energy as m g h and kinetic energy as (1/2) m v², conservation of mechanical energy (neglecting non-conservative work) yields (1/2) m v² = m g h, so v = sqrt(2 g h) = 15.5 m/s at the reference leve... | K + U = constant (conservative systems); U_g = m g h; K = (1/2) m v^2 | newton_second_law; work-energy theorem | Apply conservation of mechanical energy to free-fall motion. |
5,586 | physics | mechanics | mechanical_energy | 4 | worked_example | Conservation of mechanical energy: drop from height 5.243 m | An object of mass 0.8929 kg is released from rest at height 5.243 m above a reference level. Taking gravitational potential energy as m g h and kinetic energy as (1/2) m v², conservation of mechanical energy (neglecting non-conservative work) yields (1/2) m v² = m g h, so v = sqrt(2 g h) = 10.14 m/s at the reference le... | K + U = constant (conservative systems); U_g = m g h; K = (1/2) m v^2 | newton_second_law; work-energy theorem | Apply conservation of mechanical energy to free-fall motion. |
5,587 | physics | mechanics | mechanical_energy | 4 | worked_example | Conservation of mechanical energy: drop from height 18.55 m | An object of mass 3.008 kg is released from rest at height 18.55 m above a reference level. Taking gravitational potential energy as m g h and kinetic energy as (1/2) m v², conservation of mechanical energy (neglecting non-conservative work) yields (1/2) m v² = m g h, so v = sqrt(2 g h) = 19.07 m/s at the reference lev... | K + U = constant (conservative systems); U_g = m g h; K = (1/2) m v^2 | newton_second_law; work-energy theorem | Apply conservation of mechanical energy to free-fall motion. |
5,588 | physics | mechanics | mechanical_energy | 4 | worked_example | Conservation of mechanical energy: drop from height 39.34 m | An object of mass 18.66 kg is released from rest at height 39.34 m above a reference level. Taking gravitational potential energy as m g h and kinetic energy as (1/2) m v², conservation of mechanical energy (neglecting non-conservative work) yields (1/2) m v² = m g h, so v = sqrt(2 g h) = 27.78 m/s at the reference lev... | K + U = constant (conservative systems); U_g = m g h; K = (1/2) m v^2 | newton_second_law; work-energy theorem | Apply conservation of mechanical energy to free-fall motion. |
5,589 | physics | mechanics | mechanical_energy | 4 | worked_example | Conservation of mechanical energy: drop from height 5.253 m | An object of mass 18.31 kg is released from rest at height 5.253 m above a reference level. Taking gravitational potential energy as m g h and kinetic energy as (1/2) m v², conservation of mechanical energy (neglecting non-conservative work) yields (1/2) m v² = m g h, so v = sqrt(2 g h) = 10.15 m/s at the reference lev... | K + U = constant (conservative systems); U_g = m g h; K = (1/2) m v^2 | newton_second_law; work-energy theorem | Apply conservation of mechanical energy to free-fall motion. |
5,590 | physics | mechanics | mechanical_energy | 4 | worked_example | Conservation of mechanical energy: drop from height 38.12 m | An object of mass 12.81 kg is released from rest at height 38.12 m above a reference level. Taking gravitational potential energy as m g h and kinetic energy as (1/2) m v², conservation of mechanical energy (neglecting non-conservative work) yields (1/2) m v² = m g h, so v = sqrt(2 g h) = 27.34 m/s at the reference lev... | K + U = constant (conservative systems); U_g = m g h; K = (1/2) m v^2 | newton_second_law; work-energy theorem | Apply conservation of mechanical energy to free-fall motion. |
5,591 | physics | mechanics | mechanical_energy | 4 | worked_example | Conservation of mechanical energy: drop from height 35.27 m | An object of mass 15.9 kg is released from rest at height 35.27 m above a reference level. Taking gravitational potential energy as m g h and kinetic energy as (1/2) m v², conservation of mechanical energy (neglecting non-conservative work) yields (1/2) m v² = m g h, so v = sqrt(2 g h) = 26.3 m/s at the reference level... | K + U = constant (conservative systems); U_g = m g h; K = (1/2) m v^2 | newton_second_law; work-energy theorem | Apply conservation of mechanical energy to free-fall motion. |
5,592 | physics | mechanics | mechanical_energy | 4 | worked_example | Conservation of mechanical energy: drop from height 7.396 m | An object of mass 18.65 kg is released from rest at height 7.396 m above a reference level. Taking gravitational potential energy as m g h and kinetic energy as (1/2) m v², conservation of mechanical energy (neglecting non-conservative work) yields (1/2) m v² = m g h, so v = sqrt(2 g h) = 12.04 m/s at the reference lev... | K + U = constant (conservative systems); U_g = m g h; K = (1/2) m v^2 | newton_second_law; work-energy theorem | Apply conservation of mechanical energy to free-fall motion. |
5,593 | physics | mechanics | mechanical_energy | 4 | worked_example | Conservation of mechanical energy: drop from height 29.13 m | An object of mass 2.429 kg is released from rest at height 29.13 m above a reference level. Taking gravitational potential energy as m g h and kinetic energy as (1/2) m v², conservation of mechanical energy (neglecting non-conservative work) yields (1/2) m v² = m g h, so v = sqrt(2 g h) = 23.9 m/s at the reference leve... | K + U = constant (conservative systems); U_g = m g h; K = (1/2) m v^2 | newton_second_law; work-energy theorem | Apply conservation of mechanical energy to free-fall motion. |
5,594 | physics | mechanics | mechanical_energy | 4 | worked_example | Conservation of mechanical energy: drop from height 35.51 m | An object of mass 9.214 kg is released from rest at height 35.51 m above a reference level. Taking gravitational potential energy as m g h and kinetic energy as (1/2) m v², conservation of mechanical energy (neglecting non-conservative work) yields (1/2) m v² = m g h, so v = sqrt(2 g h) = 26.39 m/s at the reference lev... | K + U = constant (conservative systems); U_g = m g h; K = (1/2) m v^2 | newton_second_law; work-energy theorem | Apply conservation of mechanical energy to free-fall motion. |
5,595 | physics | mechanics | mechanical_energy | 4 | worked_example | Conservation of mechanical energy: drop from height 11.29 m | An object of mass 13.96 kg is released from rest at height 11.29 m above a reference level. Taking gravitational potential energy as m g h and kinetic energy as (1/2) m v², conservation of mechanical energy (neglecting non-conservative work) yields (1/2) m v² = m g h, so v = sqrt(2 g h) = 14.88 m/s at the reference lev... | K + U = constant (conservative systems); U_g = m g h; K = (1/2) m v^2 | newton_second_law; work-energy theorem | Apply conservation of mechanical energy to free-fall motion. |
5,596 | physics | mechanics | mechanical_energy | 4 | worked_example | Conservation of mechanical energy: drop from height 22.75 m | An object of mass 0.6467 kg is released from rest at height 22.75 m above a reference level. Taking gravitational potential energy as m g h and kinetic energy as (1/2) m v², conservation of mechanical energy (neglecting non-conservative work) yields (1/2) m v² = m g h, so v = sqrt(2 g h) = 21.12 m/s at the reference le... | K + U = constant (conservative systems); U_g = m g h; K = (1/2) m v^2 | newton_second_law; work-energy theorem | Apply conservation of mechanical energy to free-fall motion. |
5,597 | physics | mechanics | mechanical_energy | 4 | worked_example | Conservation of mechanical energy: drop from height 27.83 m | An object of mass 1.453 kg is released from rest at height 27.83 m above a reference level. Taking gravitational potential energy as m g h and kinetic energy as (1/2) m v², conservation of mechanical energy (neglecting non-conservative work) yields (1/2) m v² = m g h, so v = sqrt(2 g h) = 23.36 m/s at the reference lev... | K + U = constant (conservative systems); U_g = m g h; K = (1/2) m v^2 | newton_second_law; work-energy theorem | Apply conservation of mechanical energy to free-fall motion. |
5,598 | physics | mechanics | mechanical_energy | 4 | worked_example | Conservation of mechanical energy: drop from height 16.2 m | An object of mass 4.466 kg is released from rest at height 16.2 m above a reference level. Taking gravitational potential energy as m g h and kinetic energy as (1/2) m v², conservation of mechanical energy (neglecting non-conservative work) yields (1/2) m v² = m g h, so v = sqrt(2 g h) = 17.83 m/s at the reference leve... | K + U = constant (conservative systems); U_g = m g h; K = (1/2) m v^2 | newton_second_law; work-energy theorem | Apply conservation of mechanical energy to free-fall motion. |
5,599 | physics | mechanics | mechanical_energy | 4 | worked_example | Conservation of mechanical energy: drop from height 1.968 m | An object of mass 7.673 kg is released from rest at height 1.968 m above a reference level. Taking gravitational potential energy as m g h and kinetic energy as (1/2) m v², conservation of mechanical energy (neglecting non-conservative work) yields (1/2) m v² = m g h, so v = sqrt(2 g h) = 6.213 m/s at the reference lev... | K + U = constant (conservative systems); U_g = m g h; K = (1/2) m v^2 | newton_second_law; work-energy theorem | Apply conservation of mechanical energy to free-fall motion. |
5,600 | physics | mechanics | mechanical_energy | 4 | worked_example | Conservation of mechanical energy: drop from height 18.9 m | An object of mass 8.763 kg is released from rest at height 18.9 m above a reference level. Taking gravitational potential energy as m g h and kinetic energy as (1/2) m v², conservation of mechanical energy (neglecting non-conservative work) yields (1/2) m v² = m g h, so v = sqrt(2 g h) = 19.25 m/s at the reference leve... | K + U = constant (conservative systems); U_g = m g h; K = (1/2) m v^2 | newton_second_law; work-energy theorem | Apply conservation of mechanical energy to free-fall motion. |
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