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
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301 | physics | mechanics | newton_second_law | 3 | worked_example | Newton's second law: mass 46.13 kg, acceleration 7.673 m/s² | A net force acting on a mass of 46.13 kg produces an acceleration of 7.673 m/s². By Newton's second law, F_net = m a = 46.13 × 7.673 = 354 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. |
302 | physics | mechanics | newton_second_law | 3 | worked_example | Newton's second law: mass 44.03 kg, acceleration 12.97 m/s² | A net force acting on a mass of 44.03 kg produces an acceleration of 12.97 m/s². By Newton's second law, F_net = m a = 44.03 × 12.97 = 571.2 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. |
303 | physics | mechanics | newton_second_law | 3 | worked_example | Newton's second law: mass 14.17 kg, acceleration 11.87 m/s² | A net force acting on a mass of 14.17 kg produces an acceleration of 11.87 m/s². By Newton's second law, F_net = m a = 14.17 × 11.87 = 168.3 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. |
304 | physics | mechanics | newton_second_law | 3 | worked_example | Newton's second law: mass 21.04 kg, acceleration 14.02 m/s² | A net force acting on a mass of 21.04 kg produces an acceleration of 14.02 m/s². By Newton's second law, F_net = m a = 21.04 × 14.02 = 295 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. |
305 | physics | mechanics | newton_second_law | 3 | worked_example | Newton's second law: mass 25.63 kg, acceleration 12.33 m/s² | A net force acting on a mass of 25.63 kg produces an acceleration of 12.33 m/s². By Newton's second law, F_net = m a = 25.63 × 12.33 = 316 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. |
306 | physics | mechanics | newton_second_law | 3 | worked_example | Newton's second law: mass 14.5 kg, acceleration 4.548 m/s² | A net force acting on a mass of 14.5 kg produces an acceleration of 4.548 m/s². By Newton's second law, F_net = m a = 14.5 × 4.548 = 65.96 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. |
307 | physics | mechanics | newton_second_law | 3 | worked_example | Newton's second law: mass 29.55 kg, acceleration 14.98 m/s² | A net force acting on a mass of 29.55 kg produces an acceleration of 14.98 m/s². By Newton's second law, F_net = m a = 29.55 × 14.98 = 442.8 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. |
308 | physics | mechanics | newton_second_law | 3 | worked_example | Newton's second law: mass 24.74 kg, acceleration 2.314 m/s² | A net force acting on a mass of 24.74 kg produces an acceleration of 2.314 m/s². By Newton's second law, F_net = m a = 24.74 × 2.314 = 57.24 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. |
309 | physics | mechanics | newton_second_law | 3 | worked_example | Newton's second law: mass 27.16 kg, acceleration 5.242 m/s² | A net force acting on a mass of 27.16 kg produces an acceleration of 5.242 m/s². By Newton's second law, F_net = m a = 27.16 × 5.242 = 142.4 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. |
310 | physics | mechanics | newton_second_law | 3 | worked_example | Newton's second law: mass 27.82 kg, acceleration 8.197 m/s² | A net force acting on a mass of 27.82 kg produces an acceleration of 8.197 m/s². By Newton's second law, F_net = m a = 27.82 × 8.197 = 228 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. |
311 | physics | mechanics | newton_second_law | 3 | worked_example | Newton's second law: mass 23.04 kg, acceleration 4.894 m/s² | A net force acting on a mass of 23.04 kg produces an acceleration of 4.894 m/s². By Newton's second law, F_net = m a = 23.04 × 4.894 = 112.8 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. |
312 | physics | mechanics | newton_second_law | 3 | worked_example | Newton's second law: mass 9.838 kg, acceleration 10.49 m/s² | A net force acting on a mass of 9.838 kg produces an acceleration of 10.49 m/s². By Newton's second law, F_net = m a = 9.838 × 10.49 = 103.2 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. |
313 | physics | mechanics | newton_second_law | 3 | worked_example | Newton's second law: mass 28.8 kg, acceleration 3.58 m/s² | A net force acting on a mass of 28.8 kg produces an acceleration of 3.58 m/s². By Newton's second law, F_net = m a = 28.8 × 3.58 = 103.1 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. |
314 | physics | mechanics | newton_second_law | 3 | worked_example | Newton's second law: mass 38.89 kg, acceleration 0.7503 m/s² | A net force acting on a mass of 38.89 kg produces an acceleration of 0.7503 m/s². By Newton's second law, F_net = m a = 38.89 × 0.7503 = 29.18 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. |
315 | physics | mechanics | newton_second_law | 3 | worked_example | Newton's second law: mass 37.36 kg, acceleration 10.61 m/s² | A net force acting on a mass of 37.36 kg produces an acceleration of 10.61 m/s². By Newton's second law, F_net = m a = 37.36 × 10.61 = 396.3 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. |
316 | physics | mechanics | newton_second_law | 3 | worked_example | Newton's second law: mass 40.66 kg, acceleration 5.853 m/s² | A net force acting on a mass of 40.66 kg produces an acceleration of 5.853 m/s². By Newton's second law, F_net = m a = 40.66 × 5.853 = 238 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. |
317 | physics | mechanics | newton_second_law | 3 | worked_example | Newton's second law: mass 33.35 kg, acceleration 12.33 m/s² | A net force acting on a mass of 33.35 kg produces an acceleration of 12.33 m/s². By Newton's second law, F_net = m a = 33.35 × 12.33 = 411.2 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. |
318 | physics | mechanics | newton_second_law | 3 | worked_example | Newton's second law: mass 49.05 kg, acceleration 7.48 m/s² | A net force acting on a mass of 49.05 kg produces an acceleration of 7.48 m/s². By Newton's second law, F_net = m a = 49.05 × 7.48 = 366.9 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. |
319 | physics | mechanics | newton_second_law | 3 | worked_example | Newton's second law: mass 2.332 kg, acceleration 7.584 m/s² | A net force acting on a mass of 2.332 kg produces an acceleration of 7.584 m/s². By Newton's second law, F_net = m a = 2.332 × 7.584 = 17.69 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. |
320 | physics | mechanics | newton_second_law | 3 | worked_example | Newton's second law: mass 29.71 kg, acceleration 13.06 m/s² | A net force acting on a mass of 29.71 kg produces an acceleration of 13.06 m/s². By Newton's second law, F_net = m a = 29.71 × 13.06 = 388 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. |
321 | physics | mechanics | newton_second_law | 3 | worked_example | Newton's second law: mass 43.77 kg, acceleration 6.661 m/s² | A net force acting on a mass of 43.77 kg produces an acceleration of 6.661 m/s². By Newton's second law, F_net = m a = 43.77 × 6.661 = 291.5 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. |
322 | physics | mechanics | newton_second_law | 3 | worked_example | Newton's second law: mass 26.53 kg, acceleration 6.908 m/s² | A net force acting on a mass of 26.53 kg produces an acceleration of 6.908 m/s². By Newton's second law, F_net = m a = 26.53 × 6.908 = 183.3 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. |
323 | physics | mechanics | newton_second_law | 3 | worked_example | Newton's second law: mass 36.26 kg, acceleration 6.209 m/s² | A net force acting on a mass of 36.26 kg produces an acceleration of 6.209 m/s². By Newton's second law, F_net = m a = 36.26 × 6.209 = 225.1 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. |
324 | physics | mechanics | newton_second_law | 3 | worked_example | Newton's second law: mass 32.91 kg, acceleration 2.4 m/s² | A net force acting on a mass of 32.91 kg produces an acceleration of 2.4 m/s². By Newton's second law, F_net = m a = 32.91 × 2.4 = 78.99 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. |
325 | physics | mechanics | newton_second_law | 3 | worked_example | Newton's second law: mass 23.74 kg, acceleration 14.54 m/s² | A net force acting on a mass of 23.74 kg produces an acceleration of 14.54 m/s². By Newton's second law, F_net = m a = 23.74 × 14.54 = 345.2 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. |
326 | physics | mechanics | newton_second_law | 3 | worked_example | Newton's second law: mass 17.26 kg, acceleration 10.42 m/s² | A net force acting on a mass of 17.26 kg produces an acceleration of 10.42 m/s². By Newton's second law, F_net = m a = 17.26 × 10.42 = 179.9 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. |
327 | physics | mechanics | newton_second_law | 3 | worked_example | Newton's second law: mass 32.67 kg, acceleration 12.79 m/s² | A net force acting on a mass of 32.67 kg produces an acceleration of 12.79 m/s². By Newton's second law, F_net = m a = 32.67 × 12.79 = 417.9 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. |
328 | physics | mechanics | newton_second_law | 3 | worked_example | Newton's second law: mass 42.69 kg, acceleration 12.9 m/s² | A net force acting on a mass of 42.69 kg produces an acceleration of 12.9 m/s². By Newton's second law, F_net = m a = 42.69 × 12.9 = 550.9 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. |
329 | physics | mechanics | newton_second_law | 3 | worked_example | Newton's second law: mass 19.31 kg, acceleration 4.818 m/s² | A net force acting on a mass of 19.31 kg produces an acceleration of 4.818 m/s². By Newton's second law, F_net = m a = 19.31 × 4.818 = 93.04 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. |
330 | physics | mechanics | newton_second_law | 3 | worked_example | Newton's second law: mass 36.08 kg, acceleration 11.42 m/s² | A net force acting on a mass of 36.08 kg produces an acceleration of 11.42 m/s². By Newton's second law, F_net = m a = 36.08 × 11.42 = 411.8 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. |
331 | physics | mechanics | newton_second_law | 3 | worked_example | Newton's second law: mass 43.68 kg, acceleration 0.6349 m/s² | A net force acting on a mass of 43.68 kg produces an acceleration of 0.6349 m/s². By Newton's second law, F_net = m a = 43.68 × 0.6349 = 27.73 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. |
332 | physics | mechanics | newton_second_law | 3 | worked_example | Newton's second law: mass 3.887 kg, acceleration 9.504 m/s² | A net force acting on a mass of 3.887 kg produces an acceleration of 9.504 m/s². By Newton's second law, F_net = m a = 3.887 × 9.504 = 36.94 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. |
333 | physics | mechanics | newton_second_law | 3 | worked_example | Newton's second law: mass 46.09 kg, acceleration 14.96 m/s² | A net force acting on a mass of 46.09 kg produces an acceleration of 14.96 m/s². By Newton's second law, F_net = m a = 46.09 × 14.96 = 689.5 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. |
334 | physics | mechanics | newton_second_law | 3 | worked_example | Newton's second law: mass 37.46 kg, acceleration 6.566 m/s² | A net force acting on a mass of 37.46 kg produces an acceleration of 6.566 m/s². By Newton's second law, F_net = m a = 37.46 × 6.566 = 246 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. |
335 | physics | mechanics | newton_second_law | 3 | worked_example | Newton's second law: mass 5.373 kg, acceleration 9.543 m/s² | A net force acting on a mass of 5.373 kg produces an acceleration of 9.543 m/s². By Newton's second law, F_net = m a = 5.373 × 9.543 = 51.27 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. |
336 | physics | mechanics | newton_second_law | 3 | worked_example | Newton's second law: mass 43.69 kg, acceleration 6.711 m/s² | A net force acting on a mass of 43.69 kg produces an acceleration of 6.711 m/s². By Newton's second law, F_net = m a = 43.69 × 6.711 = 293.2 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. |
337 | physics | mechanics | newton_second_law | 3 | worked_example | Newton's second law: mass 34.85 kg, acceleration 13.56 m/s² | A net force acting on a mass of 34.85 kg produces an acceleration of 13.56 m/s². By Newton's second law, F_net = m a = 34.85 × 13.56 = 472.6 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. |
338 | physics | mechanics | newton_second_law | 3 | worked_example | Newton's second law: mass 2.777 kg, acceleration 11.96 m/s² | A net force acting on a mass of 2.777 kg produces an acceleration of 11.96 m/s². By Newton's second law, F_net = m a = 2.777 × 11.96 = 33.21 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. |
339 | physics | mechanics | newton_second_law | 3 | worked_example | Newton's second law: mass 15.02 kg, acceleration 5.685 m/s² | A net force acting on a mass of 15.02 kg produces an acceleration of 5.685 m/s². By Newton's second law, F_net = m a = 15.02 × 5.685 = 85.4 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. |
340 | physics | mechanics | newton_second_law | 3 | worked_example | Newton's second law: mass 7.706 kg, acceleration 8.014 m/s² | A net force acting on a mass of 7.706 kg produces an acceleration of 8.014 m/s². By Newton's second law, F_net = m a = 7.706 × 8.014 = 61.76 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. |
341 | physics | mechanics | newton_second_law | 3 | worked_example | Newton's second law: mass 28.51 kg, acceleration 11.91 m/s² | A net force acting on a mass of 28.51 kg produces an acceleration of 11.91 m/s². By Newton's second law, F_net = m a = 28.51 × 11.91 = 339.6 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. |
342 | physics | mechanics | newton_second_law | 3 | worked_example | Newton's second law: mass 8.914 kg, acceleration 1.277 m/s² | A net force acting on a mass of 8.914 kg produces an acceleration of 1.277 m/s². By Newton's second law, F_net = m a = 8.914 × 1.277 = 11.38 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. |
343 | physics | mechanics | newton_second_law | 3 | worked_example | Newton's second law: mass 43.61 kg, acceleration 9.334 m/s² | A net force acting on a mass of 43.61 kg produces an acceleration of 9.334 m/s². By Newton's second law, F_net = m a = 43.61 × 9.334 = 407 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. |
344 | physics | mechanics | newton_second_law | 3 | worked_example | Newton's second law: mass 12.42 kg, acceleration 13.7 m/s² | A net force acting on a mass of 12.42 kg produces an acceleration of 13.7 m/s². By Newton's second law, F_net = m a = 12.42 × 13.7 = 170.2 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. |
345 | physics | mechanics | newton_second_law | 3 | worked_example | Newton's second law: mass 7.584 kg, acceleration 6.971 m/s² | A net force acting on a mass of 7.584 kg produces an acceleration of 6.971 m/s². By Newton's second law, F_net = m a = 7.584 × 6.971 = 52.87 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. |
346 | physics | mechanics | newton_second_law | 3 | worked_example | Newton's second law: mass 13.07 kg, acceleration 3.904 m/s² | A net force acting on a mass of 13.07 kg produces an acceleration of 3.904 m/s². By Newton's second law, F_net = m a = 13.07 × 3.904 = 51.04 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. |
347 | physics | mechanics | newton_second_law | 3 | worked_example | Newton's second law: mass 0.9652 kg, acceleration 12.09 m/s² | A net force acting on a mass of 0.9652 kg produces an acceleration of 12.09 m/s². By Newton's second law, F_net = m a = 0.9652 × 12.09 = 11.67 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. |
348 | physics | mechanics | newton_second_law | 3 | worked_example | Newton's second law: mass 45.11 kg, acceleration 10.2 m/s² | A net force acting on a mass of 45.11 kg produces an acceleration of 10.2 m/s². By Newton's second law, F_net = m a = 45.11 × 10.2 = 460 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. |
349 | physics | mechanics | newton_second_law | 3 | worked_example | Newton's second law: mass 8.32 kg, acceleration 6.682 m/s² | A net force acting on a mass of 8.32 kg produces an acceleration of 6.682 m/s². By Newton's second law, F_net = m a = 8.32 × 6.682 = 55.59 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. |
350 | physics | mechanics | newton_second_law | 3 | worked_example | Newton's second law: mass 17.61 kg, acceleration 8.855 m/s² | A net force acting on a mass of 17.61 kg produces an acceleration of 8.855 m/s². By Newton's second law, F_net = m a = 17.61 × 8.855 = 155.9 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. |
351 | physics | mechanics | newton_second_law | 3 | worked_example | Newton's second law: mass 32.13 kg, acceleration 6.422 m/s² | A net force acting on a mass of 32.13 kg produces an acceleration of 6.422 m/s². By Newton's second law, F_net = m a = 32.13 × 6.422 = 206.3 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. |
352 | physics | mechanics | newton_second_law | 3 | worked_example | Newton's second law: mass 12.88 kg, acceleration 12.7 m/s² | A net force acting on a mass of 12.88 kg produces an acceleration of 12.7 m/s². By Newton's second law, F_net = m a = 12.88 × 12.7 = 163.5 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. |
353 | physics | mechanics | newton_second_law | 3 | worked_example | Newton's second law: mass 10.36 kg, acceleration 5.832 m/s² | A net force acting on a mass of 10.36 kg produces an acceleration of 5.832 m/s². By Newton's second law, F_net = m a = 10.36 × 5.832 = 60.43 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. |
354 | physics | mechanics | newton_second_law | 3 | worked_example | Newton's second law: mass 24.42 kg, acceleration 3.634 m/s² | A net force acting on a mass of 24.42 kg produces an acceleration of 3.634 m/s². By Newton's second law, F_net = m a = 24.42 × 3.634 = 88.75 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. |
355 | physics | mechanics | newton_second_law | 3 | worked_example | Newton's second law: mass 28.81 kg, acceleration 8.665 m/s² | A net force acting on a mass of 28.81 kg produces an acceleration of 8.665 m/s². By Newton's second law, F_net = m a = 28.81 × 8.665 = 249.6 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. |
356 | physics | mechanics | mechanical_energy | 4 | worked_example | Conservation of mechanical energy: drop from height 12.16 m | An object of mass 19.86 kg is released from rest at height 12.16 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) = 15.44 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. |
357 | physics | mechanics | mechanical_energy | 4 | worked_example | Conservation of mechanical energy: drop from height 26.5 m | An object of mass 19.56 kg is released from rest at height 26.5 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) = 22.8 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. |
358 | physics | mechanics | mechanical_energy | 4 | worked_example | Conservation of mechanical energy: drop from height 22.85 m | An object of mass 5.635 kg is released from rest at height 22.85 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) = 21.17 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. |
359 | physics | mechanics | mechanical_energy | 4 | worked_example | Conservation of mechanical energy: drop from height 29.91 m | An object of mass 13.78 kg is released from rest at height 29.91 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) = 24.22 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. |
360 | physics | mechanics | mechanical_energy | 4 | worked_example | Conservation of mechanical energy: drop from height 24.45 m | An object of mass 1.171 kg is released from rest at height 24.45 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) = 21.9 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. |
361 | physics | mechanics | mechanical_energy | 4 | worked_example | Conservation of mechanical energy: drop from height 36.21 m | An object of mass 10.04 kg is released from rest at height 36.21 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) = 26.65 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. |
362 | physics | mechanics | mechanical_energy | 4 | worked_example | Conservation of mechanical energy: drop from height 32.05 m | An object of mass 5.867 kg is released from rest at height 32.05 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) = 25.07 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. |
363 | physics | mechanics | mechanical_energy | 4 | worked_example | Conservation of mechanical energy: drop from height 14.42 m | An object of mass 12.22 kg is released from rest at height 14.42 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) = 16.82 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. |
364 | physics | mechanics | mechanical_energy | 4 | worked_example | Conservation of mechanical energy: drop from height 25.03 m | An object of mass 12.81 kg is released from rest at height 25.03 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) = 22.15 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. |
365 | physics | mechanics | mechanical_energy | 4 | worked_example | Conservation of mechanical energy: drop from height 28.98 m | An object of mass 13.62 kg is released from rest at height 28.98 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) = 23.84 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. |
366 | physics | mechanics | mechanical_energy | 4 | worked_example | Conservation of mechanical energy: drop from height 33.61 m | An object of mass 13.25 kg is released from rest at height 33.61 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) = 25.68 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. |
367 | physics | mechanics | mechanical_energy | 4 | worked_example | Conservation of mechanical energy: drop from height 36.18 m | An object of mass 12.64 kg is released from rest at height 36.18 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) = 26.64 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. |
368 | physics | mechanics | mechanical_energy | 4 | worked_example | Conservation of mechanical energy: drop from height 12.7 m | An object of mass 13 kg is released from rest at height 12.7 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) = 15.78 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. |
369 | physics | mechanics | mechanical_energy | 4 | worked_example | Conservation of mechanical energy: drop from height 23.39 m | An object of mass 8.928 kg is released from rest at height 23.39 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) = 21.42 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. |
370 | physics | mechanics | mechanical_energy | 4 | worked_example | Conservation of mechanical energy: drop from height 4.06 m | An object of mass 14.7 kg is released from rest at height 4.06 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) = 8.924 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. |
371 | physics | mechanics | mechanical_energy | 4 | worked_example | Conservation of mechanical energy: drop from height 30.03 m | An object of mass 6.043 kg is released from rest at height 30.03 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) = 24.27 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. |
372 | physics | mechanics | mechanical_energy | 4 | worked_example | Conservation of mechanical energy: drop from height 5.72 m | An object of mass 3.678 kg is released from rest at height 5.72 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) = 10.59 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. |
373 | physics | mechanics | mechanical_energy | 4 | worked_example | Conservation of mechanical energy: drop from height 38.87 m | An object of mass 10.88 kg is released from rest at height 38.87 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) = 27.61 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. |
374 | physics | mechanics | mechanical_energy | 4 | worked_example | Conservation of mechanical energy: drop from height 36.58 m | An object of mass 10.71 kg is released from rest at height 36.58 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) = 26.79 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. |
375 | physics | mechanics | mechanical_energy | 4 | worked_example | Conservation of mechanical energy: drop from height 10.65 m | An object of mass 16.64 kg is released from rest at height 10.65 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) = 14.45 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. |
376 | physics | mechanics | mechanical_energy | 4 | worked_example | Conservation of mechanical energy: drop from height 19.53 m | An object of mass 16.53 kg is released from rest at height 19.53 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) = 19.57 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. |
377 | physics | mechanics | mechanical_energy | 4 | worked_example | Conservation of mechanical energy: drop from height 29.99 m | An object of mass 16.17 kg is released from rest at height 29.99 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) = 24.25 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. |
378 | physics | mechanics | mechanical_energy | 4 | worked_example | Conservation of mechanical energy: drop from height 5.049 m | An object of mass 6.907 kg is released from rest at height 5.049 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) = 9.951 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. |
379 | physics | mechanics | mechanical_energy | 4 | worked_example | Conservation of mechanical energy: drop from height 6.06 m | An object of mass 19.27 kg is released from rest at height 6.06 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) = 10.9 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. |
380 | physics | mechanics | mechanical_energy | 4 | worked_example | Conservation of mechanical energy: drop from height 34.48 m | An object of mass 19.34 kg is released from rest at height 34.48 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 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. |
381 | physics | mechanics | mechanical_energy | 4 | worked_example | Conservation of mechanical energy: drop from height 39.21 m | An object of mass 14.54 kg is released from rest at height 39.21 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) = 27.73 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. |
382 | physics | mechanics | mechanical_energy | 4 | worked_example | Conservation of mechanical energy: drop from height 32.28 m | An object of mass 19.35 kg is released from rest at height 32.28 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) = 25.16 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. |
383 | physics | mechanics | mechanical_energy | 4 | worked_example | Conservation of mechanical energy: drop from height 31.73 m | An object of mass 7.442 kg is released from rest at height 31.73 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) = 24.95 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. |
384 | physics | mechanics | mechanical_energy | 4 | worked_example | Conservation of mechanical energy: drop from height 21.69 m | An object of mass 0.4756 kg is released from rest at height 21.69 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) = 20.63 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. |
385 | physics | mechanics | mechanical_energy | 4 | worked_example | Conservation of mechanical energy: drop from height 27.08 m | An object of mass 9.205 kg is released from rest at height 27.08 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) = 23.04 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. |
386 | physics | mechanics | mechanical_energy | 4 | worked_example | Conservation of mechanical energy: drop from height 23.59 m | An object of mass 13.51 kg is released from rest at height 23.59 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) = 21.51 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. |
387 | physics | mechanics | mechanical_energy | 4 | worked_example | Conservation of mechanical energy: drop from height 37.64 m | An object of mass 16.48 kg is released from rest at height 37.64 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) = 27.17 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. |
388 | physics | mechanics | mechanical_energy | 4 | worked_example | Conservation of mechanical energy: drop from height 9.736 m | An object of mass 2.345 kg is released from rest at height 9.736 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) = 13.82 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. |
389 | physics | mechanics | mechanical_energy | 4 | worked_example | Conservation of mechanical energy: drop from height 35.43 m | An object of mass 0.6955 kg is released from rest at height 35.43 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) = 26.36 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. |
390 | physics | mechanics | mechanical_energy | 4 | worked_example | Conservation of mechanical energy: drop from height 36.65 m | An object of mass 11.32 kg is released from rest at height 36.65 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) = 26.81 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. |
391 | physics | mechanics | mechanical_energy | 4 | worked_example | Conservation of mechanical energy: drop from height 2.997 m | An object of mass 4.583 kg is released from rest at height 2.997 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) = 7.667 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. |
392 | physics | mechanics | mechanical_energy | 4 | worked_example | Conservation of mechanical energy: drop from height 36.42 m | An object of mass 16.51 kg is released from rest at height 36.42 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) = 26.73 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. |
393 | physics | mechanics | mechanical_energy | 4 | worked_example | Conservation of mechanical energy: drop from height 16.63 m | An object of mass 6.183 kg is released from rest at height 16.63 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) = 18.06 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. |
394 | physics | mechanics | mechanical_energy | 4 | worked_example | Conservation of mechanical energy: drop from height 37.88 m | An object of mass 2.968 kg is released from rest at height 37.88 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) = 27.26 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. |
395 | physics | mechanics | mechanical_energy | 4 | worked_example | Conservation of mechanical energy: drop from height 19.96 m | An object of mass 6.226 kg is released from rest at height 19.96 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) = 19.79 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. |
396 | physics | mechanics | mechanical_energy | 4 | worked_example | Conservation of mechanical energy: drop from height 35.55 m | An object of mass 2.124 kg is released from rest at height 35.55 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) = 26.4 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. |
397 | physics | mechanics | mechanical_energy | 4 | worked_example | Conservation of mechanical energy: drop from height 18.42 m | An object of mass 2.886 kg is released from rest at height 18.42 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) = 19.01 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. |
398 | physics | mechanics | mechanical_energy | 4 | worked_example | Conservation of mechanical energy: drop from height 29.85 m | An object of mass 13.48 kg is released from rest at height 29.85 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) = 24.2 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. |
399 | physics | mechanics | mechanical_energy | 4 | worked_example | Conservation of mechanical energy: drop from height 17.06 m | An object of mass 18.93 kg is released from rest at height 17.06 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) = 18.29 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. |
400 | physics | mechanics | mechanical_energy | 4 | worked_example | Conservation of mechanical energy: drop from height 6.604 m | An object of mass 14.9 kg is released from rest at height 6.604 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) = 11.38 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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