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Question

# The following data was obtained for a body of mass 1 kg dropped from a height of 5 metres: Distance above ground Velocity 5 m 0 m/s 3.2 m 6 m/s 0 m 10 m/s Show by calculations that the above data verifies the law of conservation of energy (Neglect air resistance). (g = 10 m/s2).

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Solution

## Law of conservation of energy says if there are no external forces working on the object then at any point of time the total energy of the object remains conserved. Mass of the body = 1 kg Case I: Velocity of the body at height 5 m = 0 m/s The potential energy at the height of 5 meters = mgh = 1 kg$×$ 10 ms-2 $×$ 5 m = 50 J Kinetic energy at a height of 5 meters = 0 The total energy of the body at a height of 5 meters = 50 J +0 = 50 J Case II: The velocity of the body at the height of 3.2 meters = 6 m/s The potential energy at the height of 3.2 meters = mgh = 1 kg$×$ 10 ms-2 $×$ 3.2 m = 32 J Kinetic energy at a height of 5 meters = $\frac{1}{2}m{v}^{2}=\frac{1}{2}×1×{6}^{2}=18\mathrm{J}$ The total energy of the body at a height of 3.2 meters = 32 J + 18 J = 50 J Case III: The velocity of the body at the bottom = 10 m/s The potential energy at the bottom = mgh = 1 kg$×$ 10 ms-2 $×$ 0 m = 0 Kinetic energy at the bottom = $\frac{1}{2}m{v}^{2}=\frac{1}{2}×1×\left(10{\right)}^{2}=50\mathrm{J}$ The total energy of the body at the bottom = 0 J + 50 J = 50 J Hence, we can see through all the three cases that the total energy of the body was conserved throughout the motion.

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