1. A stone of mass 5.0 kg is dropped from a height of 2.0 m above the ground and

Question

1. A stone of mass 5.0 kg is dropped from a height of 2.0 m above the ground and falls
freely under the action of gravity. Neglect the air resistance acting on the stone.
A. Calculate the potential energy (PE), kinetic energy (KE) and the total mechanical
energy of the stone at the point of release.
B. The same (that is, PE, KE and the total energy) of the stone at a height of 0.5 m
above the ground as it falls.
C. Using the conservation of mechanical energy calculate the speed of the stone just
before it strikes the ground. What is the total mechanical energy of the stone in this
case? Does it verify the principle of conservation of energy?
D. Calculate the work done by the force of gravity

Explanation / Answer

Mass, m = 5 kg Height, h = 2 m A) PE = m g h = 5 * 9.8 * 2 = 98 J KE = (1/2) m v^2 = (1/2) * 5 * 0 = 0 J TE = PE + KE = 98 + 0 = 98 J B) Velocity at a height of 0.5 m above the ground, V = sqrt ( Vi ^2 + 2gs ) = sqrt [ 2 * 9.8 * ( 2 - 0.5 ) ] = 5.42 m/s PE = m g h = 5 * 9.8 * 0.5 = 24.5 J KE = (1/2) m v^2 = 0.5 * 5 * 5.42^2 = 73.5 J TE = PE + KE = 98 J c) Velocity on reaching the ground, V = sqrt ( Vi ^2 + 2gs ) = sqrt [ 2 * 9.8 * ( 2 ) ] = 6.27 m/s PE = m g h = 5 * 9.8 * 0 = 0 J KE = (1/2) m v^2 = 0.5 * 5 * 6.27^2 = 98 J TE = PE + KE = 98 J As the total energy is same in the cases of A, B, and C, the principal of conervation of energy is verified. D) Work done = F * S = mgh = 5 * 9.8 * 2 = 98 J From work-energy theorem, Work done, W = Change in kinetic energy = (1/2) m [ Vf^2 - Vi^2 ] =(1/2) * 5 * [ 6.27^2 - 0^2 ] = 98 J Hence, work energy theorem is verified.

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