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A body of mass 2 kg kept at the top of a building of height 10 m possess a potential energy of 200 J. What will be the kinetic energy of this body when it hits the ground? (Neglect air resistance)

This question was previously asked in
RRB Technician 2026 Grade 3 Question Paper (7-Oct-2026) (Shift 3)
The correct answer is

200 J

By the law of conservation of mechanical energy, when air resistance is neglected, the total mechanical energy (potential energy + kinetic energy) of a freely falling body stays constant during the fall.

At the top, the body is at rest, so its kinetic energy is zero and its total energy is all potential energy, 200 J. Just as it hits the ground, its height is zero, so its potential energy is zero and all of the energy has become kinetic energy.

Therefore, kinetic energy on hitting the ground = potential energy at the top = 200 J.

Check: with \(g = 10\) m/s2, the potential energy is \(mgh = 2 \times 10 \times 10 = 200\) J. The speed on reaching the ground is given by \(v^{2} = 2gh = 2 \times 10 \times 10 = 200\), so \(KE = \frac{1}{2}mv^{2} = \frac{1}{2} \times 2 \times 200 = 200\) J, the same value.

Hence, 200 J.

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