There is a ball of mass 320 g. It has 625 J potential energy when released freely from a height. The speed with which it will hit the ground is

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Question

This question was previously asked in

NDA-II 2024 (GAT) Official Paper (Held On: 01 Sept, 2024)

Answer 1

CONCEPT:

Kinetic Energy and Potential Energy

When an object is at a height, it possesses potential energy due to its position in the gravitational field.
When the object is released and falls freely under gravity, this potential energy is converted into kinetic energy.
The potential energy (PE) at a height can be given by the formula:
PE = mgh
where m is the mass, g is the acceleration due to gravity, and h is the height.
When the object hits the ground, all the potential energy is converted into kinetic energy (KE), given by:
KE = 0.5 x m x v²
where v is the velocity with which the object hits the ground.
By equating the potential energy to the kinetic energy at the ground, we can find the velocity:
mgh = 0.5 x m x v²

EXPLANATION:

Given:
- Mass (m) = 320 g = 0.32 kg
- Potential Energy (PE) = 625 J
- Acceleration due to gravity (g) = 9.8 m/s² (standard value)
We know that the potential energy at height h is converted to kinetic energy at the ground:
mgh = 0.5 x m x v²
Since PE = mgh, we can solve for v by setting PE equal to KE:
625 J = 0.5 x 0.32 kg x v²
Simplifying for v:
- 625 J = 0.16 kg x v²
- v² = 625 J / 0.16 kg
- v² = 3906.25 m²/s²
- v = √3906.25 m²/s²
- v ≈ 62.5 m/s

Therefore, the speed with which the ball will hit the ground is 62.5 m/s.