Question

A metal ball of mass 2 kg is allowed to fall freely from rest from a height of 5 m above the ground.

(a) Taking g=10 m $s^{-2}$,calculate:

(i) the potential energy possessed by the ball when it is initially at rest.

(ii) the kinetic energy of the ball just before it hits the ground?

(b) What happens to the mechanical energy after the ball hits the ground and comes to rest?

Answer 1

Given,

m = 2 kg

height, h = 5 m

g = $10m{s}^{-2}$

(a)

(i) potential energy, $U=mgh=2\times 10\times 5=100J$

(ii) Let v = final velocity before it hits the ground

u = initial velocity = 0

a = -g = $10m{s}^{-2}$

S = -10 m

we know $v^2=u^2+2aS{v}^2=0+2\times (-10)\times (-5)=100$

Kinetic energy, $K=\frac{1}{2}m{v}^2=\frac{1}{2}\times 2\times 100=100J$

(b) The total mechanical energy when the body was at rest was equal to potential energy as the body was at rest. When it starts to fall, PE starts to decrease and KE starts to increase. The change in KE is equal to change in PE. When the body hits the ground, PE is zero and KE is maximum or total mechanical energy is contributed by KE. When it hits the ground and comes to rest both KE and PE becomes zero and in turn, mechanical energy is also zero. The energy it possessed is dissipated as sound when it hit the ground.