Question

A 320 g rubber ball is dropped from 2.0 m above the ground, bounces, and returns to a maximum height 1.2 m above the ground. If air resistance does –1.8 J of work on the ball, how much energy goes into heating the ground and the ball when it bounces?

Answer 1

Answer:

0.71 J

Explanation:

320 g = 0.32 kg

According to law of energy conservation, the energy loss to external environment (air, ground) can be credited to the change in mechanical energy of the ball.

As the ball was dropped at H = 2 m above the ground then later reaches its maximum height at h = 1.2m, tts instant speed at those 2 points must be 0. So the kinetic energy at those points are 0 as well. The change in mechanical energy is the change in potential energy.

Let g = 9.81 m/s2

$\Delta E_p = mgH - mgh = mg(H - h) = 0.32*9.81*(2 - 1.2) = 2.51 J$

Since 1.8J of 2.51 J is due to work by air resistance, the rest of the energy (2.51 - 1.8 = 0.71 J) is would go to heating in the ground and ball when it bounces.