Question

A 0.513-kg basketball is dropped out of a window that is 5.29 m above the ground. The ball is caught by a person whose hands are 2.06 m above the ground. (a) How much work is done on the ball by its weight? What is the gravitational potential energy of the basketball, relative to the ground, when it is (b) released and (c) caught? (d) What is the change (PEf - PE0) in the ball's gravitational potential energy?

Answer 1

The work done on the ball by the force of gravity as it travels from the window (a distance h0 above the ground) to your friend's hands (a distance hf above the ground) may be determined by
W = Fg(h0 − hf) cos 0°
= mg(h0 − hf)
= (0.513 kg)(9.80 m/s2)(5.29m − 2.06 m)
= 16.24 J.

(b) The potential energy of the ball-Earth system relative to the ground when released is:
U0 = mgh0
= (0.513 kg)(9.80 m/s2)(5.29m)
= 26.595 J.

(c) The potential energy of the ball-Earth system relative to the ground when caught is:
Uf = mghf
= (0.513 kg)(9.80 m/s2)(2.06 m)
=10.35 J.

(d) Since the force of gravity is a conservative force, the work done on the ball by the force of gravity is the negative of the change in gravitational potential energy of the ball-Earth system. This may be expressed as:
W = −ΔU
which leads to the ratio
ΔU /W = −1.
In other words: (Uf - U0)/W = (10.356-26.595)/16.24 = -1