Question

While leaning out a window that is 6.4 m above the ground, you drop a 0.60-kg basketball to a friend at ground level. Your friend catches the ball at a height of 1.4 m above the ground. Determine the following.

(a) the amount of work done by the force of gravity on the ball
(b) the gravitational potential energy of the ball-earth system, relative to the ground when it is released
(c) the gravitational potential energy of the ball-earth system, relative to the ground when it is caught
(d) the ratio of the change (PEf − PE0) in the gravitational potential energy of the ball-earth system to the work done on the ball by the force of gravity

Answer 1

Step-by-step explanation:

Given that,

Height of window (h)=6.4m

Mass of basketball M=0.6kg

If the ball is cough at a height of 1.4m

Let g=9.81m/s²

a. Work done by gravity is given by

weight =mg

Work done by the weight of the body is weight of the body times height of the body from the ground

Wg=w•h

Wg=mgh

The work is done to where the ball was caught is height i.e h=6.4-1.4

h=5m

Wg=0.6×9.81×5

Wg=29.43J

b. Gravitation potential energy relative to the ground is give as

Wp=mgh

This is the potential energy over the whole height i.e h=6.4m

Wp=0.6×9.81×6.4

Wp=37.67J

c. Gravitational potential energy relative to when the ball was caught

The height when the ball was caught is 1.4m

Therefore, potential energy is given as

Wp=mgh

Wp=0.6×9.81×1.4

Wp=8.24J

d. Ratio of change in potential energy to the work done

∆P.E= P.E(final)-P.E(initial)

∆P.E= 8.24-37.67

∆P.E= -29.43

The ratio will be ∆P.E/Wg

Ratio= -29.43/29.43

Ratio= -1

Which is right

Because work is conservative and

W=-∆U

Then

W/∆U=-1