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Question

A 0.60 kg basketball is dropped out of the window that is 6.3 m above the ground. The ball is caught by a person whose hands are 1.3 m above the ground. How much work is done on the ball by its weight?
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What is the gravitational potential energy of the basketball, relative to the ground when it is released?
J
What is the gravitational potential energy of the basketball when it is caught?
J
How is the change (PEf - PE0) in the ball's gravitational potential energy related to the work done by its weight?

Answer 1

1) W = F*d

here w = work, F = force , d = distance.

We know that from Newton's second law

gravity force F = mg

m = mass = 0.60kg,

g = 9.8m/sec^2

F = 0.6*9.8

F = 5.88N

distance = 6.3-1.3 = 5m

W = 5.88*5 =29.4J

2) potential energy = mgh

Pef = 0.6*9.8*6.3

= 37.044J

3) potential energy Pe0 = mgh

= 0.6*9.8*1.3

= 7.644J

4) the change in gravitational potential energy is the effect of the force of gravity actng through the

distance the object has dropped.

In other words the change in gravitational potential energy is equilent to work done by its weight.

W = Pef-Pe0

= 37.044-7.6

= 29.4J