A 0.60 kg basketball is dropped out of the window that is 6.4 m above the ground
ID: 2100141 • Letter: A
Question
A 0.60 kg basketball is dropped out of the window that is 6.4 m above the ground. The ball is caught by a person whose hands are 1.8 m above the ground. How much work is done on the ball by itsweight?
J
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?
Explanation / Answer
work done by ball by its weight = mgh = 0.6*9.8*(6.4-1.8) = 27.048 J
gravitational potential energy when released = mgh = 0.6*9.8*6.4 = 37.632 J
gravitational potential energy when caught = mgh = 0.6*9.8*1.8 = 10.584 J
change (PEf - PE0) in the ball's gravitational potential energy = work done by its weight
(that is the definition of gravitational potential energy)
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