A 0.60-kg basketball is dropped out of the window that is 6.2 m above the ground
ID: 1466588 • Letter: A
Question
A 0.60-kg basketball is dropped out of the window that is 6.2 m above the ground. The ball is caught by a person whose hands are 1.7 m above the ground 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 released? What is the gravitational potential energy of the basketball when it is caught? How is the change (PE_f - PE_d) in the ball's gravitational potential energy related to the work done by its weight?Explanation / Answer
a) Work = Force*displacement = -mg*(h2-h1) = 0.60*9.8*(-1.7-(-6.2) = 26.46 J
b) potential energy when it is released = mgh1 = 0.60*9.8*6.2 = 36.45 J
c) potential energy when it is caught = mgh1 = 0.60*9.8*1.7 = 9.99 J
d) PEf - PEi = 9.99-36.45 = - 26.46 J
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