A 0.60-kg basketball is dropped out of the window that is 5.9 m above the ground

Question

A 0.60-kg basketball is dropped out of the window that is 5.9 m above the ground. The ball is caught by a person whose hands are 1.0 m above the ground.

(a) How much work is done on the ball by its weight?
J

(b) What is the gravitational potential energy of the basketball, relative to the ground when it is released?
J

(c)What is the gravitational potential energy of the basketball when it is caught?
J

(d) How is the change (PEf - PE0) in the ball's gravitational potential energy related to the work done by its weight?

Explanation / Answer

a) work is done on the ball by its weight = -mg(Hfinal - Hinitial ) = 0.60*9.8(5.9-1) = 28.81 J

b) gravitational potential energy of the basketball, relative to the ground when it is released = mgH = 0.60*9.8(5.9) = 34.69 J

c) gravitational potential energy of the basketball when it is caught = 5.88 J

d)  (PEf - PE0) = mg(Hfinal - Hinitial )= 0.60*9.8(1-5.9) = -28.81 J

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