A 0.529-kg basketball is dropped out of a window that is 5.03 m above the ground

Question

A 0.529-kg basketball is dropped out of a window that is 5.03 m above the ground. The ball is caught by a person whose hands are 1.73 m above the ground. (a) 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 (b) released and (c) caught? (d) What is the change (PEf - PE0) in the ball's gravitational potential energy?

(a) Number____ Units___

(b) Number_____ Units___

(c) Number_____ Units____

(d) Number_____ Units___

Explanation / Answer

Here ,
mass of ball , m = 0.529 Kg
h1 = 5.03 m
h2 = 1.73 m
a) work done on the ball by weight = m * g * (h2 - h1)
work done on the ball by weight = 0.529 * 9.8 * (5.03 - 1.73)
work done on the ball by weight = 17.11 J

b)
now , for the gravitational potential energy where it is released
the gravitational potential energy = m * g * h1
the gravitational potential energy = 0.529 * 9.8 * 5.03
the gravitational potential energy = 26.1 J

c)
at the point where is caught
the gravitational potential energy = m * g * h2
the gravitational potential energy = 0.529 * 9.8 * 1.73
the gravitational potential energy = 8.97 J

d)
change in gravitational potential energy = 8.97 - 26.1
change in gravitational potential energy = -17.11 J
change in gravitational potential energy is -17.11 J

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