While leaning out a window that is 6.2 m above the ground, you drop a 0.60-kg ba

Question

While leaning out a window that is 6.2 m above the ground, you drop a 0.60-kg basketball to a friend at ground level. Your friend catches the ball at a height of 1.7 m above the ground. Determine the following (a) the amount of work done by the force of gravity (b) the gravitational potential energy of the ball, relative to the ground when it is released (c) the gravitational potential energy of the ball relative to the ground when it is caught (d) the ratio of the change (PE_f - PE_0) in the ball's gravitational potential energy to the work done on the ball by the force of gravity

Explanation / Answer

(A) Work done by force = F.d

= m g deltaH

Wg = (0.60 x9.81)(6.2 - 1.7)

Wg = 26.5 J

(B) change in PE = - Work done by gravity

= - 26.5 J

(c) PE = m g (h - h0) = 0.60 x 9.81 x (1.7 - 0)

= 10 J

(d) change in PE = - Work done by gravity

so ratio = -1

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