You drop a 2.40 kg book to a friend who stands on the ground at distance D = 12.

Question

You drop a 2.40 kg book to a friend who stands on the ground at distance D = 12.0 m below. If your friend's outstretched hands are at distance d = 1.40 m above the ground (see the figure), (a) how much work Wg does the gravitational force do on the book as it drops to her hands? (b) What is the change ?U in the gravitational potential energy of the book-Earth system during the drop? If the gravitational potential energy U of that system is taken to be zero at ground level, what is U (c) when the book is released and (d) when it reaches her hands? Now take U to be 100 J at ground level and again find (e) Wg, (f ) ?U, (g) U at the release point, and (h) U at her hands.

Explanation / Answer

a)

The gravitational force on the book = Fg =mg

here m = 2.4kg

g=9.8 m/s2

hence =  Fg = 23.52 N

distance moved by book = a = 12 - 1.4 = 10.6 m

hence W= force * distance = 249.312 J

b) the change in gravitational potential energy of the book-Earth system during the drop = -W

= -249.315 J

c) at the release point U ground = 0

hence U at height 12 m = mgh = 2.4*9.8*12 = 282.24 J

d) when reaches her hands h = 1.4 now

hence U = 2.4*9.8*1.4 = 32.928 J

e) even if we take U = 100 J at ground level the Wg will not change it will be  249.312 J

f) similarlly U = -Wg = -249.312 J

g) now U at release point will be = 100 J + 282.24 J = 382.24 J

h ) now U at hands = 100J +  32.928 J = 132.928 J

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