you drop a 1.30 kg book to a friend who stands on the ground at distanceD = 12.0

Question

you drop a 1.30 kg book to a friend who stands on the ground at distanceD = 12.0 m below. If you friend's outstretched hands are at distance d = 1.30 m above the ground (see the figure),(a) how much work w_g does the gravitational force do on the book as it drops to her hands? (b) what is the change Deltau 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) w_g' (f) DeltaU, (g) U at the release point, and (h) U at her hands.

Explanation / Answer

a)

vertical displancement = 12 - 1.3 = 10.7m

W = mg*d = 1.3 * 9.8 * 10.7 = 136.318 J

part b )

U = mgy

dU = mg(yf - yi) = 1.3 * 9.8 ( 1.3 - 12) = -136.318J

part c )

Ui = mgyi

yi = 12m

Ui = 1.3 x 9.8 x 12 = 152.88 J

part d )

Uf = mgyf

yf = 1.3m

Uf = 1.3 * 1.3 * 9.8 = 16.562 J

part e )

Wg is remain same = 136.318 J

part f )

dU = -Wg = -136.318 J

part g )

now we have initiaaly Uo = 100J

U = Uo+ mgyi = 100 + 1.3 * 9.8 *12

U = 252.88 J

part f )

U = Uo + mgyf

U = 116.562 J

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