You drop a 3.00 kg book to a friend who stands on the ground at distance D = 15.

Question

You drop a 3.00 kg book to a friend who stands on the ground at distance D = 15.0 m below. If your friend's outstretched hands are at distance d = 1.20 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) Work = FS
Work = mg(h) = 3.0kg(9.8)(15.00m -1.20m) = 405.72 J

(b)Relative to ground level, PE = mg(15) =(3)(9.8)(15) = 441 J
As it drops PE became PE = mg(1.2) = 35.28 J
Change in PE = 441 – 35.28 = 405.72 J (equal to work done in dropping)
(c) 441 J
(d) 35.28N-m
(e)Work Wg will not change because it is the difference in PE between two points, still Wg = 405.72 N-m
(f) DU between what two points? (computer says not specified)
(h) PE at her hands = 35.28 + 100J = 135.28 J
(g)U = 441 N-m + 100J = 541 J

Hope this helps.

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