problem07_02

University Physics with Modern Physics with Mastering Physics (11th Edition)

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7.2: a) For constant speed, the net force is zero, so the required force is the sack’s weight, N. 49 ) s m kg)(9.80 00 . 5 ( 2 = b) The lifting force acts in the same direction as the sack’s motion, so the work is equal to the weight times the distance,
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Unformatted text preview: J; 735 m) (15.0 N) 00 . 49 ( = this work becomes potential energy. Note that the result is independent of the speed, and that an extra figure was kept in part (b) to avoid roundoff error....
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