Ch07-p017 - 17(a We use F to denote the upward force exerted by the cable on the astronaut The force of the cable is upward and the force of

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= mg /10, so F = 11 mg /10. Since the force G F and the displacement G d are in the same direction, the work done by G F is 2 4 11 11 (72 kg)(9.8 m/s )(15 m) 1.164 10 J 10 10 F mgd WF d == = = × which (with respect to significant figures) should be quoted as 1.2 × 10 4 J. (b) The force of gravity has magnitude mg and is opposite in direction to the displacement. Thus, using Eq. 7-7, the work done by gravity is 24 (72 kg)(9.8 m/s )(15 m) 1.058 10 J g Wm g d =− × which should be quoted as – 1.1 × 10 4 J. (c) The total work done is W =×−×= × 1164 . 10 J 1.058 10 J 1.06 10 J 44 3 . Since the astronaut started from rest, the work-kinetic energy theorem tells us that this (which we
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This note was uploaded on 05/17/2011 for the course PHY 2049 taught by Professor Any during the Spring '08 term at University of Florida.

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