=mg/10, so F= 11 mg/10. Since the force GFand the displacement Gdare in the same direction, the work done by GFis 241111 (72 kg)(9.8 m/s )(15 m)1.164 10 J1010FmgdWFd====×which (with respect to significant figures) should be quoted as 1.2 ×104J. (b) The force of gravity has magnitude mgand 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 JgWmgd=−×which should be quoted as – 1.1 ×104J. (c) The total work done is W=×−×=×1164.10 J 1.058 10 J1.06 10 J443. 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.