ch07-p017

# 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

This preview shows page 1. Sign up to view the full content.

= 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
This is the end of the preview. Sign up to access the rest of the document.

## 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.

Ask a homework question - tutors are online