Physics Technology Update (4th Edition)

Published by Pearson
ISBN 10: 0-32190-308-0
ISBN 13: 978-0-32190-308-2

Chapter 8 - Potential Energy and Conservation of Energy - Problems and Conceptual Exercises - Page 249: 74

Answer

(a) $0.53J$ (b) $-0.53J$ (c) zero (d) zero

Work Step by Step

(a) We can find the required work done as follows: $\Delta U=mgL(cos\theta-1)$ We plug in the known values to obtain: $\Delta U=(0.25Kg)(9.81m/s^2)(1.2m)(cos35^{\circ}-1)$ $\Delta U=-0.53J$ Now $W_{C, A to B}=-\Delta U$ $W_{C, A to B}=-(-0.53J)=0.53J$ (b) The work done by gravity from B to A is given as $W_{C,Bto A}=-W_{C,Ato B}$ $W_{C,Bto A}=-(0.53J)=-0.53J$ (c) The work done by the string is zero as the force exerted by the string is always perpendicular to the motion. (d) In this case the work done is zero as well because the force exerted is perpendicular to the motion.
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