Physics for Scientists and Engineers: A Strategic Approach with Modern Physics (4th Edition)

Published by Pearson
ISBN 10: 0133942651
ISBN 13: 978-0-13394-265-1

Chapter 9 - Work and Kinetic Energy - Exercises and Problems - Page 229: 36


(a) $W = 9.8\times 10^5~J$ (b) $P = 1.96\times 10^4~watts$

Work Step by Step

(a) Since the elevator moves at a constant speed, the tension in the elevator cable is equal in magnitude to the weight of the elevator. We can find the work done by the motor as; $W = T~d$ $W = (mg)~d$ $W = (1000~kg)(9.80~m/s^2)(100~m)$ $W = 9.8\times 10^5~J$ (b) We can find the power supplied by the motor; $P = \frac{W}{t}$ $P = \frac{9.8\times 10^5~J}{50~s}$ $P = 1.96\times 10^4~watts$
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