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 230: 60


$P = 463~watts$

Work Step by Step

The work $W$ that the firefighter does is equal to the change in potential energy. $W = PE$ $W = mgh$ $W = (90~kg+40~kg)(9.80~m/s^2)(20~m)$ $W = 25,480~J$ We can find the required power as; $P = \frac{Work}{time}$ $P = \frac{25,480~J}{55~s}$ $P = 463~watts$
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