Physics: Principles with Applications (7th Edition)

Published by Pearson
ISBN 10: 0-32162-592-7
ISBN 13: 978-0-32162-592-2

Chapter 6 - Work and Energy - General Problems - Page 168: 89


(a) $W = 8.9\times 10^5~J$ (b) $P_o = 54 ~W$ $P_o = 0.072 ~hp $ (c) $P_I = 360~W$ $P_I = 0.48~hp$

Work Step by Step

(a) The work done by the hiker will be equal to the change in gravitational potential energy. $W = mgh$ $W = (65~kg)(9.80~m/s^2)(1400~m)$ $W = 8.9\times 10^5~J$ (b) The average power output $P_o$ is equal to the total work done by the hiker divided by the total time of the hike. $P_o = \frac{W}{t} = \frac{8.9\times 10^5~J}{(4.6~h)(3600~s/h)}$ $P_o = 54~W$ We can convert the average power output to units of horsepower. $P_o = (54~W)(\frac{1~hp}{746~W}) = 0.072~hp$ (c) We can calculate the required rate of energy input $P_I.$ Note that only 15% of the power input by the hiker is used as power output. $0.15\times P_I = P_o$ $P_I = \frac{P_o}{0.15}$ $P_I = \frac{54~W}{0.15} = 360~W$ We can convert the power input $P_I$ to units of horsepower. $P_I = (360~W)(\frac{1~hp}{746~W}) = 0.48~hp$
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