Physics for Scientists and Engineers: A Strategic Approach with Modern Physics (3rd Edition)

Published by Pearson
ISBN 10: 0321740904
ISBN 13: 978-0-32174-090-8

Chapter 11 - Work - Exercises and Problems - Page 304: 33


The motor pumps 55,000 liters of water in one hour.

Work Step by Step

We can convert the power of the motor to units of watts. $P = (2.0~hp)(746~W/hp)$ $P = 1492~W$ We then find the total energy supplied by the motor in one hour. $E = P~t$ $E = (1492~W)(3600~s)$ $E = 5.37\times 10^6~J$ The energy supplied by the motor will be equal to the change in potential energy of the water. Let $M$ be the mass of the water. $PE = 5.37\times 10^6~J$ $Mgh = 5.37\times 10^6~J$ $M = \frac{5.37\times 10^6~J}{gh}$ $M = \frac{5.37\times 10^6~J}{(9.80~m/s^2)(10~m)}$ $M = 55,000~kg$ Since the density of water is 1.0 kg/liter, the motor pumps 55,000 liters of water in one hour.
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