Engineering Mechanics: Statics & Dynamics (14th Edition)

Published by Pearson
ISBN 10: 0133915425
ISBN 13: 978-0-13391-542-6

Chapter 14 - Kinetics of a Particle: Work and Energy - Section 14.4 - Power and Efficiency - Problems - Page 209: 43



Work Step by Step

We can determine the required power input as follows: $\epsilon=\frac{Power\space output}{power \space input}$ This can be rearranged as: $Power\space input=\frac{Power\space output}{\epsilon}$ $Power\space input=\frac{Fv}{\epsilon}$ We plug in the known values to obtain: $Power\space input=\frac{(300)(5)}{0.65}$ $\implies Power\space input=2307.7lb\cdot ft/s=4.20hp$
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