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: 45


$P=8.32\times 10^3hp$

Work Step by Step

We can determine the required power as follows: $v=(600mi/h)(\frac{5280ft/mi}{3600s/h})=880ft/s$ Now $P=Fv=Tv$ We plug in the known values to obtain: $P=(5200)(800)$ $\implies P=4576000lb\cdot ft/s\frac{1}{550}$ $\implies P=8.32\times 10^3hp$
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