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: 46



Work Step by Step

We can determine the required time as follows: $v=35mi/h(\frac{5280ft/min}{3600s/h})=51.33ft/s$ The kinetic energy is given as $KE=\frac{1}{2}mv^2=\frac{1}{2}(\frac{5000}{32.2})(51.33)^2=204.59\times 10^3lb\cdot ft$ We know that $P=(100W)(\frac{(550lb\cdot ft/s)/hp}{746W/hp})=73.37lb\cdot ft/s$ Now $t=\frac{KE}{P}$ We plug in the known values to obtain: $t=\frac{204.59\times 10^3}{73.73}=2774.98s=46.2min$
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