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 211: 56



Work Step by Step

The required power can be determined as follows: $\Sigma U_{1\rightarrow 2}=-10(4-1.732)+25(5-3.464)=15.72lb\cdot ft$ We know that $0+15.72=\frac{1}{2}(\frac{10}{32.2})v^2$ $\implies v_2=10.06ft/s$ Now, $P=F\cdot v$ $\implies P=Fvcos\theta$ We plug in the known values to obtain: $P=25(10.06)cos 60=125.75lb\cdot ft/s$ $P=\frac{125.75}{550}hp=0.229~hp$
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