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: 57



Work Step by Step

We can determine the required power as follows: $\Sigma F_y=ma_y=0$ $F(\frac{4}{5})-W=0$ $\implies F(\frac{4}{5})=10$ $\implies F=12.5lb$ Now $P=F\cdot v$ $\implies P=Fv(\frac{4}{5})$ We plug in the known values to obtain: $P=12.5\times 2(\frac{4}{5})$ $\implies P=20lb\cdot ft/s=\frac{20}{550}hp=0.0364hp$
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