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 210: 53



Work Step by Step

We can determine the required power supplied as follows: $\Sigma F_x=ma_x$ $\implies F-F_D=ma$ We plug in the known values to obtain: $F-(0.3)(28)^2=2300(5)$ $\implies F=11735.2N$ The efficiency of the motor is given as $\epsilon=\frac{P_o}{P_i}$ We plug in the known values to obtain: $0.68=\frac{328.58}{P_i}$ This simplifies to: $P_i=483KW$
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