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: 54



Work Step by Step

We can determine the required power as follows: $\Sigma F_x=ma_x$ $F-(10v)=2300(6)$ $\implies F=13800+10v~~~$eq(1) We know that $v=v_{\circ}+at$ $v=0+(6)(5)=30m/s$ From eq(1), $F=13800+10(30)=14100N$ The power output is given as $P_o=Fv=14100\times 30=423KW$ The efficiency of the motor is $\epsilon=\frac{P_o}{P_i}$ $\implies 0.68=\frac{423}{P_i}$ $\implies P_i=622KW$
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