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: 55



Work Step by Step

We can determine the power supplied as follows: $\Sigma F_y=0$ $\implies T_C+3T-m_Eg=0$ $\implies m_Cg+3T=m_Eg$ $\implies (60)(9.81)+3T=400(9.81)$ $\implies T=1111.8N$ We know that $2v_E+(v_E-v_p)=0$ $\implies v_p=3v_E=3(4)=12m/s$ The power output is $P_o=Fv=1111.8\times 12=13.34KW$ The efficiency of the motor is $\epsilon=\frac{P_o}{P_i}$ We plug in the known values to obtain: $0.6=\frac{13.34KW}{P_i}$ This simplifies to: $P_i=22.2KW$
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