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: 51



Work Step by Step

The required power can be determined as follows: $Sigma F_y=ma_y$ $\implies 3(500)-1000=\frac{1000}{32.2}\times a$ $\implies a=1.6ft/s^2$ We know that $v^2=v_{\circ}^2+2a(S-S_{\circ})$ $\implies v^2=(0)^2+(2\times 16(15-0))=21.98ft/s$ The power output is given as $P_o=Fv=3\times 500\times 21.98=32970lb\cdot ft/s$ The efficiency of the motor is $\epsilon=\frac{P_o}{P_i}$ $\implies 0.65=\frac{32970}{P_i}$ $\implies P_i=50723lb\cdot ft/s=\frac{50723}{500}hp=92.2hp$
Update this answer!

You can help us out by revising, improving and updating this answer.

Update this answer

After you claim an answer you’ll have 24 hours to send in a draft. An editor will review the submission and either publish your submission or provide feedback.