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 209: 48



Work Step by Step

The required time can be determined as follows: The work done by the man is given as $U=Wh$ $\implies U=150(15)=2250lb\cdot ft$ The power generated by the man is $P_m=\frac{U}{t}$ $\implies P_m=\frac{2250}{4}=562.5\times \frac{1}{550}=1.02hp$ Now, the power of the bulb is given as $P_b=100\times (\frac{550bl\cdot ft/s}{746W})=73.73lb\cdot ft/s$ We know that $t=\frac{U}{P_b}$ We plug in the known values to obtain: $t=\frac{2250}{73.73}$ $\implies t=30.5s$
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.