# Chapter 14 - Kinetics of a Particle: Work and Energy - Section 14.4 - Power and Efficiency - Problems - Page 210: 50

$P_o=4.36hp$

#### Work Step by Step

We can determine the required power output as follows: We know that $2v_p+v_M=0$ $\implies v_M=-2v_p=-2(-4)=8ft/s$ Now $P_o=Fv$ We plug in the known values to obtain: $P_o=(\frac{600}{2})(8)$ $\implies P_o=2400lb\cdot ft/s=\frac{2400}{550}hp=4.36hp$

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