## Engineering Mechanics: Statics & Dynamics (14th Edition)

$P_i=4.20hp$
We can determine the required power input as follows: $\epsilon=\frac{Power\space output}{power \space input}$ This can be rearranged as: $Power\space input=\frac{Power\space output}{\epsilon}$ $Power\space input=\frac{Fv}{\epsilon}$ We plug in the known values to obtain: $Power\space input=\frac{(300)(5)}{0.65}$ $\implies Power\space input=2307.7lb\cdot ft/s=4.20hp$