Answer
$P=0.229~hp$
Work Step by Step
The required power can be determined as follows:
$\Sigma U_{1\rightarrow 2}=-10(4-1.732)+25(5-3.464)=15.72lb\cdot ft$
We know that
$0+15.72=\frac{1}{2}(\frac{10}{32.2})v^2$
$\implies v_2=10.06ft/s$
Now, $P=F\cdot v$
$\implies P=Fvcos\theta$
We plug in the known values to obtain:
$P=25(10.06)cos 60=125.75lb\cdot ft/s$
$P=\frac{125.75}{550}hp=0.229~hp$
