Answer
$(\frac{M}{2}+m)Va+mg$
Work Step by Step
Please see the attached image first.
First of all let's apply equation F = ma to the bucket.
$\uparrow F=ma$
Let's plug known values into this equation.
$T-mg=ma$
$T=m(g+a)-(1)$
Let's apply equation $\tau=I\alpha$ to the pulley.
$\tau=I\alpha$
$FR-TR=I\alpha$
$(F-T)R= \frac{1}{2}MR^{2}(\frac{a}{R})$
$F=T+\frac{Ma}{2}-(2)$
$(1)=\gt (2)$
$F=m(g+a) +\frac{Ma}{2}$
$F=a(\frac{M}{2}+m)+mg$
We know that the Power (P) = FV
So we can write,
$P=[a(\frac{M}{2}+m)+Mg]V$
$P=(\frac{M}{2}+m)Va+mgV$
