# Power needed to raise a bucket at a constant speed by a pulley

This is the last section of this pulley exercise:

The bucket in is filled with water so its total mass is now $45$ kg. The bucket is raised at a constant speed of $2.0$ $ms^{-1}$ using an electric motor attached to the cylinder. Calculate the power output of the motor.

From previous section I know that the torque is $\tau=20.2$, radius is $r=0.2$, mass of the bucket is $24$ kg and mass of the pulley is $36$ kg.

My approach was to use $P=Fv$ the solution in the textbook uses that too. However, I think that $F$ should be the resultant force acting on the bucket i.e. $F=am=mg-T$ where $T$ is the tension of the rope from the pulley above, so the power should be $P=(mg-T)*v$. However, the textbook writes only $P=mg*v$, omitting tension. Why?

If the water bucket is not accelerating there should not be any resultant force on it. The bucket is moving at constant speed, so all forces acting on it cancel: it has gravity acting with $mg$ on it and tension equal and opposite to gravity. But, F in the equation for power is not the net force, it is the driving force: $$P = F_{driving} v$$, i.e. the force that keeps it moving, the force that is doing the work. The motor is doing the work, so the driving force, in this case, is tension. Which is equal to $mg$, so you get your answer $$P = mgv$$