Velocity reference point when calculating power 
As shown in the picture, a person (80 kg) is on a platform (20 kg) that's suspended by a rope which is on a pulley. As they pull on the other side of the rope, they reach a constant velocity of 0.4 m/s. What is the power output of the person?
I've been confused as to whether I should use the velocity as 0.8 m/s or 0.4 m/s; although the rope is moving at 0.8 m/s through the person's hands, the rope is still only moving at 0.4 m/s from a stationary reference. I'm not sure which one to use when finding power, since 0.8 m/s seems correct, since the person is pulling 0.8 meters of rope in one second. However, the rope is only moving 0.4 meters in one second.
Please advise.
 A: Power is energy transferred per unit time. $\frac{dE}{dt}$
So that is what you have to do, that is the concept and I think it is enough for you to solve. If you need a deeper explanation you probably violate HW questions policy.
A: Don't worry about the rope. The key to this exercise is the kinetic energy. In the person's frame, they have zero velocity. To determine the energy, and thence the power, consider the kinetic energy of the person plus the frame relative to the ground.
A: The mechanical power of a force, $\vec F$, is given by $ P=\vec F \cdot \vec v$ where $\vec v$ is the velocity of the material at the point of application of the force. Just apply this consistently.
In this case the reference frame is implicitly specified as the one in which the platform is going up at 0.4 m/s. In other words, the reference frame is the earth’s frame. In that frame the person’s hand is going down at 0.4 m/s. So the appropriate velocity is $\vec v=-0.4\text{ m/s }\hat i$ where $\hat i $ is the upward-pointing unit vector.
Note that power is frame variant. If you use the reference frame of the platform instead of the earth’s frame then you will get a different power. Energy will be conserved either way, but the power will be different.
