First, if we want to solve this problem properly with respect to a "variable-mass" system, then it is totally wrong to do so by assuming that such a system obeys
F=dp/dt.
This is readily seen, because with variable mass, this equation is not Galilean invariant, as it must be.
If you want to do it this way, you might want to look at my thread:
https://www.physicsforums.com/showthread.php?p=538334#post538334
In this thread, I'll follow another another approach, namely expressing the work done by F to the mechanical energy the WHOLE rope has gained from its position of rest&zero potential energy.
The first thing we need to clarify however, is what approximation we need to establish in order to find a "solution".
Clearly, the real situation is basically impossible to find a solution to, where the rope curves a bit between the part lying at rest on the table, and the part moving upwards with velocity V due to F.
Thus, we simplify the situation by saying that as time goes, a tiny piece of rope which was at rest INSTANTANEOUSLY begins moving upwards with velocity V, i.e, joining the common motion of the upwards-moving part.
We have therefore split the rope into TWO parts: the part at rest, and the part moving upwards with velocity V.
Before proceeding with our analysis, we really should determine under which condition this approximation can be regarded as a GOOD approximation:
Let us split the rope into 3 parts:
1. The part moving with velocity V, of length x(t)
2. The part of the rope strictly at rest, with length r(t)
3. The JOINING curve C, whose vertical velocity varies from 0 to V, and whose length can be called c.
Thus, we have: r(t)=L-x-c, where L is the total rope length.
Now, if we look at the segment with length r(t)+c (i.e, the part of the rope not moving with uniform vertical velocity V), when is it a good approximation to say that the center of mass of this segment has a NEGLIGIBLE vertical velocity?
Evidently, since the vertical velocity at C is less than or equal to V, we have that the center of mass' vertical velocity obeys the inequality:
0<{v}_{C.M}<\frac{\rho{c}V}{\rho{(L-x)}}=\frac{c}{L-x}V
Requiring v_{C.M}<<V
yields:
c<<L-x\to{x}<<L-c\to{x}<<L
that is, we can only expect us to find a reasonable approximation if the vertically moving part of the rope is a tiny fraction of the whole rope!
With this in mind, let us state the mechanical energy equation between the time when F has done no work on the whole rope to the time when the upper end has reached position "x":
W=\oint_{R}\rho\vec{g}\cdot{d}\vec{s}+K
where W is F's work, the first term on the right-hand side is the total rope's gain in gravitational potential energy, and K is the total kinetic energy gained by the rope, conveniently written in parametric form as:
K=\int_{s_{0}}^{s_{1}}\frac{\rho}{2}v^{2}(s)ds, s_{0}\leq{s}\leq{s}_{1}
Now, only the vertical part of the rope has any potential&kinetic energy, and we'll let the turning point of the table be at s=0, and the upper end at s=x.
However, due to the discontinuous velocity jump happening at the "lower" side of 0, we'll need to be careful:
At time t, the particle p situated at s=0^{-} has v_{p}=0, whereas at time t+dt, that particle has moved to s=0^{+}, with new velocity v_{p}=V
Thus, our work-energy relation reads:
W=\int_{0}^{x}\rho{gs}ds+\frac{\rho}{2}v_{p}^{2}dx+\int_{0^{+}}^{x}\frac{\rho}{2}V^{2}ds
Differentiating with respect to time we get, by remembering \frac{dx}{dt}=V everywhere we find it, the power is found by:
P=\frac{dW}{dt}=\rho{g}xV+\frac{\frac{\rho}{2}(V^{2}-0^{2})dx}{dt}+\frac{\rho}{2}V^{3}=\rho{V}(gx+V^{2})
(the force is therefore, of course F=\rho(gx+V^{2}))
(\rho\equiv\frac{M}{L} is the density)
Now, this happens to coincide with the result you'd get out of the flawed F=dp/dt (flawed, that is, in the variable-mass system analysis).
The only reason why that method happens to get the correct result, is that we are looking at the problem in the table's rest frame.
