Classical Physics Question?

Question

I can not figure out this problem:

"A uniformly dense rope of length "b" and mass density "u" (mu, but u will work) is coiled on a smooth table. One end is lifted by hand with aconstant velocity "v". Find the force of the rope held by the hand when the rope is a distance "a" above the table (b>a). "

I just said that F(a) = u*a*g but apparently the answer is u*a*g*(1+v^2/(a*g))

Answer 1

Newton's 2nd Law : Sum of all forces = Rate of momentum change produced on the object.
A constant velocity does not mean net force is zero. Mass can also change.

Let F be your force. Then, F - Weight = Rate of momentum change.

Momentum = u*a*v

Momentum change rate = d(uav)/dt = uvdv/dt = uv^2

Therefore, F - uag = uv^2. Hence the result.

Answer 2

I got an answer is one factor is off by 2.

Consider an small mass dm at height x. dm = u dx

total energy of dm at that height, dE= 1/2 (dm) v^2 + (dm)g x

Total energy of the rope in motion = Integrate dE from 0 to a

= Integrate (1/2 v^2 u dx + g u x dx) = 1/2 u a v^2 + 1/2 gu a^2

You have to integrate because the force F is to hold up a total mass in motion, not just dm

Since F = partial dE/dx , or look at F*dx as work done at x=a, you get close to the answer. You find that force at a particular height (not a small mass dm) by taking derivative of the total energy.

I can't pinpoint the discrepany of the factor of 2