Work, Power and Energy — Lifting a coiled rope up off a surface

[haruspex]

The detailed analysis of internal energy within the rope is outside the scope of the this homework.

Unfortunately it is not.

You made the force/momentum approach produce the work conservation answer by assuming that a perfectly elastic string of very little extensibility pulled up at a steady rate would not acquire significant persistent vertical oscillations. It is unclear to me whether that is the case.
Even then, to get your answer, the significant lateral KE which would come from pulling up from a horizontal loop (as discussed above) has to be ignored. I have not checked whether including that gives one of the listed answers.

Other valid models can produce other answers.

E.g. consider the model of a chain in which each segment, length L, starts standing vertically. The segments have no width. As each segment is lifted to height L it engages the next segment. If that is a coalescence then the work required is doubled. Even if it is perfectly elastic, would there not be persistent vertical oscillations?
Or start with zero width segments lying horizontally in a stack in a zigzag arrangement. As a segment is lifted it rotates about its lower end. Will the energy that goes into that rotation be recovered in helping to lift the next segment, or will there be residual horizontal wiggles?

Such considerations are only outside the scope of the question if those models are somehow excluded.

 

[Lnewqban]

There is also a twisting result as the rope goes up, unless the free end can rotate.
It is easily visible in construction sites when pulling an extension cord, or wire-rope, or hose that have been resting on the ground as a coil.
Same happens in reverse, when forming the coil on the ground.

 

[haruspex]

More thoughts…
If we model the rope as flexing freely down to some minimum radius of curvature, r, then the portion just above the ground consists of a quadrant of a circle of that radius. The acceleration of each bit of it is towards the centre of curvature, and that must be provided by tension in the rope.
Ignoring gravity, this makes the tension equal along the arc, including where it joins the part that is still horizontal. Specifically, $T=\lambda v^2$.
The rate of doing work in lifting new rope is therefore $\lambda v^3$.

Looking at it from work conservation instead, each dt a mass segment $\lambda\cdot dx$ of rope finishes with a vertical speed v and a horizontal speed v for a total KE of $\lambda dx .v^2=\lambda dt .v^3$, hence the rate of acquiring KE is $\lambda v^3$, agreeing with the result above.