Work done in raising a rope at constant kinetic energy? Usually the question I had was work done in raising a rope at constant speed. Obviously as you raise the rope, the mass changes over time. It'll be a simple integral.
But what about work done in raising a rope at constant kinetic energy?Assuming rope length is y with mass m. I did it but I'm not sure if I am right. Can anyone see if I am right?
.$W = ΔE$
.$W = ΔGPE + ΔKE$
Since constant kinetic energy, then KE doesn't change.
.$W = GPE_f - GPE_i$
.$ W = mgy - 0 = mgy$
Is it that simple? I have a feeling it is not.
 A: Is it that simple?
No. It's even simpler. Replace your rope with a motionless brick, lift it up, and put it on a shelf. It started with no kinetic energy, and it finished with no kinetic energy. But you did work on it, you added energy to it, and we call it gravitational potential energy*. As a result the mass of the brick increased a little. Then when you nudge the brick off the shelf it falls to Earth. Potential energy, or internal kinetic energy, is converted into macroscopic kinetic energy, and once this is dissipated you're left with a mass deficit. Have a look at mass in general relativity and you will appreciate that this simple little situation is lugging some baggage: invariant mass varies! And not only due to gravity: the mass of a hydrogen atom is less than the mass of a free electron plus the mass of a proton. The electron, and to a lesser extent the proton, has lost mass.   
*Strictly speaking we should talk about the Earth-brick system, but the motion of the Earth is not detectable, so we simplify matters by focussing on the brick. 
A: Well your answer is wrong.
Since you have tagged it as homework-and-exercise question I cannot give the full answer according to terms and conditions.
HINT: Assume all mass is concentrated at the centre of mass (i.e. for uniform rope l/2). Then treat it like point mass and apply $\Delta U = m \cdot g \cdot h$
here h is height you raised the centre of mass.
