tiny-tim said:
Hi Abdul!
Interesting!
When the hill is stationary, the normal reaction force is (obviously) parallel to the displacement , and so does no work.
But when the hill is moving with speed √(2gh), the normal reaction force is
not parallel to the displacement, and so the work done is equal to …
erm 
…
ooh! …
your problem …
you do the maths!
I don't believe this solves the problem because we could remove the hill entirely and be left with basically the same problem.
The important thing to realize is that conservation of energy and conversation of momentum are closely connected when switching frames. It is conservation of momentum that guarantees conservation of energy holds when you shift frames of reference. It becomes extremely important that you keep track of what's happening to the Earth if it is moving.
The reason you don't normally need to think about the Earth moving, but you do in this case, is the nonlinear nature of kinetic energy. The kinetic energy is:
\frac{1}{2}mv^{2}
Conservation of momentum tells us that if the Earth is stationary and a small body changes its velocity by \Delta v, then
m(v+\Delta v) + Mv_{e}=mv \rightarrow v_{e}=-\frac{m}{M}\Delta v \equiv -\delta \Delta v
(M=mass of earth, m=mass of object, v_e=final velocity of earth, \delta=m/M)
The kinetic energy of the Earth is then given by
\frac{1}{2}Mv_{e}^2 = \frac{1}{2}M\delta^2(\Delta v)^2 = \frac{1}{2}m\delta(\Delta v)^2
assuming that the velocity of the Earth is 0 initially. This quantity is negligibly small because of the small parameter \delta that has no large factor like the mass of the Earth to counter its effect. Now assume, as in your example that the Earth starts with velocity v
i, and look at the change in kinetic energy of the earth. Of course the change in velocity of the Earth will be the same when we shift reference frames, so let's still call that change v
e, which will have the same value as before:
\Delta KE = \frac{1}{2}M(v_{i} + v_{e})^2 - \frac{1}{2}Mv_{i}^2=\frac{1}{2}Mv_{i}^2 + Mv_{e}v_{i} + \frac{1}{2}Mv_{e}^2 - \frac{1}{2}Mv_{i}^2 = Mv_{e}v_{i}
Now plug in the value of v
e from conservation of momentum:
\Delta KE = Mv_{i}\delta v_{i} + \frac{1}{2}M\delta^2v_{e}^2= mv_{i}^2 + \frac{1}{2}m\delta v_{e}^2\approx 2(\frac{1}{2}mv_{i}^2)= mgh + \frac{1}{2}mv_{i}^2
I again neglected the small term with a \delta still in it. In the last equality I assumed we were working with your example where the initial velocity was chosen so that the kinetic energy of the small mass was the same as the potential energy. As you see, this is exactly the energy you started with in the shifted frame, and which seemed to go to 0 when you looked only at the small mass. Now it is clear what happened: in this frame the energy gets transferred to the earth.
The main point is that you can't neglect the change in kinetic energy of a very large body when it doesn't start from rest. Since kinetic energy is proportional to the square of velocity, and the derivative of kinetic energy is therefore proportional to velocity, changes in kinetic energy of the large body will be negligibly small when they change from rest, but the same change in v will produce a considerable change in kinetic energy when it does not.