Shouldn't the change in kinetic energy be more in a moving elevator from a stationary frame of reference? Consider an elevator moving down with uniform velocity. A person standing inside watches an object fall from the ceiling of the elevator to the floor. Say the height of the elevator is $h$. Then the work done by gravity in that frame of reference should be $mgh$. But consider this same event being watched by someone else in the stationary frame of reference. In his reference frame, the object travels a larger distance as it falls from the ceiling to the floor of the elevator because the floor itself is moving downwards (one can calculate this extra distance covered to be $u \sqrt{\frac{2h}{g}} $) and hence the change in kinetic energy should be more in that frame than in the moving frame!
I just can't seem to figure out where I'm going wrong here. I'm probably missing something very obvious. 
So I would be very grateful if anyone could explain this to me.
Edit: Okay, let's say the object is a clay ball and it collides with the floor inelastically such that it's kinetic energy is converted into heat. In the moving frame of reference the heat would be simply equal to $\frac{1}{2}mv^2$ which is equal to $mgh$. In the stationary frame of reference it would be equal to $\frac{1}{2}mv^2-\frac{1}{2}mu^2$ since the ball after colliding is moving with speed $u$. This can be calculated to be equal to $mgu\sqrt{\frac{2h}{g}} + mgh$ which is clearly greater than the heat produced in the frame attached to the elevator and this is a contradiction because the heat measured in any frame should be the same.
 A: Boy, this was tricky, but the secret is in conservation of momentum.
See, you are assuming that, after the collision, the velocity of the ball-elevator ensemble is $u$, but this is not fully true: it will be $u' = u + \frac{m}{m+M}\sqrt{2gh}$, $M$ being the mass of the elevator. Of course if $M \to \infty$ that reduces to $u' = u$, but when computing the KE, something funny happens:
$$\frac{1}{2}(m+M)u'^2 = \frac{1}{2}(m+M)u^2 + \frac{m^2}{m+M}gh + um\sqrt{2gh}$$
That last term which does not depend on $M$ is the key here. Of course the first term, with the $(m+M)$ dominates the others, but it will be cancelled out by identical terms in the KE before the collision. But if you assume that because $M \to \infty$ you can take $u' = u$, you will be missing this last term, which exactly cancels out that extra energy.
Doing the math for a finite elevator mass, and using conservation of momentum to compute the final velocity, you eventually get to energy lost in an inellastic  collision to be $\frac{1}{2}\frac{mM}{m+M}(u-v)^2$, which for $M \to \infty$ reduces to $\frac{1}{2}m(u-v)^2$, as Johannes already pointed out.
A: There is no such thing as conservation of energy between intertial reference frames. (The kinetic energy of a car is larger in any intertial frame that is not it's own restframe)
Considering the observer inside the elevation, the free fall takes $t_f=\sqrt{\frac{2h}{g}}$, after which is has velocity $gt_f$, and thus kinetic energy $mgh$ (which is cheating, as this is what you used to calculate $t_f$ in the first place. However, the total energy of the particle itself is conserved within this frame, between two times.
Now consider the external observer. It sees an increase in kinetic energy of
$$ \Delta K = \frac{1}{2}m(u+\sqrt{2gh})^2-\frac{1}{2}mu^2 $$
Which simplifies to:
$$ \Delta K = \frac{1}{2}m\left(u^2+2u\sqrt{2gh}+2gh\right)-\frac{1}{2}mu^2$$
$$ = mu\sqrt{2gh}+mgh$$
Where the first term is related precisely to the additional difference in height that you calculated!
A: I think I have the answer. The change in kinetic energy of the ball after impact in the moving elevator is more. But it is wrong to assume that all of it will be converted into heat. a part of it may be transferred and added to the KE of the elevator. Now what was the KE of the elevator? As we assumed it to be very heavy as compared to the falling object, elevator's mass was infinite. So with any finite velocity, it's KE becomes infinite. We will never get to know how much of falling objects KE got transferred to the moving elevator as it's KE was infinite before and after the impact. In order to find the amount of KE converted into heat, we must get into a reference frame where KE of the elevator is not infinite. and with infinite mass, only if the velocity of the elevator is zero, we will get a finite (zero) KE of the elevator. that's why we must calculate the KE lost in a frame of reference where the elevator is at rest. 
We've been doing this all the time. if the elevator is at rest, it's sitting on the Earth, which is a huge ball with infinite mass. so we calculate the KE lost of small falling object from a reference frame in which the Earth is at rest. consider a frame of reference in which the earth is moving at 5m/s and hits a ball at rest (perfectly inelastic). change in KE of the earth will be zero (before and after the impact, KE is infinite), but the KE of ball will increase to some finite value. and it'll also get some heat. where did that energy come from? 
A: An observer in the lift and an observer at rest in the building observe the same energy being transferred in heat:
A clay ball with mass $m$ drops from the ceiling in an elevator and hits the floor. The elevator has height $h$ and moves downward with uniform speed $u$. An observer at the ground floor observes the clay ball to fall with speed $u+g t$. When it hits the floor, the clay ball has spend a time $\sqrt{\frac{2h}{g}}$ falling and moves with speed $v = u+\sqrt{2hg}$. The lift floor moves with speed u, and the energy dissipated in the inelastic collision corresponds to the velocity difference between the two: $\frac{1}{2} m (v-u)^2 = \frac{1}{2} m (2hg) = mgh$. The velocity $u$ has dropped from this equation, expressing the fact hat he energy transferred is independent of the frame of reference chosen.
A: I think the thing to realise here is that changes in kinetic energy aren't independent of the reference frame. 
To see this, consider a mass of $2\,\text{kg}$ that accelerates from rest to a speed of $1\,\text{ms}^{-1}$ over some period of time. Its kinetic energy has changed by $\frac{1}{2}m(1^2-0^2)=1\,\text{J}$. But now consider the same mass viewed by someone travelling at $10\,\text{ms}^{-1}$ in the opposite direction. Now the mass's kinetic energy has changed by $\frac{1}{2}m(11^2-10^2)=21\,\text{J}$.
This is what's happening when you consider the two different reference frames regarding the object falling in the lift. It does indeed gain more kinetic energy in the moving frame, and I'm sure that if you calculated $\frac{1}{2}m(v_1^2-v_0^2)$ in the moving versus the stationary frames of reference, the difference would also come out to $u\sqrt{\frac{2h}{g}}$. This is counterintuitive, but it has to be that way in order for energy to be conserved.
