Conservation of energy in a different frame of reference Consider a rollercoaster that goes down a slope:

At the higher level it has speed $v_0$, then it goes down a slope and at the end it has speed $v_0 + \Delta v$. The carriage is not powered and has negligible friction. 
So by conservation of energy we should, after eliminating mass $m$ that appears in all terms, have
$$
 \frac{1}{2}v_0^2 + gh = \frac{1}{2}(v_0 + \Delta v)^2 \\
 \frac{1}{2}v_0^2 + gh = \frac{1}{2}v_0^2 + v_0\Delta v + \frac{1}{2}\Delta v^2 \\
 gh = \frac{1}{2}\Delta v^2 + v_0\Delta v
$$
(left side is energy at the beginning, right side energy at the end).
But what if we consider the same situation in a reference frame moving left with speed $v_0$. Then we should get just
$$ gh = \frac{1}{2}\Delta v^2 $$
These differ by a $v_0\Delta v$ term, so solving for $\Delta v$ gives different result. But obviously the speeds should be the same. So where is the error?
 A: I believe the difference comes from the fact that forces can do different amounts of work in  different reference frames. In particular, the normal force by the ramp does no work in the "lab" frame, but does do work in the moving frame (since there is a component of velocity that is now parallel to the normal force). I don't think you accounted for this work when you changed frames.
Frame-dependent work can cause you to calculate different changes in kinetic energy, for example, in different frames. But the changes in velocity ought to be the same (in different inertial frames) if everything is done correctly.
A: Suppose you have a constant angle $\theta$ slope in the original frame (we suppose that the transitions from horizontal movements to the slope is quasi-instantaneous). Call $x$ and $x'$ the horizontal displacements in the original and moving frame. Call $T$ the total time for going to $z=h$ to $z=0$. Then you have $x'= x- v_0 T$
With $ x= h \cot \theta$, The angle of the slope in the moving frame is given by : 
$$h \cot \theta' = h \cot \theta - v_o T \tag{1}$$
The coordinates of the normal force (in the original and moving frames) are $ \vec N = mg (- \sin \theta, \cos \theta)$. The unit displacement vector, in the moving frame, is $\vec n = (- \cos \theta', - \sin \theta')$
The total supplementary work is : 
$$W_{supp} = \vec N.\vec n \quad \dfrac{h}{\sin \theta'} = mgh\,(\sin\theta \cot \theta' - cos \theta) \tag{2}$$
Using $(1)$, we get : $$W_{supp} = -mg v_0 T \tag{3}$$
The work due to the gravity force  is :  $W = (mg \sin \theta') \dfrac{h}{\sin \theta'} = mgh$
So, finally, in the moving frame, we have  : 
$$ mgh - mgv_0T = \frac{1}{2} m(\Delta v)^2\tag{4}$$
However, $gT$ is nothing else than $\Delta v$. The work, in the LHS of the above equation, does not depend on the slope, so we may imagine a  quasi-instantaneous $90$ degrees turn from horizontal to vertical, then a quasi-vertical slope, followed by a quasi-instantaneous $90$ degrees turn from vertical to horizontal. During the quasi-instantaneous turns, the modulus of the speed is conserved (because energy is conserved and no work is done). So, considering a vertical movement, it is obvious that $\Delta v= gT$
So finally you have, skipping the overall $m$ factor : 
$$ gh - v_0 \Delta v = \frac{1}{2} (\Delta v)^2\tag{5}$$
