Problem with Kinetic Energy in Inertial Frames (I previously posted this on https://math.stackexchange.com/questions/2165992/problem-with-kinetic-energy-in-inertial-frames and it has been suggested I ask it here).
Suppose I slide a ball (in response to a now-deleted comment, it's an idealized ball which is frictionless and therefore slides rather than rolling so there is no rotational energy) down the passage and up a ramp at the end. If I release the ball with speed $v$ it acquires kinetic energy $T = (1/2) m v^2$ and, by conservation of energy. when it reaches the ramp it rises to a height $ h = T/mg = (1/2g)(v_{initial}^2 - v_{final}^2 )$ where $v_{final} = 0$ .So far so good.
Putting some numbers into this, I impart a speed of $2 m/s$ and $g = 9.81 m/s^2$. Then the ball will rise up approximately $h =  0.2m$ on the ramp.
What I didn't mention is that I am actually doing this in a train (if trains were good enough for Einstein then they're good enough for me). I'm sliding the ball in the forward direction and the train is moving at $20$ m/s. So now when I calculate the rise I get $ h = (1/2g)(22^2 - 20^2 ) = 4.81m$. How high does the ball actually rise ?
I can see that I probably need to impart more energy to the ball to accelerate it from $20 $ to $ 22 m/s$ than from $0 $ to $ 2$. What bothers me is that this seems to contradict the equivalence of inertial frames, and the speeds involved are hardly relativistic

As noted in the approved answer,the calculation above neglects conservation of momentum. So, take it that the train is moving at $20 m/s$ after I release the ball (otherwise there is an interaction to be considered when I launch the ball)  and let the train  have mass $M$.
Then the final speed of ball and train is given by $ u=\dfrac{20M + 22 m}{M+m}$ The energy equation is now the potential energy gained by the ball, $V = $ initial kinetic energy - final kinetic energy $ = 1/2 (M20^2 + m22^2) - 1/2(Mu^2 + m u^2)  $ If one then slogs through the arithmetic (which I have on paper but don't have the persistence to type in) one ends with 
$V = (m/2)(22 - 20)^2 M/(M+m) $
One should then note that the "stationary" situation actually needs a slight correction too for the momentum imparted to the world as the ball moves up the slope. In both cases, the mass of the world and the train are very large compared to the ball, and so $M/(M+m) \to 1$ and the PE gained by the ball is the same in both cases $= m.2^2/2$
A final thought on this is that when I launch the ball in the first place, conservation of momentum determines a small reduction in speed of the train (or the world) and I expect a detailed calculation would show that the amount of energy I need to impart would be the same in both cases.
 A: What you have neglected is the fact that momentum has to be conserved in the horizontal plane and as the ball rises it imparts some of its horizontal momentum to the carriage and the missing kinetic energy is the extra kinetic energy that the carriage has collected.  
So the final velocity of the ball and the carriage is not $20$ m/s it is more than that.  
If the mass of the train is $M$, the mass of the ball is $m$ and the final horizontal velocity of the train and ball is $V$ then applying conservation of momentum in the horizontal plane:  
$M20+m22 = (M+m)V \Rightarrow V=\dfrac{20M + 22 m}{M+m}= 20\left( 1+\dfrac{11m}{10M}\right )\left(1-\dfrac mM\right )^{-1}>20$
As an relatively easy to analyse illustration of what is happening consider the ball hitting the end of carriage wall and suffering an elastic collision.
You will find that the rebound speed of the ball relative to the ground not $18$ m/s but greater than that and the final speed of the carriage relative to the ground is greater than $20$ m/s.  
A: The general assumption that the energy of the ball is conserved is incorrect. The train takes up energy and momentum. The energy it takes is equal to:
$Mv\delta v + 1/2 M \delta v^2 $. 
the reason you can ignore this amount of energy sometimes is that due to conservation of momentum $\delta v$ goes like 1/M for large masses and hence the second term is 0 and only the first term contributes for large M than. This is however 0 in the rest frame only, which leads to the apparent contradiction above. So as soon as the mass of the ramp/train/earth is not infinite OR you leave the rest frame of the ramp/train/earth your system (the ball) loses energy. Importantly either one suffices. To see that it all works out correctly you than have to do the explicit calculation you have done above. 
