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I have had trouble reconciling the quadratic form of kinetic energy and reference frames traveling at different velocities for some times now. I'll give an example that confuses me for some illustration: A stationary rock (mass = 1kg) is dropped some height a gains potential energy = 4.5J In the reference frame of the rock it looses some amount of gravitational potential energy and gains kinetic energy equal to that (neglecting other lossy mechanisms)
\begin{equation} \Delta v = \sqrt{2\Delta PE / m} = 3ms^{-1} \end{equation} Now imagine the same scenario observed in a reference frame traveling at -4m/s \begin{equation} KE_{before} = 4^2/2 = 8 \end{equation} \begin{equation} KE_{after} \frac{4^2}{2} + 4.5= 8 + 4.5 = 12.25 \end{equation} \begin{equation} \Delta v = 1ms \end{equation} How is it possible that conserving energy results in different observed changes in speed?

I was under the impression the laws of physics were supposed to hold in any inertial frame, this appears to imply that there is a "right frame" which i know to be not true.

Duke of Sam
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From the moving frame of reference, the rock is moving upward with an initial velocity of 4 m/s, and, at the end of the same time interval, it is moving upward at 1 m/s. But now, as reckoned from the moving frame of reference, the change in potential energy is going to be different. This is because the datum for potential energy is moving downward, so, as reckoned from the moving frame, the potential energy increases (rather than decreases as in the case where the rock is falling). Potential energy is a frame-dependent quantity. In this problem, instead of the change in potential energy being -4.5 J after 0.3 sec., it is +7.5 J, as reckoned from the moving frame. The difference in potential energy here is 12.0 J, and is equal to the difference in elevation (4 m/s times 0.3 sec = 1.2 m) times g.

Here are some calculations to illustrate:

$$KE_1+PE_1=KE_2+PE_2$$

Taking the datum for PE for each of the reference frames the original elevation of the rock at time zero,

$$0+0=\frac{(-3)^2}{2}-4.5\tag{fixed frame of reference}$$ $$\frac{4^2}{2}+0=\frac{1^2}{2}+7.5\tag{moving reference frame}$$ Note that, when the moving reference frame moves downward by 1.2 meters during the 0.3 seconds, then, relative to an observer in this frame, the rock has moved upward 1.2 meters. So, according to this observer, the potential energy of the rock has increased by 12 J more than observed by the person in the fixed frame.

Chet Miller
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You are ignoring the change in the kinetic energy of the Earth as it and the rock accelerate towards their common centre of mass. If the rock speeds up in your moving frame then the Earth slows down in the moving frame and you have to consider both changes for the energy to balance. Obviously the change in the Earth's kinetic energy is tiny, but you need to include it to make the calculation work.

If you take the Earth as fixed then this implies there is an external force acting on the Earth that stops it accelerating towards the rock. You need to include this external force in your calculations of work done.

The calculation is not dissimilar to the one described in How am I able to stand up and walk down the aisle of a flying passenger jet?. In that case the energy discrepancy comes from ignoring the airplane moving in response to the force exerted by the walking passenger.

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John Rennie
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