Paradox in the definition of work

So, you have discovered something that is actually very important, and if I could have my way, it would be present in every starting mechanics textbook.

It is that energy is only a pseudo-stuff.

If you have conventional stuff in a box, then it has two properties:

1. if the amount of stuff in the box changes, it has to be that stuff has either flowed through the sides, or teleported in there from some other “outside” (so, it is “conserved”), and
2. if you set that box into motion, every observer agrees how much stuff is in the box (it is “frame-independent”).

Energy obeys the first, it is conserved, so you can think about it as a sort of "stuff." But it is not frame-independent, so it is not quite a stuff in the way that, say, sugar molecules are a stuff. If you transition into a reference frame where you are parachuting down on to Stonehenge, it appears to be moving upwards at constant velocity, and thus has a kinetic energy. That kinetic energy is not changing, because it exists in a state of force balance, so it is not accelerating at all, but it's also not zero.

When forces are not balanced, everyone agrees on changes in velocity. But kinetic energy is quadratic in velocity, so one in the same change in velocity maps to a much larger kinetic energy change, in those reference frames where the thing is already moving in that direction really fast, $$ \frac12 ~m~ (u + v)^2 = \frac12 ~m~ u^2 + m~u~v + \frac12~m~v^2.$$ So a train starts moving faster by $v=1\text{ km}/\text{hr}.$ In the co-moving frame where $u=0$ the change in kinetic energy is some amount $E=\frac12mv^2$. But in the frame where that train is moving forward at $u=100\text{ km}/\text{hr}$, that quadratic law means that it has gained $201 E$ of kinetic energy. The baseline energies are not the same (it started with $10000~E$ of energy), and also the changes are not the same.

The solution is to multiply each force by the velocity that it's acting on, to get the rate that it is changing the kinetic energy. This is called the power that that force exerts at any instant, and it is a dot product, so there is also a cosine of the angle between the force and the velocity, $P = \vec F \cdot \vec v = F~v~\cos\theta.$

If you integrate the net power exerted by all forces, over time, you get the total change in kinetic energy. But if the force is constant, then you are just multiplying a velocity times a time, and that gives you the displacement of the object.

So the work-energy theorem still applies in the new reference frame, it's just that the changes in kinetic energy are all different, so the works must also be different for the math to still equate the work with the change in kinetic energy.

You have noticed something very important about work, something that is almost never explicitly stated in introductory physics classes. Specifically: work is frame variant. Meaning that the amount of work done depends on the reference frame.

So, the immediate reaction to this observation is usually something along the lines of “but if work depends on the reference frame then energy can only be conserved in certain reference frames”. The key to understanding is to remember that momentum is also conserved, and to work through the consequences of momentum conservation.

Suppose you have a person firing a rifle from the bed of a pickup truck. For simplicity we will say that the truck weighs 1000 kg and has frictionless wheels, the bullet weighs 5 g, and after firing the bullet has a velocity of 200 m/s in the frame where the truck and bullet were initially at rest. In that frame the bullet has gained 100 J of energy, and has momentum of 1 kg m/s. Due to conservation of momentum the truck has a momentum of -1 kg m/s and therefore a negligible velocity change of -1 mm/s.

Now, consider the same situation in the frame where the truck is initially moving at 20 m/s. The bullet starts with 1 J and finishes with 121 J. So almost 20 J seem to have appeared out of nowhere since the chemical energy is still just 100 J. However, looking at the truck, we see that the truck started with 200 kJ, but even though the velocity change is only -1 mm/s, that small difference in velocity reduces the KE of the truck by 20 J. So the source of the additional KE of the bullet is actually the KE of the truck, transferred to the bullet by the conservation of momentum.

Notice a couple of important points. To work this out you need to consider both energy and momentum. Because energy is proportional to velocity squared, a small change in velocity starting from zero gives negligible change in energy, but the same change in velocity starting from a higher velocity is a larger change in energy. Finally, note that the KE is different in different frames, but energy is still conserved in each frame.