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Imagine a car A is accelerating.

Two observers at constant speed, B and C, analyse the change of A's kinetic energy over a same time interval.

B sees A going from 10 to 30 m/s C sees A going from 100 to 120 m/s

As both see A having the same mass (ignore relativistic effects), and as the kinetic energy depends on the square of the velocity, C sees A having a greater change on its kinetic energy than B does.

This also means C sees that the total amount of work produced by the engine of A's car over that same time interval was greater, and thus its power was greater too.

That's what doesn't make sense for me.

Because, imagine that A's engine was a spiral spring. You can calculate the amount of potential energy released over that interval irrespective to it's velocity, just by knowing the spring properties and its initial and final state.

So, if you were on another referential with a great velocity relative to A, does this mean that the change on A's kinetic energy could be greater than the amount of potential energy released from the string?

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The car is pushing against the Earth. Calculate $K_{tot} = K_{Earth} + K_{car}.$ You will find that $K_{tot}$ changes by the same amount in both frames.

As a simpler example, consider two boxes, masses 10 and 1, attached by a compressed spring. The spring is released, pushing the masses apart.

In frame A, they are both initially at rest, then the larger mass gets a velocity of -1 and the smaller +10. $K_{tot} = \frac{1}{2} 10\times (-1)^2 + \frac{1}{2}1\times 10^2 = 55,$ a gain of 55.

In frame B, they are both initially moving with velocity $v=1,$ so $K_{tot}=\frac{1}{2}11 \times 1^2=5.5.$ After the spring is released, the large mass is at rest, and the smaller mass moves with $v=11,$ so $K_{tot}=0+\frac{1}{2}1\times 11^2=60.5,$ again a gain of 55.

In other words, both frames A and B measure the same gain in $K_{tot},$ which equals the potential energy that was stored in the spring.

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  • $\begingroup$ And this amount is the potential energy of the spring? $\endgroup$ – user2934303 Aug 15 '18 at 7:36
  • $\begingroup$ Yes. (Updating my answer.) $\endgroup$ – bernander Aug 15 '18 at 8:16

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