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All the 'proofs' of the Work–Energy Theorem that I have seen show that the work done by the resultant force acting on a body is equal to $\Delta \left(\tfrac 12 m v^2)\right)$ for that body. [It's easy to modify the proof to show that, for speeds not negligible compared with $c$, $\tfrac 12 m v^2$ needs to be replaced by $mc^2 (\gamma (v)-1)$.]

Such a derivation can correctly be called 'a proof'. But the usual practice is to call it a proof that the work done by the resultant force is equal to the increase in kinetic energy. This seems questionable; for what right do we have to call $\tfrac 12 m v^2$ or $mc^2 (\gamma (v)-1)$, 'kinetic energy'?

This particular objection disappears if kinetic energy is defined as $\tfrac 12 m v^2$ or $mc^2 (\gamma (v)-1)$, independently of any connection with work (to avoid circularity). But surely we don't make such an unsupported definition.

Arguably a better alternative is to define a body's kinetic energy as the amount of work it does coming to rest, or as the amount of work needed to bring it to its present speed from rest. In which case the role of the Work–Energy theorem is to establish the formula for kinetic energy, rather than to show that work done = change in kinetic energy.

I'd be interested in others' views as to the correct interpretation.

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  • $\begingroup$ So kinetic energy should be defined in terms of the work done to change the state of an object from a standard reference state as per the last para. I want to ask, If I am correct, this approach agrees with how the potential energy of a charge is defined as the amount of work done to bring it to a point from infinity or how entropy is defined as integral of dQ/T right from T=0? $\endgroup$
    – Kutsit
    Commented Jul 20, 2022 at 10:27
  • $\begingroup$ For potential energy – yes, there's a close similarity in the definition, though the point at which there is zero PE is an arbitrary choice. For entropy I'm too rusty to comment. [Many years ago students up to first year university level were taught that energy is the capacity to do work. Clearly there are problems with this (for example zero point energy of an oscillator and irreversibility in many-particle systems) but in abandoning this definition of energy altogether, I think we've thrown out the baby with the bathwater.] $\endgroup$ Commented Jul 20, 2022 at 11:32
  • $\begingroup$ I think part of the issue is that there is no unique axiomatic formulation of classical physics (Feynman has a nice exposition on why physicists typically don't worry about such things). So I think there's probably at least two consistent answers to your question... (1) the work-energy "theorem" is actually a way of deriving the expression for kinetic energy (as you said), and (2) you could start by showing that energy is conserved, define kinetic energy, then derive the work energy theorem. $\endgroup$
    – Andrew
    Commented Jul 20, 2022 at 12:14
  • $\begingroup$ I'm grateful for comments received. Message to those who closed the question: I'm asking for help with an issue of interpretation. I don't see why this makes the question 'opinion-based', but I've made some alterations to the wording to remove first person references, in case it was the style of writing that caused the problem. $\endgroup$ Commented Jul 20, 2022 at 13:46
  • $\begingroup$ I don’t understand why the definition you reference is in any way “unsupported.” Clearly the quantity in question is useful - it is a conserved quantity in the absence of external forces, or when the external forces applied to a body compensate for each other (in the sense that the net work is zero). I don’t see why giving it a relevant name and then exploring its relationship with other dynamical quantities is unreasonable. $\endgroup$
    – J. Murray
    Commented Jul 20, 2022 at 14:24

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