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What is special about $(1/2)m v^2$ that makes physicists believe that it is a representation of kinetic energy?

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  • $\begingroup$ It's not defined that way. Work is defined as $dW=Fds$, from which it follows after a short derivation like this: www2.physics.umd.edu/~alaporta/PHYS171_f12/lectures/… $\endgroup$
    – CuriousOne
    Jan 5, 2015 at 9:34
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    $\begingroup$ Possible duplicates: physics.stackexchange.com/q/535/2451 , physics.stackexchange.com/q/27847/2451 and links therein. $\endgroup$
    – Qmechanic
    Jan 5, 2015 at 9:34
  • $\begingroup$ @CuriousOne, I always thought that the definition of work was derived from d/dt(KE). If in fact the opposite is true, what gives us confidence in the definition of work? $\endgroup$
    – user46268
    Jan 5, 2015 at 9:36
  • $\begingroup$ The same as anything in physics: experiments. None of this descends from some primordial axioms. Work is a preserved quantity for primitive machines, which "follows" from energy conservation, which "follows" from Noether's theorem. Historically we began by experimenting with and reasoning about primitive machines and worked our way up to Lie groups. That's still a good way of teaching physics to kids. If all high school kids were mathematically seriously gifted we could start with relativistic representations of Lie groups and derive work as a conserved quantity for primitive machines, too. $\endgroup$
    – CuriousOne
    Jan 5, 2015 at 9:42
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    $\begingroup$ You can find a detailed explanation here, and definition of work is explained in the given link and in the other recent answers $\endgroup$
    – bobie
    Jan 5, 2015 at 9:44

1 Answer 1

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It's the work done to accelerate a particle from rest to a final speed $v$, i.e. the energy "put in" to initiate motion.

Work $W \equiv \int_0 ^{v} \,F \, dr$, with $F = ma = m\frac{dv}{dt} = m \frac{dv}{dr}\frac{dr}{dt} = mv\frac{dv}{dr}$, so:

$$W = \int_0 ^{v} \,F \, dr=\int_0 ^{v} \, mv'\frac{dv'}{dr} dr = \int_0 ^{v} \, mv'\,dv' = \frac{1}{2}mv^2$$

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    $\begingroup$ I don't like this answer, since it uses the Work-Energy-Theorem when saying that kinetic energy is "the work done to accelerate a particle from rest to a final speed" and the WET is deduced by knowing the kinetic energy is $1/2 m v^2$. So the argument is circular. $\endgroup$
    – mdcq
    Jun 10, 2018 at 16:05

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