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What is the proof for kinetic energy $= \dfrac{mv^2}{2}$?

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marked as duplicate by Brandon Enright, Qmechanic Nov 17 '16 at 11:37

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According to work energy theorem we know Work done = change in kinectic energy so elementary work done=$$F×ds$$ where $ ds$ is the elementary displacement.Futher we know Force =$$m×a=m×dv/dt$$ Thus work done=$$ m×a×ds=m×dv/dt×ds=m×dv×ds/dt$$$$=m×v×dv $$integrating within limits from 0 to v we get total work done $=\dfrac{mv^2}{2}.$

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  • $\begingroup$ But the WET is derived by knowing that K.E. = $1/2 m v^2$. So using the WET to deduce that K.E. = $1/2 m v^2$ is a circular argument. $\endgroup$ – philmcole Jun 10 '18 at 16:27
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Work done= mass × acceleration × displacement $$=m\times a\times s$$

From physics principal, Relation of initial velocity$(u)$ & final velocity$(v)$

$$v^2 = u^2 + 2as $$
where $a$= acceleration & $s$ = displacement

But initial velocity is zero.

so $$v^2 = 2\times a\times s$$

put value of acceleration $a=\dfrac{v^2}{2s}$ into Work done equation

$$=m\times\dfrac{v^2}{2\not s}\times \not s$$ $$W.D.=\dfrac{1}{2}mv^2$$

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  • $\begingroup$ Your second equation $v^2=... $ is specific to constant acceleration. $\endgroup$ – EL_DON Nov 17 '16 at 7:16

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