What is the proof for kinetic energy $= \dfrac{mv^2}{2}$?
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1Possible duplicates: http://physics.stackexchange.com/q/535/2451 , http://physics.stackexchange.com/q/27847/2451 and links therein. – Qmechanic Nov 17 '16 at 06:34
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And also? http://physics.stackexchange.com/q/27847/ – Farcher Nov 17 '16 at 08:31
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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 $=\dfrac{mv^2}{2}.$
Pink
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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. – mdcq 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
$$=m\times\dfrac{v^2}{2\not s}\times \not s$$ $$W.D.=\dfrac{1}{2}mv^2$$
Deepak Suwalka
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