# What is the proof for kinetic energy $= mv^2/2$? [duplicate]

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

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 $=\dfrac{mv^2}{2}.$

• 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. Jun 10, 2018 at 16:27

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

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

• Your second equation $v^2=...$ is specific to constant acceleration. Nov 17, 2016 at 7:16