What is the proof for kinetic energy $= \dfrac{mv^2}{2}$?
$\begingroup$
$\endgroup$
2
-
1$\begingroup$ Possible duplicates: physics.stackexchange.com/q/535/2451 , physics.stackexchange.com/q/27847/2451 and links therein. $\endgroup$– Qmechanic ♦Commented Nov 17, 2016 at 6:34
-
$\begingroup$ And also? physics.stackexchange.com/q/27847 $\endgroup$– FarcherCommented Nov 17, 2016 at 8:31
Add a comment
|
2 Answers
$\begingroup$
$\endgroup$
1
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 $=\dfrac{mv^2}{2}.$
-
$\begingroup$ 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. $\endgroup$– mdcqCommented Jun 10, 2018 at 16:27
$\begingroup$
$\endgroup$
1
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
$$=m\times\dfrac{v^2}{2\not s}\times \not s$$ $$W.D.=\dfrac{1}{2}mv^2$$
-
$\begingroup$ Your second equation $v^2=... $ is specific to constant acceleration. $\endgroup$– EL_DONCommented Nov 17, 2016 at 7:16