Question

Work energy theorem for variable force derivation

Answer 1

Change in kinectic energy is equal to work done...
It is work energy theorem

Answer 2

$\huge {\underline {\overline{\boxed {\frak{Answer:}}}}}$

$\large \bf{Work \: energy \: theorem \:for\:variable\:force-}$

$\sf{Work \: done \: by \: the \: net \: force \: acting} \\ \sf{on \: a \: body \: is \: equal \: to \: the \: changed} \\ \sf{produced \: in \: kinetic \: energy \: of \: the} \\ \sf{body.}$

$\longrightarrow \sf{Let \: F \: be \: the \: variable \: force.}$

$\sf{∴ Work \: done \: by \: the \: variable \: force , } \\ \sf{W = \int \limits_{x_i}^{x_f}F•dx}$

$\sf{where \: x_i \: is \: the \: initial \: position \: and \: x_f}$

$\sf{is \: the \: final \: position.}$

$\bf \underline{{Kinetic \: energy \: of \: an \: object, K= \frac{1}{2} m{v}^{2}  }}$

$\longmapsto \sf{ \frac{dK}{dt} = mv \frac{dv}{dt} }$

$\longmapsto \sf{ \frac{dK}{dt} = ma \frac{dx}{dt} }$

$\longmapsto \sf{ \frac{dK}{dt} F\frac{dx}{dt} }$

$\longmapsto \sf{{dK} = F \times {dx}}$

$\sf{ \int \limits_{K_i}^{K_f}dK = \int \limits_{x_i}^{x_f}F•dx}$

$\leadsto\sf{ \triangle{K = W}}$

$\sf{Where , {\triangle{K \: is \: the \: change \: in \:kinetic \: energy .}}}$

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