Physics, asked by peeyush4, 1 year ago

how can we deduce work energy theorem for a variable force

Answers

Answered by MegaRayquaza16
1
http://ask.learncbse.in/t/prove-work-energy-theorem-for-a-variable-force/5773

I have no idea how to answer the question, but these guys might.
Answered by shaktisrivastava1234
4

  \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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