Question

Derive work-energy theorem

Answer 1

Answer:

The derivation of the work-energy theorem is provided here. The work-energy theorem also known as the principle of work and kinetic energy states that the total work done by the sum of all the forces acting on a particle is equal to the change in the kinetic energy of that particle. This explanation can be extended to rigid bodies by describing the work of rotational kinetic energy and torque.

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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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