Question

Derive the work energy theorem for a variable force exerted on a body in one dimension. ​

Answer 1

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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Answer 2

Answer:

$w = K_{f} -K_{j}$

Explanation:

Let a body of mass 'm' moving with a velocity 'u' and a force $\vec{F}$,

body be displaced by $\vec{ds}$ and its speed to 'v '

Now , we know that work done = force × distance travel

∴ $dw = \vec{F} . \vec{ds}$ ...........(i)

and  we also know that

$\vec{F} = m . \vec{a}$             (where 'm' is mass and 'a 'stand for acceleration )

⇒ $\vec{F} = m(\frac{\vec{dv}}{dt} )$                    ($'a' is \frac{dv}{dt}$ velocity by time )

Now put the value of $\vec{F}$ in equation (i)

∴ $dw = m(\frac{\vec{dv}}{dt} ). \vec{ds}$  which can be written as

$dw = m(\frac{\vec{ds}}{dt} ). \vec{dv}$  = $m.v.\vec{dv}$         (velocity (v) = $\frac{ds}{dt}$)

Now integrate both side ,

$\int_{u}^{v}dw = m\int_{u}^{v}v \vec{dv}$

⇒ w = $m(\frac{v^{2} -u^{2} }{2} )$

⇒ w = $\frac{1}{2}mv^{2} - \frac{1}{2}mu^{2}$  = $K_{f} - K_{j}$

($\frac{1}{2}mv^{2} - \frac{1}{2}mu^{2}$ this is called change in kinetic energy .)

So , $w = K_{f} -K_{j}$

Hence , here we proved the work energy theorem .

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