A particle moves in a straight line with retardation proportional to its displacement. Its loss of kinetic energy for any displacement $x$ is proportional to

x

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e^{x}

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x^{2}

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l o g_{e} x

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Solution

The correct option is C $x^{2}$
Given : retardation $\propto$ displacement
i.e.,
$a = - K x$ , $K$ = Constant
We know that, $a = v \frac{d v}{d x}$
$∴ \frac{v d v}{d x} = - K x$
$\int_{v_{1}}^{v_{2}} v d v = - K \int_{0}^{x} x d x$
$\frac{(v_{2}^{2} - v_{1}^{2})}{2} = - K \frac{x^{2}}{2}$
$\Rightarrow \frac{1}{2} m (v_{2}^{2} - v_{1}^{2}) = m (\frac{- K x^{2}}{2})$
$∴$ Loss in kinetic energy $= (\frac{m K x^{2}}{2}) \propto x^{2}$

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