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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

Option 1)
$x^{2}$
Option 2)
$e^{x}$

Option 3)
$x$
Option 4)
$\log_{e}x.$

Answers (2)

best_answer

As we learnt in

Net work done by all the forces give the change in kinetic energy -

$W=\frac{1}{2}mv^{2}-\frac{1}{2}mv{_{0}}^{2}$

$W= k_{f}-k_{i}$

- wherein

$m=mass \: of\: the\: body$

$v_{0}= initial\: velocity$

$v= final\: velocity$

Retardation $a\ \alpha-x$

$\therefore\ \frac{dv}{dt}=-kx$ [k is proportionality constant]

$\Rightarrow\ v\frac{dv}{dx}=-kx$

$\int_{v_{1}}^{v_{2}}v\ dv=-k\int_{0}^{x}x\ dx$

$\Rightarrow\ \frac{1}{2}(v_{2}^{2}-{v}_{1}^{2})=-\frac{1}{2}kx^{2}$

Therefore loss in kinetic energy $=\frac{1}{2}m({v}_{2}^{2}-{v}_{1}^{2})=-\frac{1}{2}kx^{2}$

Therefore loss in kinetic energy $\alpha\ {x}^{2}$

Correct answer is 1

Option 1)

$x^{2}$

This is the correct option.

Option 2)

$e^{x}$

This is an incorrect option.

Option 3)

$x$

This is an incorrect option.

Option 4)

$\log_{e}x.$

This is an incorrect option.

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perimeter

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