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)

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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