Test Index

UPSEE 2019 Solved Paper

© examsnet.com

Question : 7 of 150

Marks: +1, -0

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

x

is proportional to

$\frac{1}{x}$
$x$
$x^2$
$e^{x}$

Solution:

Given, retardation

a=-k x

Here,

x

is the displacement of the particle.

As we know that,

a=\frac{dv}{dt}=\frac{dv}{dx} \cdot \frac{dx}{dt}=-k x

So,

v \cdot \frac{dv}{dx} = -k x \left[\because \frac{dx}{dt}=v\right]

Let for any displacement from 0 to

x

, velocity changes from

v_0

to

v

.

\Rightarrow \int\limits_{v_0}^{v} v\,dv = -k \int\limits_{0}^{x} x\,dx

\Rightarrow \left[ \frac{v^2}{2} \right]_{v_0}^{v} = -\frac{k x^2}{2}

\Rightarrow \frac{m(v^2 - v_0^2)}{2} = -\frac{m k x^2}{2}

\Rightarrow \Delta \text{KE} \propto x^2

Hence, the correct option is (c).

© examsnet.com

Go to Question:

Prev Question

Next Question