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Byju's Answer

Standard XII

Physics

Conservative and Non Conservative Forces

A particle, w...

Question

A particle, which is constrained to move along the $x -$ axis, is subjected to a force from the origin as $F (x) = - k x + a x^{3}$ . Here $k$ and $a$ are positive constants. For $x = 0$ , the functional form of the potential energy $U (x)$ of particle is

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Solution

The correct option is D
Slope of potential energy vs $x$ graph is force.
Given,
$F (x) = - k x + a x^{3}; F_{(x)} = 0;$
for $x = 0 x = \sqrt{\frac{k}{a}}$ ;
So, slope is zero at $x = 0 x = \sqrt{\frac{k}{a}}$
$U = - \int F d x$
$U = \int (k x - a x^{3}) d x$
$U = \frac{(2 k x^{2} - a x^{4})}{4} + C$
As $x$ increases beyond $\sqrt{\frac{k}{a}}$ potential energy contiuously reduces
only option possible is $d$

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