Question

Potential energy of a particle moving along $x-$axis is given by:
$U=\left(\frac{x^3}{3}-4x+6\right)$
Here, $U$ is in joule and $x$ is in metre.

Find the position of stable equilibrium.

• A. $X=2$
• B. $X=-2$
• C. $X=1$
• D. $X=-1$

Answer 1

The correct option is A $X=2$

$U=\frac{X^3}{3}-4x+6$
$F=-\frac{dU}{dx}=-X^2+4$
$F=0$ at $X=\pm 2m$
For $X>2m,F=-ve$
$i.e.,$displacement is in positive direction and force is negative. Therefore $X=2$ is stable equilibrium position.
For $X<-2m,F=-ve$
$i.e.,$force and displacement are in negative directions. Therefore $X=-2$ is unstable equilibrium position.