Question

Potential energy of a particle moving along $x-$ axis is given by $U=(\frac{x^3}{3}-4x+6)$.
Here, $U$ is in joules and $x$ in metres. Find the position of stable and unstable equilibriums, respectively.

• A. $x=2\text{m}$ and $x=-2\text{m}$
• B. $x=-2\text{m}$ and $x=2\text{m}$
• C. $x=0$ and $x=0$
• D. $x=1\text{m}$ and $x=-1\text{m}$

Answer 1

The correct option is A $x=2\text{m}$ and $x=-2\text{m}$

$U=\frac{x^3}{3}-4x+6$
$F=-\frac{dU}{dx}=-x^2+4$
$\Rightarrow F=0$ at $x=\pm 2\text{m}$

For $X>2\text{m},F=-$ve
i.e., displacement is in positive direction and force is negative. Therefore $X=2$ is a stable equilibrium position.

For $X<-2\text{m},F=-$ve
i.e., force and displacement are in negative directions. Therefore $X=-2$ is an unstable equilibrium position.