Question

The potential energy of a particle in a force field is given by the equation $E=\frac{x^3}{3}-4x^2+15x-6$ Choose the correct statement among the options.

• A. no points of equilibrium exist.
• B. $x=3$ and $x=5$ are points of equilibrium.
• C. There is a stable equilibrium at $x=3$ and unstable equilibrium at $x=5$.
• D. There is neutral equilibrium at $x=3$ and $x=5$.

Answer 1

The correct option is B $x=3$ and $x=5$ are points of equilibrium.

Given,
$E=\frac{x^3}{3}-4x^2+15x-6$
For equilibrium,
$\frac{dE}{dx}=0$
Therefore, we get,
$\frac{dE}{dx}=\frac{d}{dx}(\frac{x^3}{3}-4x^2+15x-6)=0\Rightarrow \frac{3x^2}{3}-8x+15=0\Rightarrow x^2-8x+15=0\Rightarrow (x-3)(x-5)=0\Rightarrow x=3orx=5$

Hence the body will be in equilibrium at $x=3$ and $x=5$.

$\mathrm{If}\frac{d^{2}E}{d{x}^2}>0,\mathrm{then}\mathrm{it}\mathrm{is}a\mathrm{stable}\mathrm{equilibrium},\mathrm{If}\frac{d^{2}E}{d{x}^2}<0,\mathrm{then}\mathrm{it}\mathrm{is}\mathrm{an}\mathrm{unstable}\mathrm{equilibrium}.$
$\frac{d^{2}E}{d{x}^2}=2x-8$
At $x=3$, $\frac{d^{2}E}{d{x}^2}=2\times 3-8=-2$

The system is in an unstable equilibrium at $x=3$

At $x=5$,
$\frac{d^{2}E}{d{x}^2}=2\times 5-8=2$
The system is in a stable equilibrium at $x=5$.