Question

The potential energy of a $2kg$ particle, free to move along the $x$-axis is given by $V(x)=\left(\frac{x^4}{4}-\frac{x^2}{2}\right)J$. The total mechanical energy of the particle is $2J$ then, the maximum speed (in $m{s}^{-1}$ ) is

Answer 1

Answer: 1.5

Total energy $E_T=2J$ is fixed.
For maximum speed, kinetic energy is maixmum.
The potential energy should, therefore, be minimum
$\therefore V(x)=\frac{x^4}{4}-\frac{x^2}{2}$
or $\frac{dV}{dx}=\frac{4x^3}{4}-\frac{2x}{2}=x\left(x^2-1\right)$
For $V$ to be minimum, $\frac{dV}{dx}=0$
$\therefore x\left(x^2-1\right)=0$,
or $x=0,\pm 1$ at $x=0,v(x)=0$
At $x=\pm 1,V(x)=-\frac{1}{4}J$
$\therefore$ (Kinetic energy) ${)}_{\max }=E_T-V_{\text{min }}$
Or $\therefore$ (Kinetic energy) ${)}_{\max }=2+\frac{1}{4}=\frac{9}{4}J$
Or $\frac{1}{2}m{v}_m^2=\frac{9}{4}$
$v_m^2=\frac{9\times 2}{m\times 4}$
or $v_m^2=\frac{9\times 2}{m\times 4}=\frac{9\times 2}{2\times 4}$
$\therefore v_m=1.5m{s}^{-}$