Question

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

Answer 1

Answer: 3

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