Question

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

Answer 1

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 $\frac{d V}{d x}=\frac{4 x^{3}}{4}-\frac{2 x}{2}=x\left(x^{2}-1\right)$
For $V$ to be minimum, $\frac{d V }{ dx }=0$
$ \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 }=1.5 m s ^{-1} $