Question

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

• A. $3/\sqrt{2}$
• B. $\sqrt{2}$
• C. $1/\sqrt{2}$
• D. $2$

Answer 1

The correct option is A $3/\sqrt{2}$

Given $V\left(x\right)=(\frac{x^4}{4}-\frac{x^2}{2})$
for maximum value of $V$
$\frac{dV}{dx}=0\Rightarrow \frac{4x^3}{4}-\frac{2x}{2}=0$
$x^3-x=0$
$x(x^2-1)=0$
$x=0,x=\pm 1$
so, $V_{min}(x=\pm 1)=\frac{1}{4}-\frac{1}{2}=\frac{-1}{4}J$
$\therefore$ Total mechanical energy $K_{max}+V_{min}$
$\Rightarrow K_{max}=2+\frac{1}{4}=\frac{9}{4}J$
$\Rightarrow \frac{m{v}^2}{2}=\frac{9}{4}=\frac{3}{\sqrt{2}}m/s$