Question

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

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

Answer 1

Answer: (A)

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