Question

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

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

Answer 1

Answer: (A)

$\frac{3}{\sqrt{2}}$

Velocity is maximum when K.E. is maximum
For minimum. P.E.,

$\frac{dV}{dx}=0\Rightarrow x^3-x=0\Rightarrow x=\pm 1$

$\Rightarrow Min.P.E.=\frac{1}{4}-\frac{1}{2}=-\frac{1}{4}J$

$K.E{.}_{\left(max\right)}+P.E{.}_{\left(min\right)}=2\left(Given\right)$

$\therefore K.E{.}_{\left(max\right)}=2+\frac{1}{4}=\frac{9}{4}$

$K.E{.}_{max}=\frac{1}{2}m{v}_{max}^2$

$\Rightarrow \frac{1}{2}\times 1\times v_{max}^2=\frac{9}{4}\Rightarrow v_{max}=\frac{3}{\sqrt{2}}$