Question

The potential energy of a 1 kg 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 2 J. Then, the maximum speed (in m/s) is

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

Answer 1

Answer: (B)

$3/\sqrt{2}$

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

$\frac{dV}{dx}$= 0 ⇒ $x^3–x=0$ ⇒ x = ± 1, ⇒ Min. P.E. =$\frac{1}{4}-\frac{1}{2}=-\frac{1}{4}J$

$K.E{.}_{\left(max\right)}+P.E{.}_{\left(min\right)}$= 2 (Given), or $K.E{.}_{(max.)}=2+\frac{1}{4}=\frac{9}{4}$

$K.E{.}_{max}=\frac{1}{2}m{\nu }_{max}^2$ , ⇒ $\frac{1}{2}\times 1\times {\nu }^{2}max=$ $\frac{9}{4}$$\Rightarrow$ ${\nu }_{max}=\frac{3}{\sqrt{2}}$