Question

The potential energy of a 1 kg particle free to move along the x-axis is given by $V \left(x\right)=\left(\frac{x^{4}}{4}-\frac{x^{2}}{2}\right) J\cdot$ 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)

3/ $\sqrt{2}$

Answer 2

$v\left(x\right)=\left(\frac{x^{4}}{4}-\frac{x^{2}}{2}\right)$ For minimum value of V, $\frac{d v}{d x}=0$ $\Rightarrow \frac{4x^{3}}{4}-\frac{2x}{2}=0 \Rightarrow x=0, x=\pm 1$ $so, \, v_{min} \left(x=\pm1\right)=\frac{1}{4}-\frac{1}{2}=\frac{-1}{4}J$ $\therefore k_{max} +v_{min.}=$ total mechanical energy $\therefore k_{max} =\left(1 /4\right)+2 \Rightarrow k_{max.} =9 /4$ $\Rightarrow\, \quad\frac{m v^{2}}{2}=\frac{9}{4} \Rightarrow v=\frac{3}{\sqrt{2}}$ m/s