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 is 2 J. Then, the maximum speed ( $\left(inm{s}^{-1}\right)$) is

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

Answer 1

Answer: (A)

$3/\sqrt{2}$

Particle will have maximum velocity when Kinetic energy is maximum.

SInce, $K.E+P.E.=T.E$

Kinetic energy will be maximum, when potential energy is minimum.

To find points where potential energy is minimum,

$\frac{dV}{dx}=0$

$x^3-x=0⟹x=0,1,-1$

Now, $\frac{d^{2}V}{d{x}^2}=3x^2-1$

• For, $x=0,\frac{d^{2}V}{d{x}^2}{|}_{x=0}=-1.$

Since, $\frac{d^{2}V}{d{x}^2}{|}_{x=0}<0$, Potential energy is maxima at $x=0$

• For, $x=1,-1⟹\frac{d^{2}V}{d{x}^2}{|}_{x=0}=2.$

Since, $\frac{d^{2}V}{d{x}^2}{|}_{x=0}>0$, Potential energy is minima at $x=1,-1$

Potential energy at $x=1,-1$, $P.E=-\frac{1}{4}J$

$(K.E)=T.E-P.E=2+\frac{1}{4}=\frac{9}{4}J$

$\frac{1}{2}m{v}^2=\frac{9}{4}⟹v^2=\frac{9}{2}$

$v=\frac{3}{\sqrt{2}}m/s$