Question

The $PE$ of a $2kg$ particle, free to move along x-axis is given by $V\left(x\right)=(\frac{x^3}{3}-\frac{x^2}{2})J$. The total mechanical energy of the particle is $4J$. Maximum speed (in $m{s}^{-1})$ is

• A. $\frac{1}{\sqrt{2}}$
• B. $\sqrt{2}$
• C. $\frac{3}{\sqrt{2}}$
• D. $\frac{5}{\sqrt{6}}$

Answer 1

The correct option is D $\frac{5}{\sqrt{6}}$

Given,

Mass $m=2kg$

Total Mechanical Energy $=4J$

Potential Energy $V\left(x\right)=(\frac{x^3}{3}-\frac{x^2}{2})$

After differentiating

$dV=d(\frac{x^3}{3}-\frac{x^2}{2})=x^2-x$

For minimum potential, $dV=0$

$x^2-x=0$

$x=0,1$

Double differentiate,

$d^{2}V=d(x^2-x)=2x-1$

At $(x=0),d^{2}v=-1$, Maximum

At $(x=1),d^{2}V=+1$, Minimum

At $(x=1)$ Minimum potential energy is $V\left(1\right)=(\frac{1^3}{3}-\frac{1^2}{2})=\frac{-1}{6}$

Total Mechanical energy = kinetic energy + potential energy

$4=\frac{1}{2}m{v}^2-\frac{1}{6}$

$v=\sqrt{\frac{\frac{25}{6}\times 2}{m}}=\sqrt{\frac{\frac{25}{3}}{2}}=2.04m/s\cong \frac{5}{\sqrt{6}}$

Maximum velocity is $\frac{5}{\sqrt{6}}m/s$