Question

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

$\begin{array}{c}v\left(x\right)=\frac{x^3}{3}-\frac{x^2}{2}\\ U=4J\end{array}$

$K.E$ will be max at minimum potential energy minimizing $v\left(x\right)$,

$\begin{array}{c}\frac{dv}{dx}=x^2-x\left(x=1,0\right)\\ \frac{d^{2}v}{d{x}^2}=2x-1\\ Hencewefindv\left(x\right)isminatx=1\\ v\left(x\right)=\frac{1}{3}-\frac{1}{2}=\frac{-1}{6}J\\ KE=4+\frac{1}{6}=\frac{25}{6}J\\ \frac{1}{2}m{v_{max}}^2=\frac{25}{6}=\frac{1}{2}\times 2\times {v_{max}}^2\\ V_{max}=\frac{5}{\sqrt{6}}\end{array}$