\[(a)\;\frac{3}{\sqrt 2}\; m/s \quad (b)\;\sqrt 2 \;m/s\quad (c)\; \frac{1}{\sqrt 2}\; m/s \quad (d)\;2\; m/s\]

Velocity is maximum when $KE$ of particle is maximum. ie potential energy of particle is minimum

$\large\frac{dV}{dx}$$=0=>x^3-x=0=>x=\pm 1$

minimum $PE$ is when $x=1$

min $PE=\large\frac{1}{4}-\frac{1}{2}=\frac{-1}{4}$$J$

Given $KE_{\large \max}$$+PE_{\large min}$$=2\;J$

$KE_{\large max }=2 +\large\frac{1}{4}=\frac{9}{4}$

$\large\frac{9}{4}=\frac{1}{2} $$mv^2_{\large max} \qquad (m=1 kg)$

$ v^2 _{\large max}=\large\frac{9}{2}$

$v_{\large max}=\large\frac{3}{\sqrt 2}$$\;m/s$

Hence a is the correct answer.

Ask Question

Tag:MathPhyChemBioOther

Take Test

...