# The potential energy of a 1 kg mass free to move along the x-axis is given by $V_{(x)}=\bigg[\large\frac{x^4}{4}-\frac{x^2}{2}\bigg]$$J The total mechanical energy of the particle is 2J then the maximum speed is? $(a)\;\frac{3}{\sqrt 2}\; m/s \quad (b)\;\sqrt 2 \;m/s\quad (c)\; \frac{1}{\sqrt 2}\; m/s \quad (d)\;2\; m/s$ ## 1 Answer Velocity is maximum when KE of particle is maximum. ie potential energy of particle is minimum \large\frac{dV}{dx}$$=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.

edited Feb 17, 2014 by meena.p