The potential energy of a $2 k g$ particle free to move along the x-axis is given by $U (x) = (\frac{x^{4}}{4} - \frac{x^{2}}{2}) J$ . The total mechanical energy of the particle is $2 J$ . Then the maximum speed (in $m / s$ ) is

see full answer

Start JEE / NEET / Foundation preparation at rupees 99/day !!

21% of IItians & 23% of AIIMS delhi doctors are from Sri Chaitanya institute !!

An Intiative by Sri Chaitanya

$\sqrt{2}$

$\frac{1}{\sqrt{2}}$

$\frac{3}{2}$

answer is D.

(Unlock A.I Detailed Solution for FREE)

Ready to Test Your Skills?

Check your Performance Today with our Free Mock Test used by Toppers!

Take Free Test

Detailed Solution

$Kinetic energy is maximum when potential energy is minimum. \frac{d U}{d x} = 0 \begin{array}{l} \Rightarrow x^{3} - x = 0 \\ \Rightarrow x (x^{2} - 1) = 0 \\ \Rightarrow x^{2} = 1 \\ Given KE + PE = 2 \\ at x = 1, P E = U (1) = (1 / 4) - (1 / 2) = - \frac{1}{4} \\ \begin{array}{l} KE = 2 - PE \\ = 2 + (1 / 4) = (9 / 4) \\ W e k n o w, K E = \frac{1}{2} m v^{2} = \frac{9}{4} \\ \Rightarrow v = \frac{3}{2} m / s \end{array} \end{array}$