The potential energy of a 1 kg particle free to move along the x -axis is given by U ( x ) = ( 4 x 4 − 2 x 2 ) J. The total mechanical energy of the particle is 2 J. Then, the maximum speed (in m/

Question

The potential energy of a 1 kg particle free to move along the x -axis is given by U ( x ) = ( 4 x 4 ​ − 2 x 2 ​ ) J. The total mechanical energy of the particle is 2 J. Then, the maximum speed (in m/s) is :-

TestHub · Accepted Answer

The kinetic energy is maximum where the potential energy is minimum.

$\frac{d U}{d x} = x^{3} - x = 0 \Rightarrow x = 0, \pm 1$

$U (0) = 0 (local maximum)$

$U (\pm 1) = \frac{1}{4} - \frac{1}{2}$

$U (\pm 1) = - \frac{1}{4} J$

This is the minimum.

Maximum kinetic energy:

$K E_{ma x} = E - U_{min}$

$K E_{ma x} = 2 + \frac{1}{4} = \frac{9}{4} J$

$\frac{1}{2} (1) v_{ma x}^{2} = \frac{9}{4}$

$v_{ma x} = \frac{9}{2} = \frac{3}{2} m/s$

Answer: option (A).

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