The potential energy of a 1 kg particle free to move along the x-axis is given by $V\left( x \right) = \left( {\frac{{{x^4}}}{4} - \frac{{{x^2}}}{2}} \right){\rm{J}}$.The total mechanical energy of particle is 2 J. Then, the maximum speed $\left( {{\rm{in\;m}}{{\rm{s}}^{ - 1}}} \right)$ is

JEE PHYSICS MCQ: WORK-ENERGY-POWER

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The potential energy of a 1 kg particle free to move along the x-axis is given by

V\left( x \right) = \left( {\frac{{{x^4}}}{4} - \frac{{{x^2}}}{2}} \right)\mathrm{J}

.The total mechanical energy of particle is 2 J. Then, the maximum speed

\left( {\mathrm{in\;m}{\mathrm{s}^{ - 1}}} \right)

View answer and explanation

Correct option: A

3/\sqrt 2

Option A (

3/\sqrt 2

) is correct because it is derived from equating the total mechanical energy to the sum of maximum kinetic energy and minimum potential energy.

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v_1