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NEET UG PHYSICS MCQ: WORK-ENERGY-POWER

1 Hard

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Application +4 | -1

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)J

The total mechanical energy of the particle is 2J. Then, the maximum speed (in m/s) is

View answer and explanation

Correct option: D

3/\sqrt 2

Option D (

3/\sqrt 2

) is correct because the maximum speed occurs when the kinetic energy is maximum. Since

E = KE + PE

,

KE = E - PE

. We need to find the minimum potential energy and then calculate the maximum kinetic energy and corresponding speed.

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