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

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Correct answer: B. 3 / 2

k E _ = E _ T - U _ U _ ( 1)=-1 / 4 ~J KE _ =9 / 4 ~J U = 3 2 ~J

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

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