The potential energy of a 1 kg particle free to move along the x -axis is given by V(x) = ≤ft( x^4/4 - x^2/2 ) J . The total mechanical energy of the - Physics MCQ AIEEE 2006 2026 | ExamTest.live

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

Solution & Step-by-step Explanation

Speed is maximum when potential energy V (x) is minimum.Find minimum of V (x) : \frac{d V}{d x} = x^{3} - x = 0 ⟹ x (x^{2} - 1) = 0 ⟹ x = 0, \pm 1 . V (0) = 0 V (1) = \frac{1}{4} - \frac{1}{2} = - 1/4 J V (- 1) = - 1/4 J Minimum potential energy U_{min} = - 1/4 J .Total Energy E = K E + V . 2 = K E_{ma x} + (- 1/4) K E_{ma x} = 2 + 1/4 = 9/4 J . \frac{1}{2} m v_{ma x}^{2} = 9/4 Since m = 1 kg : v_{ma x}^{2} = 9/2 ⟹ v_{ma x} = \frac{3}{2} m/s .

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

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