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JEE Main Physics question

PhysicsWork, Energy and PowerJEE Main

The potential energy of a 1 kg particle free to move along the x-axis is given by V(x) = (x⁴/4 − x²/2) J. The total mechanical energy of the particle is 2 J. Then, the maximum speed (in m/s) is

Answer: (D)

Solution

occurs where dV/dx = x³ − x = 0 ⟹ x(x²−1) = 0 ⟹ x = 0, ±1. Checking d²V/dx² = 3x² − 1 > 0 at x = ±1 confirms these are minima, with

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Question ID VK-154084