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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 ms⁻¹) is

Answer: (A)

Solution

V is minimum where dV/dx = x³ − x = 0 ⟹ x = 0, ±1. At x = ±1, = 1/4 − 1/2 = −1/4 J. Maximum KE occurs

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Question ID VK-153568