The potential energy of a 2 kg particle, free to move along x-axis is given by V(x) = (x³/3 − x²/2) J. The total mechanical energy of the particle is 4 J. Maximum speed (in ms⁻¹) is:

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Answer: 5/√6 — For maximum speed, potential energy must be minimum. dU/dx = x² − x = 0 ⇒ x = 0, 1. At x = 1, U is minimum and Umin = 1/3 − 1/2 = −1/6 J. So, Kmax = E − Umin = 4 + 1/6 = 25/6 J. Thus, (1/2)(2)v² = 25/

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