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A particle of mass 1 kg can move only along the x-axis, and its potential energy is V(x) = (x⁴/4 - x²/2) J. If the particle’s total mechanical energy is 2 J, what is its greatest possible speed (in m/s)?

1. 3/√(2)
2. √(2)
3. 1/√(2)
4. 2

Correct answer: 3/√(2)

Solution

V(x) = x^4/4 - x^2/2 has minima at x = +/-1 where V = -1/4 J. Max KE = E - V_min = 2 - (-1/4) = 9/4 J, so (1/2)(1)v^2 = 9/4 gives v = 3/sqrt(2) m/s.

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