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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^3-x=0 gives minima at x=+-1 with V=-1/4. Max KE = E - V_min = 2 - (-1/4) = 9/4, so (1/2)(1)v^2 = 9/4 and v = sqrt(9/2) = 3/sqrt(2) m/s.

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