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 (a) 3/√(2) (b) √(2) (c) 1/√(2) (d) 2

Answer: (a). Speed is maximum where V is minimum: dV/dx = x³ - x = 0 ⇒ x = ± 1. Vmin = 1/4 - 1/2 = -1/4 J Kmax = 2 + 1/4 = 9/4 J = 1/2(1)v² ⇒ v = 3/√(2) m/s So the answer is (a) (3)/(√ 2).

The potential energy of a 1 kg particle free to move along the x… | JEE Main 2006 (AIEEE)

Q15JEE Main 2006 · AIEEEMedium

Potential Energy & Equilibrium

The potential energy of a $1\text{ kg}$ particle free to move along the $x$ -axis is given by $V(x) = \left(\frac{x^4}{4} - \frac{x^2}{2}\right)\text{ J}$ . The total mechanical energy of the particle is $2\text{ J}.$ Then, the maximum speed (in $\text{m/s}$ ) is

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Answer: (a)

Speed is maximum where $V$ is minimum: $\frac{dV}{dx} = x^3 - x = 0 \Rightarrow x = \pm 1$ .
$V_{min} = \frac{1}{4} - \frac{1}{2} = -\frac{1}{4}\text{ J}$
$K_{max} = 2 + \frac{1}{4} = \frac{9}{4}\text{ J} = \frac{1}{2}(1)v^2 \Rightarrow v = \frac{3}{\sqrt{2}}\text{ m/s}$
So the answer is (a) $\frac{3}{\sqrt 2}$ .

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