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

JEE Main · 2006 · 30 Apr 2006 · Offline · Work, Power & Energy

The potential energy of a 1 kg particle free move along the x-axis is given by

V (x) = (\frac{x ^{4}}{4} - \frac{x ^{2}}{2}) J

The total mechanical energy of the particle 2 J. Then, the maximum speed (in

m / s

) is

(A)2
(B) $3/ 2$ ✓ Correct
(C) $2$
(D) $1/ 2$

Answer: (B)

Solution

The particle moves fastest where its kinetic energy is greatest, which is at the minimum of the potential. Setting

\frac{d V}{d x} = x^{3} - x = 0

gives

x = \pm 1

, where

V = \frac{1}{4} - \frac{1}{2} = - \frac{1}{4} J

. With a total energy of

2 J

, the greatest kinetic energy is

2 - (- \frac{1}{4}) = \frac{9}{4} J

. For the

1 kg

particle,

\frac{1}{2} v^{2} = \frac{9}{4}

gives

v = \frac{9}{2} = \frac{3}{2} m/s

JEE Main · 2006 · 30 Apr 2006 · Offline · Single correct option

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