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The potential energy of a particle under a conservative force is given by $U (x) = (x^{2} - 3 x) J .$ The equilibrium position of the particle is at

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$x = 2.5 m$

$x = 1.5 m$

$x = 3 m$

$x = 2 m$

answer is A.

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Detailed Solution

$U (x) = (x^{2} - 3 x) J$

For a conservative field, Force $F = - \frac{d u}{d x}$

$∴ F = - \frac{d}{d x} (x^{2} - 3 x) = - (2 x - 3) = - 2 x + 3$

At equilibrium position, $F = 0$
$\Rightarrow - 2 x + 3 = 0 \Rightarrow x = \frac{3}{2} m = 1.5 m$

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