Potential energy curve is shown in figure. Point at stable equilibrium is :

x = 6

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x = 4

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x = 2

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x = 3

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Solution

The correct option is A $x = 6$
$F = - \frac{d U}{d x}$

$= - (s l o p e o f U - x g r a p h)$

$X = 2 m$ is unstable equilibrium position as potential energy at this position is maximum.

$X = 6 m$ is stable equilibrium position as potential energy at this position is minimum.

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Similar questions

E = \frac{x^{3}}{3} - 4 x^{2} + 15 x - 6

x

U = 2 x^{3} - 3 x^{2}

x = 0

x = 1

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U (x) = x^{2} - 4 x

x = 2

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