The potential energy function $U (x)$ of a particle moving along $x$ direction is given by $U (x) = a x^{2} - b x$ . Find the equilibirum point $(x_{e})$ .

x_{e} = \frac{a}{2 b}

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x_{e} = \frac{b}{2 a}

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x_{e} = \frac{a^{2}}{b^{2}}

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x_{e} = \frac{2 a}{b}

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Solution

The correct option is B $x_{e} = \frac{b}{2 a}$
We know that the relation between potential energy and force for this case will be
$F = \frac{- d U}{d x}$
$= \frac{- d}{d x} (a x^{2} - b x)$
For equillibrium, $F = 0$
$0 = - (2 a x_{e} - b)$
or $x_{e} = \frac{b}{2 a}$

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Conservative Force as Gradient of Potential

Standard XII Physics

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The potential energy function U(x) of a particle moving along x direction is given by U(x)=ax2−bx. Find the equilibirum point (xe).

The correct option is B xe=b2aWe know that the relation between potential energy and force for this case will be F=−dUdx =−ddx(ax2−bx) For equillibrium, F=0 0=−(2axe−b) or xe=b2a

The potential energy function $U (x)$ of a particle moving along $x$ direction is given by $U (x) = a x^{2} - b x$ . Find the equilibirum point $(x_{e})$ .

The correct option is B $x_{e} = \frac{b}{2 a}$
We know that the relation between potential energy and force for this case will be
$F = \frac{- d U}{d x}$
$= \frac{- d}{d x} (a x^{2} - b x)$
For equillibrium, $F = 0$
$0 = - (2 a x_{e} - b)$
or $x_{e} = \frac{b}{2 a}$