A particle moves in a potential region given by $U = 8 x^{2} - 4 x + 400 J$ . Its state of equilibrium will be

x = 25 m

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x = 0.25 m

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x = 0.025 m

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x = 2.5 m

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Solution

The correct option is B
x = 0.25 m

$F = - \frac{d U}{d x} = - \frac{d}{d x} (8 x^{2} - 4 x + 400)$
For the equilibrium condition $F = - \frac{d U}{d x} = 0 \Rightarrow 16 x - 4 = 0 \Rightarrow x = \frac{4}{16} ∴ x = 0.25 m .$

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Similar questions

A particle moves in a potential region given by $U = 8 x^{2} - 4 x + 400 J$ . Its state of equilibrium will be

U = 8 x^{2} - 4 x + 400

U = 8 x^{2} - 4 x + 400 J

U = K (x^{2} + y^{2} + z^{2})

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U = (\frac{x^{3}}{3} - 4 x + 6)

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