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Standard XII

Physics

Relation between Work Done and PE

If a particle...

Question

If a particle has position-dependent potential defined by $U (x) = x^{2} - 3 x - 4$ , where will the particle be at equilibrium?

- 1 m

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0 m

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1.5 m

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2 m

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4 m

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Solution

The correct option is E $1.5 m$
Potential energy of the particle $U (x) = x^{2} - 3 x - 4$
For particle to be at equilibrium, $F = \frac{- d U}{d x} = 0$
$∴$ $\frac{d U}{d x} =$ $- 2 x + 3 = 0$
$⟹ x = 1.5 m$

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If a particle has position-dependent potential defined by U(x)=x2−3x−4, where will the particle be at equilibrium?

The correct option is E 1.5mPotential energy of the particle U(x)=x2−3x−4For particle to be at equilibrium, F=−dUdx=0∴ dUdx= −2x+3=0 ⟹x=1.5 m

If a particle has position-dependent potential defined by $U (x) = x^{2} - 3 x - 4$ , where will the particle be at equilibrium?

The correct option is E $1.5 m$
Potential energy of the particle $U (x) = x^{2} - 3 x - 4$
For particle to be at equilibrium, $F = \frac{- d U}{d x} = 0$
$∴$ $\frac{d U}{d x} =$ $- 2 x + 3 = 0$
$⟹ x = 1.5 m$