Potential energy of a particle varies with po | vedclass

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(B) The force $F$ acting on the particle is given by the negative gradient of the potential energy: $F = -\frac{dU}{dr}$.Given $U = \alpha r^{-4} - \b

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$\frac{5\beta}{4\alpha}$

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(B) The force $F$ acting on the particle is given by the negative gradient of the potential energy: $F = -\frac{dU}{dr}$.Given $U = \alpha r^{-4} - \beta r^{-5}$.Differentiating with respect to $r$:$F = -\frac{d}{dr}(\alpha r^{-4} - \beta r^{-5}) = -[\alpha(-4r^{-5}) - \beta(-5r^{-6})] = 4\alpha r^{-5} - 5\beta r^{-6}$.For the particle to be in equilibrium,the net force must be zero: $F = 0$.$4\alpha r^{-5} - 5\beta r^{-6} = 0$.$4\alpha r^{-5} = 5\beta r^{-6}$.Multiplying both sides by $r^6$:$4\alpha r = 5\beta$.Therefore,$r = \frac{5\beta}{4\alpha}$.

Potential energy of a particle varies with position $r$ as,$U = \left( \frac{\alpha}{r^4} - \frac{\beta}{r^5} \right) \text{ J}$,where $\alpha$ and $\beta$ are positive constants. The particle will be in equilibrium at $r = \dots$

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