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Physics · WORK, ENERGY AND POWER
हिन्दीA particle of mass \( m \) moves along the positive \( x \)-axis under a potential energy given by \( V(x) = \frac{\alpha}{x^3} - \frac{\beta}{x^2} \), where \( \alpha \) and \( \beta \) are positive constants. The position of stable equilibrium is at:
Correct answer · Option 1
\( x = \frac{3\alpha}{2\beta} \)
Explanation
\( F = -\frac{dV}{dx} = -(-\frac{3\alpha}{x^4} + \frac{2\beta}{x^3}) = \frac{3\alpha}{x^4} - \frac{2\beta}{x^3} = 0 \implies x = \frac{3\alpha}{2\beta} \). The second derivative is positive here, confirming stable equilibrium.
