The velocity (v) of a particle (under a force | vedclass

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(C) Given, $v = \frac{k}{\sqrt{x}} = k x^{-1/2}$, where $k$ is a constant.Acceleration $a = \frac{dv}{dt} = \frac{dv}{dx} \cdot \frac{dx}{dt} = v \fra

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$F \propto \frac{1}{x^2}$

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(C) Given, $v = \frac{k}{\sqrt{x}} = k x^{-1/2}$, where $k$ is a constant.Acceleration $a = \frac{dv}{dt} = \frac{dv}{dx} \cdot \frac{dx}{dt} = v \frac{dv}{dx}$.First, find $\frac{dv}{dx}$:$\frac{dv}{dx} = k \cdot (-\frac{1}{2}) x^{-3/2} = -\frac{k}{2x^{3/2}}$.Now, substitute $v$ and $\frac{dv}{dx}$ into the acceleration formula:$a = (k x^{-1/2}) \cdot (-\frac{k}{2x^{3/2}}) = -\frac{k^2}{2x^2}$.According to Newton's second law, $F = ma$.The magnitude of the force is $F = |ma| = m \cdot \frac{k^2}{2x^2}$.Since $m$ and $k$ are constants, $F \propto \frac{1}{x^2}$.

The velocity $(v)$ of a particle (under a force $F$) depends on its distance $(x)$ from the origin (with $x > 0$) as $v \propto \frac{1}{\sqrt{x}}$. Find how the magnitude of the force $(F)$ on the particle depends on $x$.

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