Question

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

• A. $F\propto\frac{1}{x^{\frac{3}{2}}} $
• B. $F\propto\frac{1}{x} $
• C. $F\propto\frac{1}{x^{2}} $
• D. $F\propto x $

Answer 1

Answer: (C)

According to the question,
$v \propto \frac{1}{\sqrt{x}}$
or $v=\frac{k}{\sqrt{x}} \ldots$(i)
$\frac{d v}{d t}=\frac{d}{d t} \cdot \frac{K}{\sqrt{x}}=k \cdot \frac{-1}{2} \cdot x^{-\frac{1}{2}-1} \cdot \frac{d x}{d t}$
$a=-\frac{k}{2} \cdot x^{-\frac{3}{2}} \cdot \frac{k}{\sqrt{x}}\left[\because \frac{d x}{d t}=v\right.$ and $v=\frac{k}{\sqrt{x}}$ from Eq.(i) $]$
$= k^{2} \cdot \frac{1}{x^{2}} $
$ \therefore $ Force, $ F =$ Mass $ \times $| Acceleration |
$=m \frac{k^{2}}{2} \cdot \frac{1}{x^{2}}$ (magnitude)
$F=\frac{k^{2}}{2} \cdot \frac{m}{x^{2}}$
$\therefore F \propto \frac{1}{x^{2}}$