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 $v=\frac{k}{\sqrt{x}}\dots$(i)
$\frac{dv}{dt}=\frac{d}{dt}\cdot \frac{K}{\sqrt{x}}=k\cdot \frac{-1}{2}\cdot x^{-\frac{1}{2}-1}\cdot \frac{dx}{dt}$
$a=-\frac{k}{2}\cdot x^{-\frac{3}{2}}\cdot \frac{k}{\sqrt{x}}[\because \frac{dx}{dt}=v$ 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)
$F=\frac{k^2}{2}\cdot \frac{m}{x^2}$
$\therefore F\propto \frac{1}{x^2}$