Question

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

• 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

The correct option is C $F\propto \frac{1}{x^2}$

$V\propto \frac{1}{\sqrt{x}}$
$\frac{dv}{dt}=\frac{1}{2x^{3/2}}\cdot \frac{dx}{dt}-\frac{dv}{dt}\propto \frac{1}{x^{3/2}}\times \frac{1}{x^{1/2}}$
$\frac{dv}{dt}\propto \frac{1}{x^2}\Rightarrow F\propto \frac{1}{x^2}$.