Question

A particle of mass ${10}^{-2}kg$ is moving along the positive x-axis under the influence of a force, $F\left(x\right)=\frac{-k}{2x^2}$, where $k={10}^{-2}N{m}^2$. At time $t=0$, it is at $x=1.0m$ and its velocity $v=0$. The speed of the particle when it reaches $x=05m$ is

Answer 1

Given, $F\left(x\right)=\frac{-k}{2x^2}m={10}^{-2}kgk={10}^{-2}N{m}^2$

$a=\frac{F}{m}=\frac{-k}{2x^2\times {10}^{-2}}=\frac{-{10}^{-2}}{2x^2\times {10}^{-2}}=\frac{-1}{2x^2}$

$a=\frac{dv}{dt}=\frac{dv}{dx}\times \frac{dx}{dt}=v\frac{dv}{dx}=\frac{-1}{2x^2}$

$\Rightarrow vdv=\frac{-dx}{2x^2}$

Integrating,

${\int }_0^{v}vdv={\int }_1^x\frac{-dx}{2x^2}$

(at $v=0,x=1$)

$\Rightarrow \frac{v^2}{2}=\frac{1}{2}{\left[\frac{1}{x}\right]}_1^x=\frac{1}{2}[\frac{1}{x}-1]$

At $x=0.5$,
$v^2=1⟹v=\pm 1m/s$

$\therefore |v|=1m/s$