Question

A particle of mass ${10}^{-2}\text{kg}$ is moving along the positive $x$-axis under the influence of a force $F\left(x\right)=-\frac{\alpha }{8x^2}$, where $\alpha ={10}^{-2}{\text{N.m}}^2$. At time $t=0$, it is at $x=1.0\text{m}$ and velocity is $v=0$. Find the velocity when it reaches $x=0.5\text{m}$. Consider the magnitude along with the sign to represent the velocity.

• A. $-0.5\text{m/s}$
• B. $-1\text{m/s}$
• C. $1\text{m/s}$
• D. $2.5\text{m/s}$

Answer 1

The correct option is A $-0.5\text{m/s}$

Given, $F\left(x\right)=\frac{-\alpha }{8x^2}$
Here, $\alpha$ is $+$ve and $x^2$ is always positive. Hence $F$ is always negative.

Acceleration of particle along $x$-axis;
$a\left(x\right)=\frac{F\left(x\right)}{m}$


Let velocity $v$ at point $B$ when $x=0.5\text{m}$
We have,
$a\left(x\right)=\frac{F\left(x\right)}{m}$
$\Rightarrow a\left(x\right)=\frac{-\alpha }{8m{x}^2}=\frac{-{10}^{-2}}{8\times {10}^{-2}\times x^2}=-\frac{1}{8x^2}...........\left(i\right)$

Relating acceleration with displacement by equation:

$a=v\frac{dv}{dx}.............\left(ii\right)$

From Eq. $\left(i\right)$ and $\left(ii\right)$ we get;

$v\frac{dv}{dx}=-\frac{1}{8x^2}$

$vdv=-\frac{1}{8}\frac{dx}{x^2}$

Now, integrating on both sides and putting suitable limits (i.e at $v=0$ $x=1\text{m}$ and $v=v$ at $x=0.5\text{m}$)

$\Rightarrow {\int }_0^{v}vdv=-\frac{1}{8}{\int }_{x=1}^{x=0.5}\frac{dx}{x^2}$

$\Rightarrow \frac{v^2}{2}=\frac{-1}{8}{\left[\frac{x^{-1}}{-2+1}\right]}_{x=1.0}^{x=0.5}$

$\Rightarrow \frac{v^2}{2}=\frac{1}{8}{\left[\frac{1}{x}\right]}_1^{0.5}$

$\therefore \frac{v^2}{2}=\frac{1}{8}[\frac{1}{0.5}-\frac{1}{1}]$

or $v^2=\frac{1}{4}\left[1\right]$

$\Rightarrow v^2=\frac{1}{4}\Rightarrow v=\pm \frac{1}{2}$
$v=\pm 0.5\text{m/s}$

$\therefore v=-0.5\text{m/s}$ because velocity will be along negative $x$-direction.