A particle of mass  ${10}^{-2} \text{kg}$ is moving along the positive x-axis under the influence of a force  $F\left(x\right)=-K/{\left(2x\right)}^2$ where  $K=1{\text{Nm}}^2$. At time t = 0, it is at $x=1.0 \text{m}$  and its velocity is v = 0.Then finds its velocity magnitude when it reaches $x=0.50 \text{m}$ . (in SI units)

The particle of mass $(m={10}^{-2} \text{kg})$  is moving along positive $x$ -axis
 
 $F\left(x\right)=-\frac{K}{2x^2},K={10}^{-2}{\text{Nm}}^2;$
At t = 0  $x=1.0mandv=0$
A)  $F\left(x\right)=-\frac{K}{2x^2}orm\frac{dv}{dt}=-\frac{K}{2x^2}$
or   $mv\left(\frac{dv}{dx}\right)=-\frac{K}{2x^2}$
Now integrating both side,  $m{{\int }^{\text{}}}_0^{v}dv=-{{\int }^{\text{}}}_1^x\frac{K}{2x^2}dx$
or   $\frac{m{v}^2}{2}={\left[\frac{K}{2x}\right]}_1^x=\frac{K}{2}(\frac{1}{x}-1)$
or  $v^2=\frac{K}{m}(\frac{1}{x}-1)$ or   $v=\sqrt{\frac{K}{m}(\frac{1}{x}-1)}$           (1)
When  $x=0.5m,v=\sqrt{\frac{K}{m}(\frac{1}{0.5}-1)}$
$=\sqrt{\frac{K}{m}}=\pm 10$ 
As the force is acting along negative $x$ -direction, therefore, the velocity will be in  $-x$  direction. Hence  $v=-10 \text{m}/\text{s}$ and  $\overrightarrow{v}=-10\widehat{i} \text{m}/\text{s}$