A particle of mass $m={10}^{-2}$ kg is moving along the positive x-axis under the influence of a force  $F=-\frac{k}{2x^2}$ where $k={10}^{-2}N{m}^2$ . When the particle is at $x=1.0$  m, its velocity  $\upsilon =0.$ when it reaches $x=0.5$m, the magnitude of its velocity is

$\text{F=ma = m}\frac{\text{dv}}{\text{dt}}\text{= m}\frac{\text{dv}}{\text{dx}}\text{.}\frac{\text{dx}}{\text{dt}}\text{= mv}\frac{\text{dv}}{\text{dx}}$

Given  $\text{F=-}\frac{\text{k}}{{\text{2x}}^{\text{2}}}\text{.Hence}$ 

$-\frac{k}{2x^2}=mv\frac{dv}{dx}\Rightarrow vdv=-\frac{k}{2m}{x}^{-2}dx$

Integrating. We have

$\int vdv=-\frac{k}{2m}\int x^{-2}dx$

$\Rightarrow \frac{v^2}{2}=\frac{k}{2mx}+C$

Where C is the constant of integration Given $v=0$ when x =1.0 m . 

Using this in Eq(1) we get C=-k/2m.

Equation (1) becomes 

$\frac{v^2}{2}=\frac{k}{2m}\left(\frac{1}{x}-1\right)\Rightarrow v={\left[\frac{k}{m}\left(\frac{1}{x}-1\right)\right]}^{1/2}$

Substituting $k={10}^{-2}N{m}^2,m={10}^{-2}kgandx=0.5m$ and simplifying we get ${\text{v =1.0ms}}^{\text{-1}}$