The kinetic energy $K$ of a particle moving along a circle of radius $R$ depends upon the distance $s$ as $K=a{s}^2.$ The force acting on the particle is

Given that $K=a{s}^{2}or\frac{1}{2}m{v}^2=a{s}^2$ Or $m{v}^2=2a{s}^2\mathrm{..........}\left(i\right)$

Differentiating w.r.t. time, we get $m2v\times \frac{dv}{dt}=2a\times 2s\times \frac{ds}{dt}$ But $\frac{ds}{dt}=v$

So $2m\frac{dv}{dt}=4asorm\frac{dv}{dt}=2as$

Now, $m\frac{dv}{dt}=\text{tangential\hspace{0.17em}ƒorce}=F_t, F_t=2as$

Centripetal force $=F_r=\frac{m{v}^2}{R}=\frac{2a{s}^2}{R}$ (from (i))

$F_{net}=\sqrt{F_t^2+F_r^2}=\sqrt{{\left(2as\right)}^2+{\left(\frac{2a{s}^2}{R}\right)}^2}=2as\sqrt{1+\frac{s^2}{R^2}}$