The kinetic energy K of a particle moving along a circle of radius R depends on the distance covered s as K = as² . The force acting on the particle is

We know that kinetic energy, $\mathrm{E}=\frac{1}{2}{\mathrm{mv}}^2$

Given that, KE=as²

$\therefore \frac{1}{2}{\mathrm{mv}}^2={\mathrm{as}}^2\text{ or }{\mathrm{v}}^2=\frac{2{\mathrm{as}}^2}{\mathrm{m}}$

Differentiating with respect to time, we get

$\begin{array}{rl}2\mathrm{v}\left(\frac{\mathrm{dv}}{\mathrm{dt}}\right)& =\frac{4\mathrm{a}}{\mathrm{m}}\cdot \mathrm{s}\frac{\mathrm{ds}}{\mathrm{dt}}\\ \mathrm{or} \left(\frac{\mathrm{dv}}{\mathrm{dt}}\right)& =\frac{2\mathrm{as}}{\mathrm{m}}\text{or }{\mathrm{a}}_{\mathrm{t}}=\frac{2\mathrm{as}}{\mathrm{m}} ...\left(1\right)\end{array}$

Now normal acceleration,

${\mathrm{a}}_{\mathrm{n}}=\frac{{\mathrm{V}}^2}{\mathrm{R}}=\frac{2{\mathrm{as}}^2}{\mathrm{mR}}$  …(2)

$\text{Net acceleration}=\mathrm{\alpha }=\sqrt{({\mathrm{a}}_{\mathrm{t}}^2+{\mathrm{a}}_{\mathrm{n}}^2)}$

$\begin{array}{rl}\therefore \mathrm{\alpha }& =\sqrt{\left[{\left(\frac{2\mathrm{as}}{\mathrm{m}}\right)}^2+{\left(\frac{2{\mathrm{as}}^2}{\mathrm{mR}}\right)}^2\right]}\\ & =\frac{2\mathrm{as}}{\mathrm{m}}{\left[1+{\left(\frac{\mathrm{s}}{\mathrm{R}}\right)}^2\right]}^{1/2}\end{array}$

Now $\mathrm{F}=\mathrm{m\alpha }=2\mathrm{as}{\left[1+{\left(\frac{\mathrm{s}}{\mathrm{R}}\right)}^2\right]}^{1/2}$.