According to given problem, $\frac{1}{2}m{v}^2=a{s}^2$

$\Rightarrow v=s\sqrt{\frac{2a}{m}}$

So, $a_c=\frac{v^2}{R}=\frac{2a{s}^2}{mR}$ ... (i)

Further more as

$a_t=\frac{dv}{dt}=\frac{dv}{ds}.\frac{ds}{dt}=v\times \frac{dv}{ds}$ ... (ii)

[by chain rule]

Which in light of Equation. (i) i.e. $v=s\sqrt{\frac{2a}{m}}$ yields

$a_t=\left[s\sqrt{\frac{2a}{m}}\right]\left[\sqrt{\frac{2a}{m}}\right]=\frac{2as}{m}$ ... (iii)

So that $a=\sqrt{a_R^2+a_t^2}$

$=\sqrt{{\left[\frac{2a{s}^2}{mR}\right]}^2+{\left[\frac{2as}{m}\right]}^2}$

Hence, $a=\frac{2as}{m}\sqrt{1+{\left[\frac{s}{R}\right]}^2}$

$\therefore$ Force acting on the particle $F=ma=2as\sqrt{1+{\left[\frac{s}{R}\right]}^2}$