The kinetic energy K of a particle moving along a circle of radius R depends upon the distance moved from rest s, as  $K=a{s}^2$. The force acting on the particle is (a is a positive constant)
 

Given that  $K=a{s}^{2}or\text{\hspace{0.17em}}\frac{1}{2}m{v}^2=a{s}^2$
  $orm{v}^2=2a{s}^2$                                ……………….(i)
Differentiating w.r.t time t,m  $2\nu \frac{d\nu }{dt}=2a.2s.\frac{ds}{dt}$
But,  $\frac{ds}{dt}=\nu$
$\therefore 2m\frac{d\nu }{dt}=4asorm\frac{d\nu }{dt}=2as$
$Now,m\frac{d\nu }{dt}={\text{tangential force\hspace{0.17em}=\hspace{0.17em}F}}_t$
      $\therefore F_t=2as$                                   ……………….(ii) $\text{Centripetal force\hspace{0.17em}}=F_r=\frac{m{\nu }^2}{R}=\frac{2a{s}^2}{R}$      ……………….(iii)
$\therefore F=\sqrt{F_t^2+F_r^2}=\sqrt{{\left(2as\right)}^2+{\left(\frac{2as}{R}\right)}^2}$
$=2as\sqrt{1+\frac{s^2}{R^2}}$