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(B) Given,$K = as^2$. Since $K = \frac{1}{2}mv^2$,we have $\frac{1}{2}mv^2 = as^2$,which implies $mv^2 = 2as^2$. Differentiating with respect to time

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$2as\left[ 1 + \frac{s^2}{R^2} \right]^{1/2}$

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(B) Given,$K = as^2$. Since $K = \frac{1}{2}mv^2$,we have $\frac{1}{2}mv^2 = as^2$,which implies $mv^2 = 2as^2$. Differentiating with respect to time $t$: $m(2v \frac{dv}{dt}) = 2a(2s \frac{ds}{dt})$. Since $\frac{ds}{dt} = v$,we get $2mv \frac{dv}{dt} = 4asv$. Thus,the tangential force $F_t = m \frac{dv}{dt} = 2as$. The centripetal force is $F_c = \frac{mv^2}{R} = \frac{2as^2}{R}$. The net force $F$ is given by $F = \sqrt{F_t^2 + F_c^2} = \sqrt{(2as)^2 + \left( \frac{2as^2}{R} \right)^2}$. Factoring out $2as$,we get $F = 2as \sqrt{1 + \frac{s^2}{R^2}}$.

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

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