The kinetic energy of a particle moving along | vedclass

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(B) The kinetic energy is given by $K = \frac{1}{2}mv^2 = as^2$. Thus,$v^2 = \frac{2as^2}{m}$.The tangential acceleration is $a_t = \frac{dv}{dt} = \f

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$2as\sqrt{1 + \frac{s^2}{R^2}}$

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(B) The kinetic energy is given by $K = \frac{1}{2}mv^2 = as^2$. Thus,$v^2 = \frac{2as^2}{m}$.The tangential acceleration is $a_t = \frac{dv}{dt} = \frac{dv}{ds} \cdot \frac{ds}{dt} = v \frac{dv}{ds}$.Since $v = s\sqrt{\frac{2a}{m}}$,we have $\frac{dv}{ds} = \sqrt{\frac{2a}{m}}$.So,$a_t = (s\sqrt{\frac{2a}{m}})(\sqrt{\frac{2a}{m}}) = \frac{2as}{m}$.The tangential force is $F_t = ma_t = 2as$.The centripetal force is $F_c = \frac{mv^2}{R} = \frac{m(2as^2/m)}{R} = \frac{2as^2}{R}$.The net force is $F = \sqrt{F_t^2 + F_c^2} = \sqrt{(2as)^2 + (\frac{2as^2}{R})^2}$.Factoring out $2as$,we get $F = 2as \sqrt{1 + \frac{s^2}{R^2}}$.

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

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