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(B) Given kinetic energy $k = \frac{1}{2}mv^2 = as^2$. From this,the velocity is $v = s\sqrt{\frac{2a}{m}}$. The centripetal acceleration is $a_R = \f

Vedclass · Accepted Answer

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

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(B) Given kinetic energy $k = \frac{1}{2}mv^2 = as^2$. From this,the velocity is $v = s\sqrt{\frac{2a}{m}}$. The centripetal acceleration is $a_R = \frac{v^2}{R} = \frac{2as^2}{mR}$. The tangential acceleration is $a_t = \frac{dv}{dt} = v\frac{dv}{ds}$. Substituting $v$,we get $a_t = \left(s\sqrt{\frac{2a}{m}}\right) \frac{d}{ds}\left(s\sqrt{\frac{2a}{m}}\right) = s\left(\frac{2a}{m}\right) = \frac{2as}{m}$. The net acceleration is $a_{net} = \sqrt{a_R^2 + a_t^2} = \sqrt{\left(\frac{2as^2}{mR}\right)^2 + \left(\frac{2as}{m}\right)^2}$. Simplifying,$a_{net} = \frac{2as}{m} \sqrt{\frac{s^2}{R^2} + 1}$. The net force is $F = m \cdot a_{net} = 2as \sqrt{1 + \frac{s^2}{R^2}}$.

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

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