The kinetic energy K of a particle moving alo | vedclass

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Question

(B) Given kinetic energy $K = \frac{1}{2}mv^2 = as^2$. Thus,$v^2 = \frac{2as^2}{m}$,which implies $v = s\sqrt{\frac{2a}{m}}$. The radial acceleration

Vedclass · Accepted Answer

$2as\left(1 + \frac{s^2}{R^2}\right)^{1/2}$

Vedclass · Answer

(B) Given kinetic energy $K = \frac{1}{2}mv^2 = as^2$. Thus,$v^2 = \frac{2as^2}{m}$,which implies $v = s\sqrt{\frac{2a}{m}}$. The radial acceleration is $a_R = \frac{v^2}{R} = \frac{2as^2}{mR}$. The tangential acceleration is $a_t = \frac{dv}{dt} = v\frac{dv}{ds}$. Since $v = s\sqrt{\frac{2a}{m}}$,we have $\frac{dv}{ds} = \sqrt{\frac{2a}{m}}$. Therefore,$a_t = \left(s\sqrt{\frac{2a}{m}}\right) \left(\sqrt{\frac{2a}{m}}\right) = \frac{2as}{m}$. The net acceleration is $a = \sqrt{a_R^2 + a_t^2} = \sqrt{\left(\frac{2as^2}{mR}\right)^2 + \left(\frac{2as}{m}\right)^2}$. Factoring out $\frac{2as}{m}$,we get $a = \frac{2as}{m} \sqrt{\frac{s^2}{R^2} + 1}$. The net force is $F = ma = 2as\sqrt{1 + \frac{s^2}{R^2}}$.

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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