The kinetic energy k of a particle moving alo | vedclass

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Question

According to given problem $\frac{1}{2} \mathrm{mv}^{2}=\mathrm{as}^{2}$

$\Rightarrow \mathrm{v}=\mathrm{s} \sqrt{\frac{2 \mathrm{a}}{\mathrm{m}}}$

Vedclass · Accepted Answer

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

Vedclass · Answer

According to given problem $\frac{1}{2} \mathrm{mv}^{2}=\mathrm{as}^{2}$

$\Rightarrow \mathrm{v}=\mathrm{s} \sqrt{\frac{2 \mathrm{a}}{\mathrm{m}}}$

So $a_{R}=\frac{v^{2}}{R}=\frac{2 a s^{2}}{m R}$           $...(i)$

Further more as $a_{t}=\frac{d v}{d t}=\frac{d v}{d s}  \frac{d s}{d t}=v \frac{d v}{d s} \ldots .$ $(ii)$

(By chain rule)

Which in light of equation $(i)$ i.e. $\mathrm{v}=\mathrm{s} \sqrt{\frac{2 \mathrm{a}}{\mathrm{m}}}$ yields

$a_{t}=[s \sqrt{\frac{2 a}{m}}][\sqrt{\frac{2 a}{m}}]=\frac{2 a s}{m}$      $...(iii)$

So that $a=\sqrt{a_{R}^{2}+a_{t}^{2}}=\sqrt{\left[\frac{2 a s^{2}}{m R}ight]^{2}+\left[\frac{2 a s}{m}ight]^{2}}$

Hence $a=\frac{2 a s}{m} \sqrt{1+[s / R]^{2}}$

$	herefore \mathrm{F}=\mathrm{ma}=2 \mathrm{as} \sqrt{1+[\mathrm{s} / \mathrm{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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