The kinetic energy K of a particle moving alo | vedclass

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Question

Given, $\quad \mathrm{K}=\mathrm{as}^{2}$ or $\frac{1}{2} \mathrm{mv}^{2}=\mathrm{as}^{2}$

Or          $

Vedclass · Accepted Answer

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

Vedclass · Answer

Given, $\quad \mathrm{K}=\mathrm{as}^{2}$ or $\frac{1}{2} \mathrm{mv}^{2}=\mathrm{as}^{2}$

Or          $\mathrm{mv}^{2}=2 \mathrm{as}^{2}$         $...(1)$

Differentating $w.r.t.$ time t, $\mathrm{m} .2 \mathrm{v} \cdot \frac{\mathrm{dv}}{\mathrm{dt}}=2 \mathrm{a} .2 \mathrm{s} \cdot \frac{\mathrm{ds}}{\mathrm{dt}}$

But          $\frac{\mathrm{ds}}{\mathrm{dt}}=\mathrm{v}$

${2 \mathrm{m} \frac{\mathrm{dv}}{\mathrm{dt}}=4 \mathrm{as} 	ext { or } \mathrm{m} \frac{\mathrm{dv}}{\mathrm{dt}}=2 \mathrm{as}}$

$\mathrm{Now}$        ${\mathrm{m} \frac{\mathrm{dv}}{\mathrm{dt}}=	ext { tangential force }=\mathrm{F}_{\mathrm{t}}}$

$F_{1}=2 \mathrm{as}$         $...(2)$

Centripetal force $=F_{x}=\frac{m v^{2}}{R}=\frac{2 a s^{2}}{R}$          $...(3)$

${\mathrm{F}=\sqrt{\mathrm{F}_{\mathrm{t}}^{2}+\mathrm{F}_{\mathrm{r}}^{2}}=\sqrt{(2 \mathrm{as})^{2}+\left(\frac{2 \mathrm{as}^{2}}{\mathrm{R}}ight)^{2}}}$

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

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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