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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
$2a\frac{{{s^2}}}{R}$
$2as{\left[ {1 + \frac{{{s^2}}}{{{R^2}}}} \right]^{1/2}}$
$2as$
$2a$
Solution
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 $\frac{\mathrm{ds}}{\mathrm{dt}}=\mathrm{v}$
${2 \mathrm{m} \frac{\mathrm{dv}}{\mathrm{dt}}=4 \mathrm{as} \text { or } \mathrm{m} \frac{\mathrm{dv}}{\mathrm{dt}}=2 \mathrm{as}}$
$\mathrm{Now}$ ${\mathrm{m} \frac{\mathrm{dv}}{\mathrm{dt}}=\text { tangential force }=\mathrm{F}_{\mathrm{t}}}$
$F_{1}=2 \mathrm{as}$ $…(2)$
Centripetal force $=F_{x}=\frac{m v^{2}}{R}=\frac{2 a s^{2}}{R}$ $…(3)$
${\mathrm{F}=\sqrt{\mathrm{F}_{\mathrm{t}}^{2}+\mathrm{F}_{\mathrm{r}}^{2}}=\sqrt{(2 \mathrm{as})^{2}+\left(\frac{2 \mathrm{as}^{2}}{\mathrm{R}}\right)^{2}}}$
${=2 \mathrm{as} \sqrt{1+\frac{\mathrm{s}^{2}}{\mathrm{R}^{2}}}}$
