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5.Work, Energy, Power and Collision
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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
A$2a\frac{{{s^2}}}{R}$
B$2as{\left( {1 + \frac{{{s^2}}}{{{R^2}}}} \right)^{1/2}}$
C$2as$
D$2a\frac{{{R^2}}}{s}$
Solution
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 $a=\sqrt{a_{R}^{2}+a_{t}^{2}}=\sqrt{\left[\frac{2 a s^{2}}{m R}\right]^{2}+\left[\frac{2 a s}{m}\right]^{2}}$
Hence $a=\frac{2 a s}{m} \sqrt{1+[s / R]^{2}}$
$\therefore \mathrm{F}=\mathrm{ma}=2 \mathrm{as} \sqrt{1+[\mathrm{s} / \mathrm{R}]^{2}}$
$\Rightarrow \mathrm{v}=\mathrm{s} \sqrt{\frac{2 \mathrm{a}}{\mathrm{m}}}$
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 $a=\sqrt{a_{R}^{2}+a_{t}^{2}}=\sqrt{\left[\frac{2 a s^{2}}{m R}\right]^{2}+\left[\frac{2 a s}{m}\right]^{2}}$
Hence $a=\frac{2 a s}{m} \sqrt{1+[s / R]^{2}}$
$\therefore \mathrm{F}=\mathrm{ma}=2 \mathrm{as} \sqrt{1+[\mathrm{s} / \mathrm{R}]^{2}}$
Standard 11
Physics
