The kinetic energy k of a particle moving alo | vedclass

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Question

(b) According to given problem $\frac{1}{2}m{v^2} = a{s^2}$ $ \Rightarrow v = s\sqrt {\frac{{2a}}{m}} $

So ${a_R} = \frac{{{v^2}}}{R} = \frac{{2a{s

Vedclass · Accepted Answer

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

Vedclass · Answer

(b) According to given problem $\frac{1}{2}m{v^2} = a{s^2}$ $ \Rightarrow v = s\sqrt {\frac{{2a}}{m}} $

So ${a_R} = \frac{{{v^2}}}{R} = \frac{{2a{s^2}}}{{mR}}$&hellip;$(i)$

Further more as ${a_t} = \frac{{dv}}{{dt}} = \frac{{dv}}{{ds}} \cdot \frac{{ds}}{{dt}} = v\frac{{dv}}{{ds}}$&hellip;$(ii)$   (By chain rule)

Which in light of equation $(i)$ i.e. $v = s\sqrt {\frac{{2a}}{m}} $ yields

${a_t} = \left[ {s\sqrt {\frac{{2a}}{m}} } \right]\,\left[ {\sqrt {\frac{{2a}}{m}} } \right] = \frac{{2as}}{m}$&hellip;$(iii)$

So that $a = \sqrt {a_R^2 + a_t^2} = \sqrt {{{\left[ {\frac{{2a{s^2}}}{{mR}}} \right]}^2} + {{\left[ {\frac{{2as}}{m}} \right]}^2}} $

Hence $a = \frac{{2as}}{m}\sqrt {1 + {{[s/R]}^2}} $

 $F = ma = 2as\sqrt {1 + {{[s/R]}^2}} $

The kinetic energy $k$ of a particle moving along a circle of radius $R$ depends on the distance covered $s$ as $k = a{s^2}$ where $a$ is a constant. The force acting on the particle is

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