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

\frac{2 a s^{2}}{R}

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2 a s {(1 + \frac{s^{2}}{R})}^{\frac{1}{2}}

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2 a s

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2 a \frac{R^{2}}{s}

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Solution

The correct option is B $\frac{2 a s^{2}}{R}$
$K E = a s^{2}$
$\Rightarrow \frac{1}{2} m v^{2} = a s^{2}$
$\Rightarrow \frac{m v^{2}}{R} = \frac{2 a s^{2}}{R}$

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The kinetic energy T of a particle moving along a circle of radius R depends on the distance covered as T=as2. The force acting on the particle is:

The correct option is B 2as2RKE=as2⇒12mv2=as2⇒mv2R=2as2R

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

The correct option is B $\frac{2 a s^{2}}{R}$
$K E = a s^{2}$
$\Rightarrow \frac{1}{2} m v^{2} = a s^{2}$
$\Rightarrow \frac{m v^{2}}{R} = \frac{2 a s^{2}}{R}$