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(B) The net force on the particle is given by the vector sum of the centripetal force $(F_c)$ and the tangential force $(F_t)$:$F_{Net} = \sqrt{F_c^2

Vedclass · Accepted Answer

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

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(B) The net force on the particle is given by the vector sum of the centripetal force $(F_c)$ and the tangential force $(F_t)$:$F_{Net} = \sqrt{F_c^2 + F_t^2}$ ... $(i)$Given kinetic energy $K = \frac{1}{2}mv^2 = as^2$,we have $v^2 = \frac{2as^2}{m}$.The centripetal force is $F_c = \frac{mv^2}{R} = \frac{m}{R} \left( \frac{2as^2}{m} \right) = \frac{2as^2}{R}$ ... (ii)To find the tangential force,we use $F_t = ma_t$,where $a_t = \frac{dv}{dt} = \frac{dv}{ds} \cdot \frac{ds}{dt} = v \frac{dv}{ds}$.Since $v = s \sqrt{\frac{2a}{m}}$,we have $\frac{dv}{ds} = \sqrt{\frac{2a}{m}}$.Thus,$a_t = \left( s \sqrt{\frac{2a}{m}} \right) \left( \sqrt{\frac{2a}{m}} \right) = \frac{2as}{m}$.Therefore,$F_t = m \left( \frac{2as}{m} \right) = 2as$ ... (iii)Substituting (ii) and (iii) into $(i)$:$F_{Net} = \sqrt{\left( \frac{2as^2}{R} \right)^2 + (2as)^2} = \sqrt{(2as)^2 \left( \frac{s^2}{R^2} + 1 \right)} = 2as \sqrt{1 + \frac{s^2}{R^2}}$.

$A$ particle moves along a circle of radius $R$ with kinetic energy $K = as^2$,where $s$ is the distance traveled. What is the net force acting on the particle?

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