A particle of mass 10 text{ g} moves along a | vedclass

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(C) Given: Mass $m = 10 	ext{ g} = 10^{-2} 	ext{ kg}$, Radius $r = 6.4 	ext{ cm} = 6.4 	imes 10^{-2} 	ext{ m}$, Kinetic Energy $K = 8 	imes 10^{

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$0.1$

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(C) Given: Mass $m = 10 	ext{ g} = 10^{-2} 	ext{ kg}$, Radius $r = 6.4 	ext{ cm} = 6.4 	imes 10^{-2} 	ext{ m}$, Kinetic Energy $K = 8 	imes 10^{-4} 	ext{ J}$.Since the particle starts from rest, its initial velocity $u = 0$.The distance covered in two revolutions is $s = 2 	imes (2 \pi r) = 4 \pi r$.Using the work-energy theorem, the work done by the tangential force $F_t$ equals the change in kinetic energy:$W = F_t \cdot s = \Delta K$Since $F_t = m a_t$, we have:$m a_t \cdot (4 \pi r) = K - 0$$a_t = \frac{K}{m \cdot 4 \pi r}$Substituting the values:$a_t = \frac{8 	imes 10^{-4} 	ext{ J}}{10^{-2} 	ext{ kg} 	imes 4 	imes 3.14 	imes 6.4 	imes 10^{-2} 	ext{ m}}$$a_t = \frac{8 	imes 10^{-4}}{25.6 \pi 	imes 10^{-4}} = \frac{8}{25.6 	imes 3.14} \approx 0.1 	ext{ m/s}^2$.

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