A particle of mass 10, g moves along a circle | vedclass

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(D) Given: Mass $m = 10\, g = 10^{-2}\, kg$,Radius $R = 6.4\, cm = 6.4 	imes 10^{-2}\, m$,Final kinetic energy $K_f = 8 	imes 10^{-4}\, J$,Initial k

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(D) Given: Mass $m = 10\, g = 10^{-2}\, kg$,Radius $R = 6.4\, cm = 6.4 	imes 10^{-2}\, m$,Final kinetic energy $K_f = 8 	imes 10^{-4}\, J$,Initial kinetic energy $K_i = 0$.The work done by the tangential force is equal to the change in kinetic energy,as the centripetal force does no work.Work done $W = F_t 	imes d$,where $F_t = m a_t$ and $d$ is the distance covered in two revolutions.Distance $d = 2 	imes (2\pi R) = 4\pi R$.Using the work-energy theorem: $W = K_f - K_i = K_f$.$m a_t (4\pi R) = K_f$.$a_t = \frac{K_f}{4\pi R m}$.Substituting the values: $a_t = \frac{8 	imes 10^{-4}}{4 	imes 3.14159 	imes 6.4 	imes 10^{-2} 	imes 10^{-2}}$.$a_t = \frac{8 	imes 10^{-4}}{4 	imes 3.14159 	imes 6.4 	imes 10^{-4}} = \frac{8}{25.6 	imes 3.14159} \approx \frac{8}{80.42} \approx 0.0995\, m/s^2$.Rounding to the nearest value,$a_t \approx 0.1\, m/s^2$.

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