A particle of mass 10 g moves along a circle of radius 6.4 cm with a constant tangential acceleration. What is the magnitude of this acceleration if the kinetic energy of the particle becomes equal to 8 xx 10^(-4) J by the end of the second revolution after the beginning of the motion?

A particle of mass 10 g moves along a circle of radius 6.4 cm with a c

1.	A particle of mass 10 g moves along a circle of radius 6.4 cm with a constant tangential acceleration. What is the magnitude of this acceleration if the kinetic energy of the particle becomes equal to 8 xx 10^(-4) J by the end of the second revolution after the beginning of the motion?
Answer» `0.18 m//s^2` `0.2 m//s^2` `0.1 m//s^2` `0.15 m//s^2` Solution :Here, ` m = 10 G = 10^(-2) kg` `R= 6.4 CM = 6.4 xx 10^(-2) m, K_f = 8 xx 10^(-4) J` `K_i = 0, a_t = ?` Using work energy theorem, Work done by all the force = Change in KE `W_("TANGENTIAL force") + W_("centripetal force") = K_f - K_i` `implies F_t xx s + 0 = K_f - 0 implies ma_t xx (2 xx 2pi R) = K_f` `implies a_t = (K_f)/(4 pi m) = (8 xx 10^(-4))/(4 xx 22/7 xx 6.4 xx 10^(-2) xx 10^(-2))` `= 0.99 ~~ 0.1 MS^(-2)`.