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?

Options

A. `0.18 m//s^2`
B. `0.2 m//s^2`
C. `0.1 m//s^2`
D. `0.15 m//s^2`

Answer

C

Explanation

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)`.