A particle of mass 10 g moves along a circle of radius 6.4 cm with constant tangential acceleration. What is the magnitude of this acceleration if the kinetic energy of the particle becomes equal `8xx10^(-4)J` by the end of the second revolution after the beginning of the motion?

Options

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

Answer

D

Explanation

`m=0.01kg`
`r=6.4cm`
`(1)/(2)mv^(2)=8xx10^(-4)J`
`therefore v^(2)=(16xx10^(-4))/(m)`
`=(16xx10^(-4))/(0.01)`
`therefore v^(2)=16xx10^(-2)(m//s)^(2)`
`2a_(T)S=16xx10^(-2)`
Distance covered in second revolution
`S=2pir+2pir`
`a_(T)=(16xx10^(-2))/(2S)`
`=(16xx10^(-2))/(2xx4pir)`
`=(2xx10^(-2))/(3.14xx6.4xx10^(-2))`
`=(2)/(3.14xx6.4)`
`=0.099~~0.1m//s^(2)`