A particle of mass 10 g moves along a circle of radius 64 cm with a constant tangential acceleration. What is the magnitude of this acceleration if the kinetic energy of the particle becomes equal to `8xx10^(-4)J` by the end of the second revolution after the beginning of the motion ?

Options

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

Answer

A

Explanation

Given: Mass of particle ,M=10g`=(10)/(1000)kg`
radius of circle R=6.4 cm
Kinetic energy E of particle `=8xx10^(-4)J`
acceleration `a_(t)=?`
`(1)/(2)mv^(2)=Eimplies(1)/(2)((10)/(1000))v^(2)=8xx10^(-4)`
`impliesv^(2)=16xx10^(-2)`
`impliesv=4xx10^(-1)=0.4 m//s`
Now,using
`v^(2)=u^(2)+2a_(t)(4xx(22)/(7)xx(6.4)/(100))`
`impliesa_(t)=(0.4)^(2)xx(7xx100)/(8xx22xxx6.4)=0.1m//s^(2)`