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

`1/2mv^2=Erarr1/2((10)/(1000))v^2=8xx10^(-4)`
`v^2=(8xx10^(-4))200=(16)/(100)ms^(-1)`
v=`(4)/(10)ms^(-1)`
Now applying `v^2-u^2=2as`
`((4)/(10))^2=2a(4piR),[s=4piR=2(2piR)]`
`(16)/(100)=2a(4pi(6.4)/(100))`
`a=(16)/(100)xx[(7xx100)/(8xx22xx6.4)]=0.1" "m//s^2`