A boy of mass 30 kg starts running from rest along a circular path of radius 6 m with constant tangential acceleration of magnitude `2 m s^(-2)`. After 2 s from start he feels that his shoes started slipping on th ground. The friction between his shoes and ground is
(Take `g=10 m s^(-2)`)

Answer

6

Explanation

After `2` sec speed of boy will be
`v=2xx=4m//s`
At this moment centripetal force on boy is
`F_(r)=(mv^(2))/R=(30xx16)/6=80Nt`.
Tagential force on boy is
`F_(t)=ma=30xx2=60Nt`.
Total friction action on boy is
`F=sqrt(F_(t)^(2)+f_(t)^(2))=100Nt`
At the time of slipping
`F=mug`
or `100=muxx30xx10`
`=mu=1/3`