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