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A body of mass 30 kg starts running rest...

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

A

`(1)/(2)`

B

`(1)/(3)`

C

`(1)/(4)`

D

`(1)/(5)`

Text Solution

Verified by Experts

The correct Answer is:
B
After `2 sec` speed of boy will be
`v=2xx2=4m//s`
At this moment centripetal force on boy is
`F_(r)=(mv^(2))/(R)=(30xx16)/(6)=80Nt. `
Tangential forc on boy is
`F_(t)=ma=30xx2=60Nt.`
Total friction acting on boy is
`F=sqrt(F_(r)^(2)+F_(t)^(2))=100Nt`
At the time of slipping
`F=mumg`
or `100 =mu xx 30 xx 10`
`rArr mu=(1)/(3)`
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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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