A boy is pushng a ring of mass 2kg and radius 0.5 m with a stick as shwon in figure. The stick applies a force of 2N on the ring and rolls it without slipping with an accelertaion of `0.3 m/s^2.` The coefficinet of friction between the ground and the ring is large enough that rolling always occurs and the coefficient of friction between the stick and the ring of (P/10). The value of P is

Answer

`4`

Explanation

`N_(1) = 2N`
`N_(1) - f = ma`…(i)
`(f - mu N_(1)) R = mR^(2) alpha = ma` ….(ii)
From equation (i) and (ii) we get
`N_(1) (1 - mu) = 2 ma`

`2(1 - mu) = 2 xx 2 xx 0.3`
`1 - mu = 0.6 rArr mu = 0.4`.