A block starts moving up an inclined plane of inclination `30^@` with an initial velocity of `v_0` . It comes back to its initial position with velocity `(v_0)/(2)` . The value of the coefficient of kinetic friction between the block and the inclined plane is close to `I/(1000)` , The nearest integer to I is ____________ .

Answer

346

Explanation

A to B :
`a_(1)=g sin 30^(@)+mu g cos30^(@)`
`a_(1)=(g)/(2)+(mu gsqrt(3))/(2)`, `g=10m//s^(2)`
`v_(0)^(2)-2a_(1)(s)=0`
`s=(v_(0)^(2))/(a_(1))` ...........(i)
B to A :
`a_(2)=(g)/(2)-(mu sqrt(3))/(2)g`
`((V_(0))/(2))^(2)=2a_(2)(s)`
`s=(V_(0)^(2))/(4a_(2))`..........(ii)
From equation (i) and (ii)
`(V_(0)^(2))/(a_(1))=(V_(0)^(2))/(4a_(2))impliesa_(1)=4a_(2)`
`implies5+5sqrt(3)mu=4(5-5sqrt(3)mu)`
`implies5+5sqrt(3)mu=20-20sqrt(3)muimplies25sqrt(3)mu=15`
`impliesmu=(sqrt(3))/(5)=0.346=(346)/(1000)`
So, `(I)/(1000)=(346)/(1000)` i.e., `I=346`