Two stationary blocks A and B of equal masses are released from an inclined plane of inclination 60° at t = 0 , such that block A is 2.4m behind block B. The coefficient of kinetic friction between the block A and the inclined plane is 0.4 while it is 0.6 for block B. Both the blocks meet after A has travelled a distance of how much distance ( in m) down the plane ? (Take `g = 10 m//s^(2)` and `sqrt(3) = 1.73`)

Answer

`15.95`

Explanation

Acceleration of A down the plane,
`a_(A) = g sin 60^(@) - mu_(A) cos 60^(@)`
` 10 xx sqrt(3)/2 - 0.4 xx 10 xx 1/2 = 5sqrt(3) -2`
`therefore alpha_(A) = 6.65 m//s^(2)`
Acceleration of B down the plane,
`alpha_(B) = g sin 60^(@) - mu_(B)g cos 60^(@)`
`= 10 xx sqrt(3)/2 - 0.6 xx 10 xx 1/2 = 5sqrt(3) -3`
`therefore a_(B) = 5.65 m//s^(2)`
The front face of A and B will come in line when,
`s_(A) = s_(B) + 2.4`
`therefore 1/2 a_(A)t^(2) = 1/2 a_(B)t^(2) + 2.4`
`therefore 1/2 xx 6.65 xx t^(2) = 1/2 xx 5.65 xx t^(2) + 2.4`
`therefore 0.5 t^(2) = 2.4`
`therefore t = 2.19 s`
Now, `s_(A) = 1/2 alpha_(A)t^(2) = 1/2 xx 6.65 xx (2.19)^(2)`
`therefore s_(A) = 15.95` m