Figure shown two blocks in contact sliding down an inclined surface of inclination `30^(@)`. The block of mass `2 kg `and the incline is `mu_(1) = 0.20` and that between the block of mass 4kg and the incline is `mu_(2) = 0.30` .Find the acceleration of `2.0 kg` block `g = 10 ms^(-2)`

Explanation

Since `mu_(1) lt mu_(2)` ,acceleration of `4kg` block, if allowed to move seperately. But as the `2.0kg` block is behind the `4.0kg` block both of them will move with same acceleration say `a`. Taking both the block as a single system,
Force down the plane on the system `= (4+2) g sin 30^(@)`
`= (6)(10) ((1)/(2)) =30N`
Force up the plane on the system
`= mu_(1)(2)(g) cos 30^(@) + mu_(2)(4)(g) cos 30^(@)`
`= (2mu_(1) +4 mu_(2)) g cos 30^(@)`
`= (2xx0.2+4xx0.3) (10) (0.86)`
`= 13.76N`
:. Net force down the plane is `F =30-13.76 = 16.24N`
:. Acceleration of both the block down the plane will be `a`.
`a= (F)/(4+2) (16.24)/(6) =2.7m//s^(2)`