Figure shown two blocks in contact sliding down an inclined surface of inclination `30^(@)`. The friction coefficient between 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 `2 kg` block down drow the plane will be more than the acceleration of `4 -kg` block ,if allowed to move sepseparatcly .
In this , both blocks are treated as a system of mass `(4 + 2) = 6` kg and will move down with the same acceleration . Net force down the plane is
`F = (m_(1) + m_(2)) g sin theta - mu_(1) m_(1) g cos theta - mu_(2) m_(2) g cos theta`
`=(4 + 2)g sin 30^(@) - (0.2)(2) g cos 30^(@)- (0.3) (4) g cos 30^(@)`
` = (6)(10) ((1)/(2)) - (0.4)(10) (sqrt(3)/(2)) - (1.2)(10) (sqrt(3)/(2))`
`= 30 - 13.76 = 16.24 N`
Therefore , acceleration of both the blocks down the plane will be
`a = (F)/(m_(1) + m_(2)) = (16.24)/(4 +2) = 2.7 ms^(-2)`