Find out the contact force between 2kg & 3kg block placed on the incline plane as shown in figure.

Explanation

Considering both the block as a 5kg system because both will move the same acceleration.

Now show forces on the 5kg block

`:'` Acceleration of 5kg block is down the incline.
So choose one axis down the incline and other perpendicular to it .

From Newton's second Law
`N = 5 g cos 37^(@)` ....(i)
`5 g sin 37^(@) - 20 = 5a` ....(ii)
`30 - 20 = 5a `
`a = 2 m//s^(2) ` ( down the incline )
For contact force `( N_(1))` between 2kg & 3kg block we draw F.B.D. of 3kg block
From `F_("net") = ma`
`rArr 3 g sin 37^(@) - N_(1) = 3 xx 2 `
`18 - N_(1) = 6`
`N_(1) = 12 N `