Figure shows two blocks in contact sliding down an inclined surface of inclination $30°$. The friction coefficient between the block of mass $2.0 \mathrm{kg}$ and the incline is ${\mu }_1=0.20$ and that between the block of mass $4.0 \mathrm{kg}$ and the incline is ${\mu }_2=0.30$. Find the acceleration of $2.0 \mathrm{kg}$ block.$\left(\mathrm{g}=10 \mathrm{m}/{\mathrm{s}}^2\right)$.

 Since, ${\mu }_1<{\mu }_2$, acceleration of 2kg block down the plane will be more than the acceleration of 4kg block, if allowed to move separately. But, as the 2kg block is behind the 4kg block both of them will move with same acceleration say a. Taking both the blocks as a single system:

Force down the plane on the system $=(4+2)g\sin 30°$

$=\left(6\right)\left(10\right)\left(\frac{1}{2}\right)=30 \mathrm{N}$

Force up the plane on the system

$$\begin{gathered} ={\mu }_1\left(2\right)\left(g\right)\cos 30°+{\mu }_2\left(4\right)\left(g\right)\cos 30° \\ =\left(2{\mu }_1+4{\mu }_2\right)g\cos 30° \\ =(2\times 0.2+4\times 0.3)\left(10\right)(0.86) \\ =13.76 \mathrm{N} \end{gathered}$$

$\therefore$ Net force down the plane is $F=30-13.76=16.24 \mathrm{N}$

$\therefore$ Acceleration of both the blocks down the plane will be a.

$a=\frac{F}{4+2}=\frac{16.24}{6}=2.7 \mathrm{m}/{\mathrm{s}}^2$