Figure shows two blocks in contact sliding down an inclined surface of inclination $30^{\circ}$. The friction coefficient between the block of mass $2.0 \mathrm{~kg}$ and theincline is $p$ and that between the block of mass $4.0 \mathrm{~kg}$ and the incline is $\mu$. Calculate the acceleration of the $2.0 \mathrm{~kg}$ block if $(\mathrm{a}) \mu 1=0.20$ and $\mu 2=0.30$, (b) $\mu 1=0.30$ and $\mu 2=0.20$. Take $g=10^{m} / \mathrm{s}^{2}$.
$P$ is the contact force.
$N_{1}=4 g \cos 30$
$N_{1}=4 \times 10 \times \frac{\sqrt{3}}{2}=20 \sqrt{3} \mathrm{~N}$
for $4 \mathrm{Kg}$ block
$P+4 g \sin 30-\mu_{2} N_{2}=m a$
$P=4 a+6 \sqrt{3}-20 \quad-(i)$
For $2 \mathrm{Kg}$ block
$N_{1}=2 g \cos 30=10 \sqrt{3}$
$P=2 a+10+2 \sqrt{3} \quad-(i i)$
On solving (i) and (ii)
$a=2.7 m / s^{2}$
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