Question

Figure shows two blocks in contact sliding down an inclined surface of inclination ${30}^{\circ }$. The friction coefficient between the block of mass $2.0kg$ and the incline is ${\mu }_2$, and that between the block of mass $4.0kg$ and the incline is ${\mu }_1$. Calculate the acceleration of the $2.0kg$ block if ${\mu }_1=0.30$ and ${\mu }_2=0.20$. Take $g=10m/s^2$.

• A. $2m/s^2$
• B. $3.2m/s^2$
• C. $4m/s^2$
• D. $2.7m/s^2$

Answer 1

The correct option is D $2.7m/s^2$


${\mu }_1>{\mu }_2\Rightarrow a_2>a_1$
i.e both blocks will move together with same acceleration

$f_1=0.3\times 4\times 10\cos {30}^{\circ }=6\sqrt{3}N$
$f_2=0.2\times 2\times 10\cos {30}^{\circ }=2\sqrt{3}N$

$F_{net}=Ma$ where $M$ is total mass
$Mg\sin \theta -f_1-f_2=6a$
$\Rightarrow 6g\sin {30}^{\circ }-6\sqrt{3}-2\sqrt{3}=6a$
$\Rightarrow a=2.7m{s}^{-2}$