Question

Figure shows two blocks in contact and sliding down the inclined surface of inclination ${30}^{\circ }$. Coefficient of friction between the block of mass $4\text{kg}$ and the inclined plane is ${\mu }_1=0.3$ and that between the block of mass $2\text{kg}$ and the inclined plane is ${\mu }_2=0.20$. Assume coefficients of static and kinetic friction to be equal. Find the acceleration of $2\text{kg}$ block. Take $g=10{\text{m/s}}^2$.

• A. $\frac{15-4\sqrt{3}}{3}{\text{m/s}}^2$
• B. $5-\sqrt{3}{\text{m/s}}^2$
• C. $\sqrt{15}{\text{m/s}}^2$
• D. $5{\text{m/s}}^2$

Answer 1

The correct option is A $\frac{15-4\sqrt{3}}{3}{\text{m/s}}^2$

FBD of the system:


$m_1=4\text{kg}$, $m_2=2\text{kg}$
Since there is relative motion between the blocks and incline, kinetic friction will act.
$f_1={\mu }_{1}N1={\mu }_1{m}_{1}g\cos \theta$
$f_2={\mu }_{2}N2={\mu }_2{m}_{2}g\cos \theta$

Since, ${\mu }_2<{\mu }_1$, acceleration of $2\text{kg}$ block down the plane will be more than the acceleration of $4\text{kg}$
$\Rightarrow$ Block $2\text{kg}$ will always remain in contact with block $4\text{kg}$.

Applying the equation of dynamics down the inclined plane for system of ($2\text{kg}+4\text{kg}$):
$F_{\text{net ext}}=(2+4)a_{net}$
$\Rightarrow (m_1+m_2)g\sin \theta -f_1-f_2=6a_{net}$
$\Rightarrow 6g\sin \theta -{\mu }_1{m}_{1}g\cos \theta -{\mu }_2{m}_{2}g\cos \theta =6a_{net}$
$\Rightarrow 6a_{net}=6\times 10\times \frac{1}{2}-(0.3\times 4\times 10\frac{\sqrt{3}}{2})-(0.2\times 2\times 10\times \frac{\sqrt{3}}{2})$
$\Rightarrow a_{net}=\frac{30-8\sqrt{3}}{6}$
$\therefore a_{net}=\frac{15-4\sqrt{3}}{3}{\text{m/s}}^2$