Question

Figure shows two blocks in contact sliding down an inclined surface of inclination ${30}^0$. The friction coefficient between the block of mass $2.0kg$ and the incline is ${\mu }_1=0.20$ and that between the block of mass $4.0kg$ and the incline is ${\mu }_2=0.30$. Find the acceleration of $2.0kg$ block. $g=10m/s^2$

• A. $3.8m/s^2$
• B. $5.1m/s^2$
• C. $1.2m/s^2$
• D. $2.7m/s^2$

Answer 1

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

From the free body diagram

$R=4g\cos 30=4\times 10\times \sqrt{3}/2=20\sqrt{3}.........\left(i\right)$

${\mu }_{2}R+4a-P-4g\sin 30=0$

$0.3\left(40\right)\cos 30+4a-P-40\sin 20=\mathrm{0...............}\left(ii\right)$

$P+2a+{\mu }_1{R}_1-2g\sin 30=\mathrm{0............................}\left(iii\right)$

$R_1=2g\cos 30=2\times 10\times \sqrt{3}/2=10\sqrt{3}.......\left(iv\right)$

Equation (ii), $6\sqrt{3}+4a-P-20=0$

Equation (iv), $P+2a+2\sqrt{3}-10=0$

From equation (ii) & (iv)

$6\sqrt{3}+6a-30+2\sqrt{3}=0$

$6a=30-8\sqrt{3}=30-13.85=16.15$

$a=\frac{16.15}{6}=2.69\approx 2.7m/s^2$