Question

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

• A. 5.7 ,1.2
• B. 2..3 ,2.4
• C. 2.7 ,1.2
• D. None of these

Answer 1

The correct option is D

None of these

Free Body Diagram (Assuming no contact)

Now for A and B to be in contact with each other, acceleration of B in the absence of the contact force should be greater than that of A.

As they are in motion,

$f_2={\mu }_2{N}_2{f}_2={\mu }_1{N}_1$

$={\mu }_{2}Magcos\theta ={\mu }_1{M}_{B}gcos\theta$

$0.3\times 4\times 10\times \frac{\sqrt{3}}{2}0.2\times 2\times 10\times \frac{\sqrt{3}}{2}$

$=6\sqrt{3}N=2\sqrt{3}N$

In the absence of contact ,

$M_A{a}_A=m_{A}gsin\theta -f_2$

$\Rightarrow a_A=\frac{M_{A}gsin\theta -f_2}{M_A}$

=$\frac{4\times 10\times \frac{1}{2}-6\sqrt{3}}{4}$

=5-$\frac{3}{2}\sqrt{3}m/s^2$

$M_B{a}_B=m_{B}gsin\theta -f_1$

$\Rightarrow a_B=\frac{M_{B}gsin\theta -f_1}{M_B}$

=$\frac{2\times 10\times \frac{1}{2}-2\sqrt{3}}{2}$

=5 - $\sqrt{3}m/s^2$

$\therefore a_B$ > $a_A$

This means B and A will be in contact and move with same acceleration.

$\therefore M_{A}gsin\theta +C-f_2=M_{A}a$

$M_{B}gsin\theta +C-f_1=M_{B}a$

$M_{A}gsin\theta +M_{B}gsin\theta -f_1-f_2=(m_B+m_A)a$

$\Rightarrow a=\frac{gsin\theta (m_A+m_B)-f_1-f_2}{(m_A+m_B)}$

=$\frac{10\times \frac{1}{2}(2+4)-2\sqrt{3}-6\sqrt{3}}{2+4}$

=$\frac{5\left(6\right)-8\sqrt{3}}{6}=\frac{30-8\sqrt{3}}{6}=\frac{15-4\sqrt{3}}{3}$

=5-$\frac{4}{3}\sqrt{3}=2.69\approx 2.7m/s^2$

As they will move with the same acceleration, ${\theta }_B$ = 2.7 $m/s^2$

Case III when ${\mu }_1=0.3,{\mu }_2=0.2$

In the absence of contact

$f_2={\mu }_2{m}_{A}gcos\theta f_1={\mu }_1{m}_{A}gcos\theta$

$=0.2\times 4\times 10\times \frac{\sqrt{3}}{2}=0.3\times 2\times 10\times \frac{\sqrt{3}}{2}$

$=4\sqrt{3}N=3\sqrt{3}N$

$a_A=\frac{M_{A}gsin\theta -f_2}{M_A}{a}_B=\frac{M_{B}gsin\theta -f_2}{M_A}$

$\frac{4\times 10\times \frac{1}{2}-4\sqrt{3}}{4}\frac{2\times 10\times \frac{1}{2}-3\sqrt{3}}{2}$

$=5-\sqrt{3}m/s^2=\frac{10-3\sqrt{3}}{2}$

$a_A=3.27m/s^2{a}_B=2.4m/s^2$

As $a_B$ is less than $a_A$ , A will move faster.

$\therefore$There will be no contact.