Question

Find the acceleration of the two blocks A and B respectively, assuming them to be at rest initially. Take $g=10{\text{m/s}}^2$.

• A. $4{\text{m/s}}^2,2{\text{m/s}}^2$
• B. $2{\text{m/s}}^2,4{\text{m/s}}^2$
• C. $2{\text{m/s}}^2,2{\text{m/s}}^2$
• D. $4{\text{m/s}}^2$, zero

Answer 1

The correct option is C $2{\text{m/s}}^2,2{\text{m/s}}^2$

The FBDs of the blocks are as shown
Given, the force acting on the block $A$ is $F=180\text{N}$


We have from the FBDs
$N_1=20g=200\text{N}$
$f_{\text{1 max}}={\mu }_1{N}_1=\left(0.8\right)\left(20g\right)=0.8\times 200=160\text{N}$
$N_2=N_1+10g=200+100=300\text{N}$
$f_{\text{2 max}}={\mu }_2{N}_2=\left(0.4\right)\left(300\right)=120\text{N}$

Since, $F>f_{\text{1 max}}$ and $f_{\text{1 max}}>f_{\text{2 max}}$, both the blocks will accelerate.
Now, let us assume both the blocks will move together, then
Common acceleration is given by
$a_c=\frac{180-f_{\text{2 max}}}{m_A+m_B}=\frac{180-120}{30}=2{\text{m/s}}^2$

Again for block $B$:
$f_1-f_{\text{2 max}}=m_B{a}_c\Rightarrow f_1=10\times 2+120=140\text{N}$

Since, $f_1<f_{\text{1 max}}$, our assumption is right
Hence, the friction will be static in nature & the two blocks will move together with common acceleration of $2{\text{m/s}}^2$