Question

Find the acceleration of the two bodies if the system is shown, is at rest initially.
Take $g=10{\text{m/s}}^2$.

• A. $a_A=a_B=2.5{\text{m/s}}^2$
• B. $a_A=2.5{\text{m/s}}^2,a_B=1{\text{m/s}}^2$
• C. $a_A=5.1{\text{m/s}}^2,a_B=5{\text{m/s}}^2$
• D. $a_A=a_B=1{\text{m/s}}^2$

Answer 1

The correct option is C $a_A=5.1{\text{m/s}}^2,a_B=5{\text{m/s}}^2$

The FBDs of the blocks are as shown




We have,
$N_1=10g=100\text{N}$
$N_2=N_1+10g=100+100=200\text{N}$
$f_{\text{max}}=\mu N_1=0.5\times 10g=50\text{N}$

Let us assume that both the block move together
$\Rightarrow$ Common acceleration $a_c=\frac{101}{m_A+m_B}=\frac{101}{20}=5.05{\text{m/s}}^2$

From FBD of $B:$ Required frictional force $f$ to make them move together is
$f=m_B\times a_c=10\times 5.05=50.5N$
Since, $50.5N>50N$
$\Rightarrow f>f_{\text{max}}$, it means our assumption was wrong, they will slip over each other.

As they move separately, kinetic friction is involved.
So, again we have FBDs of the blocks as



we have, $f_k=\mu N_1=0.5\times 10g=50N$

$\therefore a_A=\frac{101-f_k}{m_A}=\frac{101-50}{10}=5.1{\text{m/s}}^2$
And, $a_B=\frac{f_k}{m_B}=\frac{50}{10}=5{\text{m/s}}^2$

Note: $a_A>a_B$ as force is applied on $A$.