Question

In the figure below, masses $m_A=2kg$ and $m_B=4kg$. For what minimum value of $F$, block A starts slipping over B $(g=10m/s^2)$

• A. $24N$
• B. $36N$
• C. $12N$
• D. $20N$

Answer 1

The correct option is B $36N$

FBD of block A and B,
Maximum frictional force between A and B,
$f_1={\mu }_1{m}_{A}g=(0.2\times 2\times 10)N$
$f_1=4N$
Acceleration, $a=\frac{f_1}{m_A}=\frac{4}{2}=2m/s^2$
Now, taking (A+B) as the system,
$(f_2{)}_{max}={\mu }_2(m_A+m_B)g=24N$
$F-24=(m_A+m_B)a=6\times 2=12$
$\therefore F=36N$

The correct option is (B)