Question

The situation shown in the figure, a block $A$ of mass $2kg$ is placed inside a box $B$ of mass $2kg$. The coefficient of static and kinetic friction between the blocks and also between box and ground are ${\mu }_s=0.6$ and ${\mu }_k=0.4$. If a force $F$ is applied on the box $B$, the block $A$ begins to slide relative to the box. The value of $F$ (in newton) is____.

Answer 1

Answer: 40

The maximum static friction on block $A$ is
$f_{1\max }={\mu }_s{m}_{1}g=0.6\times 2\times 10=12N$
The maximum static friction between ground and the box $B$
$f_{2\max }={\mu }_s\left(m_1+m_2\right)g$
$=0.6\times 4\times 10=24N$
Assume both blocks move together.
$F-{\mu }_k\left(m_1+m_2\right)g=\left(m_1+m_2\right)a$
Or $F-0.4\times 4\times 10=4a$
$\because F-16=4a$
For block $A,f_1=2a=2\left(\frac{F-16}{4}\right)$
as $f_1=\frac{F-16}{2}$
For just sliding, $f_1=f_{1\max }$
$\Rightarrow \frac{F-16}{2}=12$
$\therefore F=40N$