Question

There are two blocks $A$ and $B$ of masses $m_{A}=1kg$ and $m_{B}=3kg$ kept on the table as shown in figure. The coefficient of friction between $A$ and $B$ as well as between $B$ and the surface of the table is $0.2$ . The maximum force $F$ that can be applied on $B$ horizontally so that the block $A$ does not slide over the block $B$ is:
(Take $g=10ms^{- 2}$ )

• A. $12N$
• B. $8N$
• C. $16N$
• D. $40N$

Answer 1

Answer: (C)

For $1kg$ block,
$a_{max}=\frac{\mu \left(\right. 1 \left.\right) g}{\left(\right. 1 \left.\right)}=2ms^{- 2}$
So, $F_{m a x}-f_{g r o u n d}=m_{t o t a l}\times a_{max}$
$F_{max}-\mu \left(\right.3+1\left.\right)g=\left(\right.3+1\left.\right)2$
$F_{max}=16N$