Q123 / 180 · KCET 2011 · Physics
Block $A$ of mass of $2 \mathrm{~kg}$ is placed over block $B$ of mass $8 \mathrm{~kg}$. The combination is placed over a rough horizontal surface. Coefficient of friction between $B$ and the floor is $0.5$. Coefficient of friction between blocks $A$ and $B$ is $0.4$. A horizontal force of $10 \mathrm{~N}$ is applied on
block $B$. The force of friction between blocks
$A$ and $B$ is $\left(g=10 \mathrm{~ms}^{-2}\right)$
block $B$. The force of friction between blocks $A$ and $B$ is $\left(g=10 \mathrm{~ms}^{-2}\right)$
- A$100 \mathrm{~N}$
- B$40 \mathrm{~N}$
- C$50 \mathrm{~N}$
- Correct answerDZero
Solution
Total mass of blocks $A$ and $B=2+8=10 \mathrm{~kg}$
Friction between surface and combination of $A$ and $B$
$$
\begin{aligned}
F &=\mu R \\
&=0.5 \times 10 \times 10=50 \mathrm{~N}
\end{aligned}
$$
Here applied force on box $B$ is $10 \mathrm{~N}$ that is less than $50 \mathrm{~N}$. So the system will be in rest because of this there is no friction between blocks $A$ and $B$.
Hence, the correct answer is option: 'D'.