Question

In an arrangement shown in the figure, the mass of $A=1\text{kg}$, the mass of $B=2\text{kg}$, and coefficient of friction between $A$ and $B$ is $0.2$. There is no friction between $B$ and the ground. The frictional force on $A$ is (Take $g=10{\text{m/s}}^2$)

(Assume there is no slipping between the blocks).

• A. $\frac{5}{3}\text{N}$
• B. $\frac{10}{3}\text{N}$
• C. $2\text{N}$
• D. $0\text{N}$

Answer 1

The correct option is A $\frac{5}{3}\text{N}$

As it is given that no slipping between the blocks, both the blocks move with common acceleration.

Common acceleration $\left(a\right)=\frac{F}{m_1+m_2}=\frac{5}{1+2}=\frac{5}{3}{\text{m/s}}^2$

By the FBD of $A:$

$\therefore f=m_{1}a=1\times \frac{5}{3}=\frac{5}{3}\text{N}$ as $(m_1=1\text{kg})$

$R_1=m_{1}g=10\text{N}$
$\therefore (f_s{)}_{\text{max}}={\mu }_s{R}_1=0.2\times 10=2\text{N}$
Since $f<(f_s{)}_{\text{max}}$
Frictional force acting on $A$ is static in nature and the value is $\frac{5}{3}\text{N}$

Hence option A is the correct answer