Question

Coefficient of friction between two blocks shown in figure is $\mu =0.6$. The blocks are given velocities in the direction shown in figure. Find the common velocity of the two blocks.

• A. $4$
• B. $3$
• C. $5$
• D. $2$

Answer 1

The correct option is A $4$

Given : $m_1=2kg$ and $m_2=1kg$

Friction force $f$ acting on both the blocks is shown in the figure.

Friction force $f=\mu m_{1}g$

$\therefore$ $f=\left(0.6\right)\left(2\right)\left(10\right)=12N$

Thus acceleration of 2kg block $a_1=\frac{f}{m_1}=\frac{12}{2}=6m/s^2$ (towards left)

Also, acceleration of 1kg block $a_2=\frac{-f}{m_2}=\frac{-12}{1}=-12m/s^2$ (towards left)

Let the common velocity of the blocks be $v$.

For 2kg block : $v=-u_1+a_{1}t$

Or $v=-3+6t$ ....(1)

For 1kg block : $v=u_2+a_{2}t$

Or $v=18-12t$ ....(2)

From equation $\left(2\right)+2\times$ equation $\left(1\right)$, we get $v+2v=18-6$

Or $3v=12$

$⟹$ $v=4m/s$