Question

Three blocks $m_1,m_2$ and $m_3$ of masses $8\text{kg},3\text{kg}$ and $1\text{kg}$ are placed in contact on a smooth surface. Forces $F_1=140\text{N}$ and $F_2=20\text{N}$ are acting on blocks $m_1$ and $m_3$ respectively as shown. The reaction between blocks $m_2$ and $m_3$ is

• A. $2.5\text{N}$
• B. $7.5\text{N}$
• C. $2.5\text{N}$
• D. $30\text{N}$

Answer 1

The correct option is D $30\text{N}$

We know that, the acceleration of the system will be $a_{\text{sys}}=\frac{\sum F}{\text{total mass}}$,

$a=\frac{F_1-F_2}{m_1+m_2+m_3}=10{\text{ms}}^{-2}$
and the whole system moves towards right.
Now for the reaction between $m_2$ and $m_3$, by the free body diagram


By the newton's law equation, here $R$ is the reaction force on the block $m_3$ by $m_2$

$R-F_2=m_{3}a\Rightarrow R=20+1\times 10$
$R=30\text{N}$

Hence option (d) is correct.