Physics - NLM Question with Solution | TestHub

First, consider the three blocks as a single system. The total mass is $M_{total}=m_1+m_2+m_3=2+3+5=10\text{ kg}$. Applying Newton's second law for the system, the acceleration$a$ is:$$a=\frac{F}{M_{total}}=\frac{40\text{ N}}{10\text{ kg}}=4{\text{ m/s}}^2$$

Now, consider the free-body diagram for block$m_3$. The only horizontal force acting on $m_3$ is the contact force $N_{23}$ exerted by $m_2$ on $m_3$. This force causes $m_3$ to accelerate.$$N_{23}=m_{3}a$$ Substituting the values:$m_3=5\text{ kg}$ and $a=4{\text{ m/s}}^2$.$$N_{23}=5\text{ kg}\times 4{\text{ m/s}}^2=20\text{ N}$$