If the horizontal force needed to move the top block is at least 12 N, we can calculate the static coefficient of friction with the formula below:
[tex]\begin{gathered} F_{friction}=F_{normal}\cdot\mu_s\\ \\ 12=m\cdot g\cdot\mu_s\\ \\ 12=4\cdot10\cdot\mu_s\\ \\ \mu_s=\frac{12}{40}\\ \\ \mu_s=0.3 \end{gathered}[/tex]
(a)
Let's draw the force diagram of the system:
Using the second law of Newton for the complete system (considering the two blocks as one big block), we have:
[tex]\begin{gathered} F=m\cdot a\\ \\ F_{applied}=(4+5)\cdot a\\ \\ a=\frac{F_{applied}}{9} \end{gathered}[/tex]
Now, for the top block, we have:
[tex]\begin{gathered} F=m\cdot a\\ \\ F_{applied}-F_{friction}=4\cdot a\\ \\ F_{applied}-12=\frac{4}{9}F_{applied}\\ \\ \frac{5}{9}F_{applied}=12\\ \\ F_{applied}=12\cdot\frac{9}{5}\\ \\ F_{applied}=21.6\text{ N} \end{gathered}[/tex]
(b)
The resulting acceleration will be:
