Given
Initial situation
m1: mass one
m1 = 4.8 kg
F: horizontal force
F = 12 N
Second situation
m2: mass two
m2 = 7kg
Procedure
Let's start with the free body diagram
Summation on y-axis
[tex]\begin{gathered} N-mg=0 \\ N=mg \end{gathered}[/tex]Summation on x-axis
[tex]\begin{gathered} F-F_f=0 \\ F=F_f \\ \end{gathered}[/tex]We can see that the Friction is equal to the horizontal Force. Now, we can calculate the coefficient of friction
[tex]\begin{gathered} F=\mu_sN \\ \mu_s=\frac{F}{N} \\ \mu_s=\frac{12N}{4.8\operatorname{kg}\cdot9.8} \\ \mu_s=0.255 \end{gathered}[/tex]Second part:
We can now calculate the force required to move the two blocks together
[tex]\begin{gathered} F=\mu_smg \\ F=0.255\cdot(4.8+7)\cdot9.8 \\ F=29.50\text{ N} \end{gathered}[/tex]
