Question

A 4.80-kg block is at rest on a horizontal floor. If you push horizontally on the 4.80-kg block with a force of 12.0 N, it just starts to move.A 7.00-kg block is stacked on top of the 4.80-kg block. What is the magnitude F of the force, acting horizontally on the 4.80-kg block as before, that is required to make the two blocks start to move together?

Answer 1

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

$\begin{gathered} N-mg=0 \\ N=mg \end{gathered}$

Summation on x-axis

$\begin{gathered} F-F_f=0 \\ F=F_f \\ \end{gathered}$

We can see that the Friction is equal to the horizontal Force. Now, we can calculate the coefficient of friction

$\begin{gathered} F=\mu_sN \\ \mu_s=\frac{F}{N} \\ \mu_s=\frac{12N}{4.8\operatorname{kg}\cdot9.8} \\ \mu_s=0.255 \end{gathered}$

Second part:

We can now calculate the force required to move the two blocks together

$\begin{gathered} F=\mu_smg \\ F=0.255\cdot(4.8+7)\cdot9.8 \\ F=29.50\text{ N} \end{gathered}$