91Ó°ÊÓ

First, calculate the coefficient of static friction for the surface between the two blocks. When the force applied is at a maximum, the frictional force between the two blocks must also be a maximum. Using Newton's second law, we can solve the given problem.

The system consists of two blocks, one on top of the other. If we pull the bottom block too hard, the top block will slip on the bottom one. We're interested in the maximum force that can be applied such that the two will move together. The free-body diagrams for the two blocks are shown below.

Since F_t = 12N of force has to be applied to the top block for slipping to take place, using F_t = f_s,max

$\begin{array}{rcl}{F}_t& =& {\mathrm{μ}}_s{F}_{N,t}\\ F_t& =& {\mathrm{μ}}_s{m}_{t}g\\ {\mathrm{μ}}_s& =& \frac{F_t}{m_{t}g}\\ & =& \frac{12N}{\left(4.0kg\right)\left(9.8m/s^2\right)}\\ & =& 0.31\end{array}$

Using the same reasoning, for the two masses, the force would be,

$F={\mathrm{μ}}_s\left(m_t+m_b\right)g$

Substituting the value of${\mathrm{μ}}_s$, the maximum horizontal force has a magnitude that can be calculated as:

$F={\mathrm{μ}}_s\left(m_t+m_b\right)g$

Substitute the values, and we get,

$$\begin{gathered} F=0.31\left(4.0kg+5.0kg\right)\left(9.8m/s^2\right) \\ F=27N \end{gathered}$$

Thus, the maximum horizontal force is 27N.

The maximum acceleration is,

$\begin{array}{rcl}{a}_{max}& =& \frac{F}{m_t+m_b}\\ a_{max}& =& {\mathrm{μ}}_{s}g\\ a_{max}& =& 0.31×9.8m/s^2\\ & =& 3.0m/s^2\end{array}$

Thus, the resulting acceleration of blocks is 3.0m/s².