Answer:
[tex]R = 20\,N[/tex]
Explanation:
The free body diagrams of each box are presented in the images attached below. Given the smoothness of the level floor, friction forces can be neglected and due to the contact (3rd Newton's Law) betweeen each other, both boxes experiment the same acceleration:
Box A
[tex]\Sigma F_{x} = F - R = m_{A}\cdot a[/tex]
[tex]\Sigma F_{y} = N_{A} - m_{A}\cdot g = 0[/tex]
Box B
[tex]\Sigma F_{x} = R = m_{B}\cdot a[/tex]
[tex]\Sigma F_{y} = N_{B} - m_{B}\cdot g = 0[/tex]
After some algebraic manipulation:
[tex]F = (m_{A} + m_{B})\cdot a[/tex]
The acceleration experiment for each box is:
[tex]a = \frac{F}{m_{A}+m_{B}}[/tex]
[tex]a = \frac{32\,N}{3\,kg + 5\,kg}[/tex]
[tex]a = 4\,\frac{m}{s^{2}}[/tex]
The contact between the two boxes is:
[tex]R = (5\,kg)\cdot (4\,\frac{m}{s^{2}})[/tex]
[tex]R = 20\,N[/tex]
