Answer
The force on the Wheaties box from the Cheerios box is $~~8.5~N$
Work Step by Step
We can find the acceleration of the system of two boxes:
$F-(2.0~N)-(4.0~N) = (1.0~kg+3.0~kg)~a$
$a = \frac{F-(2.0~N)-(4.0~N)}{1.0~kg+3.0~kg}$
$a = \frac{(12~N)-(2.0~N)-(4.0~N)}{1.0~kg+3.0~kg}$
$a = 1.5~m/s^2$
We can consider the forces on the Wheaties box to find $F_C$, the force on the Wheaties box from the Cheerios box:
$F_C-4.0~N = m_w~a$
$F_C = 4.0~N + m_w~a$
$F_C = 4.0~N + (3.0~kg)(1.5~m/s^2)$
$F_C = 8.5~N$
The force on the Wheaties box from the Cheerios box is $~~8.5~N$
