Suppose we now consider the rear (15 kg) box as an independent system. How can we reconcile the fact that the force applied by the person to this box is 50 N, but the 15 kg box only accelerates at a rate of 2 m/s2?
The rear box is subject to external influences from the person, the rear box, the earth (gravity) and the floor (normal force).
If the 50 N force applied by the person were the only force acting on the rear box, then the acceleration of the box would be 50 N / 15 kg = 3.3 m/s2, but we know that the box is accelerating at 2 m/s2. To reconcile this apparent contradiction, we must remember that the rear box is not only being pushed on by the person, but also by the front box. If we again label the rear box as box 1 and the front box as box 2, we can call the force from the front box on the rear box F12. It is clear from the relative positions of the boxes that the force from the front box on the rear box would point to the left in the picture above. Thus, we can write the x-component of Newton's 2nd Law for the rear box in the form:
Unknown macro: {latex}
\begin
Unknown macro: {large}
[ F - F_
Unknown macro: {12}
= m_
Unknown macro: {1}
a ]\end
We can solve this for F12 to give:
Unknown macro: {latex}
\begin
Unknown macro: {large}
[ F_
Unknown macro: {12}
= F - m_
Unknown macro: {1}
a = \mbox
Unknown macro: {50 N}
- (\mbox
Unknown macro: {15 kg}
)(\mbox
Unknown macro: {2 m/s}
^
Unknown macro: {2}
) = 20 N ]\end