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{composition-setup}{composition-setup}
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!pushing2boxes.png!
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{excerpt}A person pushes a box of mass 15 kg along a smooth floor by applying a perfectly horizontal force _F_. In the process, the 15 kg box pushes against another box with a mass of 10 kg and causes it to move.{excerpt} The boxes accelerate horizontally at a rate of 2.0 m/s{color:black}^2^{color}.
{card:label=Part A}
h3. Part A
What is the magnitude of _F_?
h4. Solution
{toggle-cloak:id=sysa} *System:* {cloak:id=sysa} Both boxes together as a single [point particle].
!pushing2boxessys1.png!
{cloak}
{toggle-cloak:id=inta} *Interactions:* {cloak:id=inta}External influences from the person (applied force) the earth (gravity) and the floor (normal force).{cloak}
{toggle-cloak:id=moda} *Model:* {cloak:id=moda}[Point Particle Dynamics].{cloak}
{toggle-cloak:id=appa} *Approach:*
{cloak:id=appa}
The word *smooth* in the problem statement is a keyword, telling us that the floor exerts no horizontal force on the box. Thus, the only external _horizontal_ force acting on the system composed of the two boxes together is that applied by the person. This implies that[Newton's 2nd Law|Newton's Second Law] for the _x_ direction can be written:
{latex}\begin{large}\[ \sum F_{x} = F = m_{\rm system}a_{x} = (\mbox{15 kg} + \mbox{10 kg})(\mbox{2 m/s}^{2}) = \mbox{50 N}\] \end{large}{latex}
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{card:label=Part B}
h3. Part B
What is the force applied on the front (10 kg) box by the rear (15 kg) box?
h4. Solution
{toggle-cloak:id=sysb} *System:* {cloak:id=sysb}This time, we will focus only on the front box.
!pushing2boxessys2.png!
{cloak}
{toggle-cloak:id=intb} *Interactions:* {cloak:id=intb}The front box is subject to external influences from the rear box, the earth (gravity) and the floor (normal force). {cloak}
{toggle-cloak:id=modb} *Model:* {cloak:id=sysb}[Point Particle Dynamics]{cloak}
{toggle-cloak:id=appb} *Approach:*
{cloak:id=appb}
For simplicity, we will refer to the front box as box 2 and the rear box as box 1. We will call the force applied by box 1 to box 2 _F_~21~. The only horizontal force experienced by the front box is the force _F_~21~ provided by the rear box. Thus, [Newton's 2nd Law|Newton's Second Law] for the _x_ direction can be written:
{latex}\begin{large}\[ \sum F_{x} = F_{21} = m_{2}a_{x} = (\mbox{10 kg})(\mbox{2 m/s}^{2}) = \mbox{20 N}\] \end{large}{latex}
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{card:label=Part C}
h3. Part C
Suppose we now consider the rear (15 kg) box as a
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