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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!
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{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=modb}[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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