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!Pushing a Box^pushingbox.png|width=400!

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h3. Part A

A person pushes a box of mass 15 kg along a floor by applying a perfectly horizontal force _F_.  The box accelerates horizontally at a rate of 2.0 m/s{color:black}^2^{color}.  {excerpt}Assuming the coefficient of kinetic friction between the box and the ground is 0.45, what is the magnitude of _F_?{excerpt}

h4. Solution

{toggle-cloak:id=sysa} *System:*  {cloak:id=sysa}Box as [point particle].{cloak}

{toggle-cloak:id=inta} *Interactions:* {cloak:id=inta} External influences from the person (applied force) the earth (gravity) and the floor (normal force and friction).{cloak}

{toggle-cloak:id=moda} *Model:* {cloak:id=moda}[Point Particle Dynamics].{cloak}

{toggle-cloak:id=appa} *Approach:*  

{cloak:id=appa}

{toggle-cloak:id=diaga} {color:red} *Diagrammatic Representation* {color}

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The free body diagram for this situation is:

!pushingboxfrfbd1.png!

{cloak:diaga}

{toggle-cloak:id=matha} {color:red} *Mathematical Representation* {color}

{cloak:id=matha}

With this free body diagram, [Newton's 2nd Law|Newton's Second Law] can be written:

{latex}\begin{large}\[ \sum F_{x} = F - F_{f} = ma_{x} \]
\[ \sum F_{y} = N - mg = ma_{y} = 0 \]\end{large}{latex}

where we have assumed that the _y_ acceleration is zero because the box is sliding along a horizontal floor, not moving upward or downward.  This realization is important, because we know _F_~f~ = μ_N_.  Thus, because the _y_ acceleration is zero, we can solve Newton's 2nd Law in the _y_ direction to yield:

{latex}\begin{large}\[ N = mg\]\end{large}{latex}

so that:

{latex}\begin{large}\[ F = ma_{x}+F_{f} = ma_{x} + \mu_{k}N = ma_{x} + \mu_{k}mg = \mbox{96 N} \] \end{large}{latex}

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{card:label=Part B}

h3. Part B

A person pushes a box of mass 15 kg along a floor by applying a perfectly horizontal force _F_.  The box moves horizontally at a constant speed of 2.0 m/s in the direction of the person's applied force.  Assuming the coefficient of kinetic friction between the box and the ground is 0.45, what is the magnitude of _F_?

{toggle-cloak:id=sysb} *System, Interactions and Model:* {cloak:id=sysb} As in Part A.{cloak}

{toggle-cloak:id=appb} *Approach:*

{cloak:id=appb}

Just as above, [Newton's 2nd Law|Newton's Second Law] can be written:

{latex}\begin{large}\[ \sum F_{x} = F - F_{f}= ma_{x} \]
\[ \sum F_{y} = N - mg = 0\] \end{large}{latex}

This time, however, the _x_ acceleration is also zero, since the box maintains a constant speed.  This implies:

{latex}\begin{large}\[ F = ma_{x} + F_{f} = F_{f} = \mu_{k}mg = \mbox{66 N} \] \end{large}{latex}

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