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Composition Setup
 Deck of Cards
idbigdeck
Image Added
 Wiki Markup
h4. Part A

!Pushing a Box^pushingbox.png|width=40%!

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}.  Assuming the coefficient of kinetic friction between the box and the ground is 0.45, what is the magnitude of _F_?

System:  Box as [point particle] subject to external influences from the person (applied force) the earth (gravity) and the floor (normal force and friction).

Model: [Point Particle Dynamics].

Approach:  The free body diagram for this situation is:

!pushingboxfrfbd1.png!

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

{latex}
 Card
labelPart A
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/s2. 
 Excerpt
Assuming the coefficient of kinetic friction between the box and the ground is 0.45, what is the magnitude of F?
Solution
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idsysa
 System: 
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Box as .
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idinta
 Interactions: 
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idinta
 External influences from the person (applied force) the earth (gravity) and the floor (normal force and friction).
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idmoda
 Model: 
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idmoda
.
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 Approach: 
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idappa
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iddiaga
  Diagrammatic Representation 
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The free body diagram for this situation is:
Image Added
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  Mathematical Representation 
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With this free body diagram, Newton's 2nd 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}
f = μN. 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}

h4. 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_?

System and Model:  As in Part A.

Approach:  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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labelPart B
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?
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 System, Interactions and Model: 
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 As in Part A.
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 Approach:
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