Versions Compared

Old Version 6

changes.mady.by.user Andrew E Pawl

Saved on

compared with

New Version Current

changes.mady.by.user Andrew E Pawl

Saved on

Key

  • This line was added.
  • This line was removed.
  • Formatting was changed.
Comment: Migration of unmigrated content due to installation of a new plugin
Composition Setup
 Deck of Cards
idbigdeck
Image Added
 Wiki Markup
{composition-setup}{composition-setup}
{table:border=1|frame=void|rules=cols|cellpadding=8|cellspacing=0}
{tr:valign=top}
{td:width=350|bgcolor=#F2F2F2}
{live-template:Left Column}
{td}
{td}

{deck:id=bigdeck}

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

{card:label=Part A}

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}

{cloak:id=diaga}

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}
 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
 Toggle Cloak
idsysa
 System: 
 Cloak
idsysa
Box as .
 Toggle Cloak
idinta
 Interactions: 
 Cloak
idinta
 External influences from the person (applied force) the earth (gravity) and the floor (normal force and friction).
 Toggle Cloak
idmoda
 Model: 
 Cloak
idmoda
.
 Toggle Cloak
idappa
 Approach: 
 Cloak
idappa
 Toggle Cloak
iddiaga
  Diagrammatic Representation 
 Cloak
iddiaga
The free body diagram for this situation is:
Image Added
 Cloak
diaga
diaga
 Toggle Cloak
idmatha
  Mathematical Representation 
 Cloak
idmatha
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}

{cloak:matha}
{cloak:appa}
{card}
{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}

{cloak}
{card}
{deck}
{td}
{tr}
{table}
{live-template:RELATE license}


 Cloak
matha
matha
 Cloak
appa
appa
 Card
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?
 Toggle Cloak
idsysb
 System, Interactions and Model: 
 Cloak
idsysb
 As in Part A.
 Toggle Cloak
idappb
 Approach:
 Cloak
idappb