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Deck of Cards
idbigdeck

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

2.

Card
labelPart A
Wiki Markup
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 {

?

Solution

Toggle Cloak

:

id

=

sysa

} *

System:
Cloak
idsysa

Box as .

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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

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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
*  {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}\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

_

Ff = μN. Thus, because the y acceleration is zero, we can solve Newton's 2nd Law in the y direction to yield:

Latex
_~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}

{cloak:matha}
{cloak:appa}
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.

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idappb
Approach:

Cloak
idappb