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{excerpt:hidden=true}Explore the reason that kinetic friction usually produces negative work.{excerpt}

Consider a box of mass _m_ moving along a rough, level surface.  The box is subject only to horizontal applied forces, gravity, normal force and kinetic friction as it moves.  

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h3. Part A

First, suppose the box is moved in one dimension from position _x_~i~ directly to position _x_~f~ as shown in the figure below.

!work1stage.png!


Assuming that the coefficient of kinetic friction is a constant μ~k~, find an expression for the work done by friction in the course of this movement.  

h4. Solution

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

{toggle-cloak:id=inta} *Interactions:* {cloak:id=inta}The box experiences [external interactions|external force] with the earth ([gravity|gravity (near-earth)]), the surface ([normal force] and [kinetic friction]) and whatever is pushing it ([applied force]).{cloak:inta}

{toggle-cloak:id=moda} *Model:* {cloak:id=moda}[Point Particle Dynamics] plus the definition of [work].{cloak:moda}

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

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

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A [free body diagram] for the box will look like:

!workfbd.png!

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

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The friction force will be:

{latex}\begin{large}\[ F_{f} = \mu_{k}N\]\end{large}{latex}

where _N_ is the normal force on the box from the floor.  However, since the box is only moving horizontally (meaning _a_~y~ is zero) the [free body diagram] implies we can write [Newton's 2nd Law|Newton's Second Law] for the _y_ direction as:

{latex}\begin{large}\[ N-mg = ma_{y} = 0\]\end{large}{latex}

This tells us that the normal force is equal in magnitude to the box's [weight] and so the [force] of [kinetic friction] has the constant magnitude:

{latex}\begin{large}\[ F_{f} =\mu_{k}mg\]\end{large}{latex}

Further, since [kinetic friction] is _always_ directed opposite to the motion, and the motion is always in the + _x_ direction, we can write:

{latex}\begin{large}\[ \vec{F}_{f} = -\mu_{k}mg\hat{x}\]\end{large}{latex}

and:

{latex}\begin{large}\[ d\vec{r} = dx\hat{x}\]\end{large}{latex}

Thus, our path integral is reduced to a one-dimensional integral of the form:

{latex}\begin{large}\[ W = \int_{x_{i}}^{x_{f}} (-\mu_{k} mg)\;dx = -\mu_{k}mg(x_{f}-x_{i})\]\end{large}{latex}

which depends only upon the endpoints...

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{card:Part A}
{card:label=Part B}

h3. Part B

Now consider the alternate path between the _same endpoints_ _x_~i~ and _x_~f~ shown here:

!work3stage.png!

What is the [work] done by [kinetic friction] on the box in the course of _this_ movement?

h4. Solution

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

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

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

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|!workfbd.png!|!workfbd2.png|width=225px!|!workfbd.png!|
||Stage 1 FBD||Stage 2 FBD||Stage 3 FBD||

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

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This path has the box first making the trip from the position _x_~i~ to a position _x_~m~ between the initial and final positions of the box, then returning to the initial position _x_~i~ and then completing the trip to the final position _x_~f~.  Just as in Part A, the box is assumed to be subject to purely horizontal applied forces so that the friction force has a constant magnitude of:

{latex}\begin{large}\[F_{f} = \mu_{k}mg\]\end{large}{latex}

To evaluate the path integral for the work, we must break the path up into three parts which consist of motion in one _direction_ only.  The parts coincide with the stages labeled 1 through 3 in the figure above.  In each stage, we must determine the _vector_ form of the friction force.  

{note}The same splitting of the path is needed for the computation of the [distance] traveled in kinematics.{note}

{latex}\begin{large}\[ \mbox{Stage 1:   }\vec{F}_{f} = - \mu_{k}mg\hat{x}\]
\[ \mbox{Stage 2:   }\vec{F}_{f} = + \mu_{k}mg\hat{x}\]
\[ \mbox{Stage 3:   }\vec{F}_{f} = - \mu_{k}mg\hat{x}\]\end{large}{latex}

{note}It is important to ensure that the friction force vector always points in the direction opposite the motion. {note}

We can now write the path integral for the work, using the fact that all motion is in the _x_ direction:

{latex}\begin{large}\[  W = \int_{x_{i}}^{x_{m}} (-\mu_{k} mg)\;dx + \int_{x_{m}}^{x_{i}} \mu_{k}mg\;dx + \int_{x_{i}}^{x_{f}}(-\mu_{k}mg)\;dx\]\end{large}{latex}

{warning}Note that although the box is moving in the --{_}x_ direction in the middle part of the path integral, the differential _dx_ remains positive.  The sign of the motion is encoded in the endpoints of the integral.  The lower limit of the integration is a larger _x_ value than the upper limit, implying that the box is moving in the --{_}x_ direction.{warning}

{info}For a conservative force such as gravity, there would be no sign flip in the middle term, so that the sum of the first two integrals would be zero, and the integration would simply be from _x_~i~ to _x_~f~, giving dependence only on the endpoints of the path.{info}

The result of the integrations is:

{latex}\begin{large}\[ W = -\mu_{k}mg (x_{m}-x_{i} + x_{m} - x_{i} + x_{f} - x_{i}) = -\mu_{k}mg(2(x_{m}-x_{i}) + (x_{f}-x_{i})) = -\mu_{k}mgd \]\end{large}{latex}

where _d_ is the total *distance* (_not_ displacement) traveled by the box.  Since distance depends on the path (it is not a function of the endpoints only) we see that the exact path traveled by the box is important.

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{card:Part B}
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