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}\begin{large}\[ W = \int_{\rm path}\vec{F}\cdot d\vec{r}\]\end{large}{latex}

h2. Importance of Path

h4. Conservative Forces treated as Potential Energy

The form of our definition of work involves a path integral.  For some forces, however, the value of the path integral is determined solely by its endpoints.  These forces are, by definition, [conservative forces].  This path-independence is the property which allows us to define a [potential energy] to associate with the force.  Thus, the work done by conservative forces will usually be ignored, since their interaction is instead expressed as a contribution to the [mechanical energy] of the system.  The two commonly considered conservative forces in introductory mechanics are:

* [gravity]
* spring forces

h4. Non-Conservative Forces

Common forces in introductory physics whose effects are usually path _dependent_ are:

* [contact force] 
* [normal force]
* [friction] 
* [tension]

For these forces, it is not possible to define a useful potential energy, and so the path of the system must be understood in order to compute the work when energy is used to describe a system subject to these interactions.

h4. Example of Friction

To see the dependence of non-conservative work on path, consider a box of mass _m_ moving along a rough, level surface.  Suppose the box is subject only to horizontal applied forces, gravity, normal force and kinetic friction as it moves.  Suppose the box is moved in one dimension from position _x_~i~ directly to position _x_~f~ as shown in the figure below.

PICTURE


If we assume that the coefficient of kinetic friction is a constant μ~k~, we can find an expression for the work done by friction in the course of this movement.  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) and the only vertical forces are gravity and the normal force, 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 friction has the constant magnitude:

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

Further, since friction is _always_ 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 seems to depend only upon the endpoints.  Consider, however, the alternate path between the *same endpoints* _x_~i~ and _x_~f~ shown here:

PICTURE

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~.  Again, 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 stages which consist of motion in one _direction_ only.  Our chosen stages are labeled 1 through 3 in the figure below.  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}

PICTURE

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