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{excerpt}An interaction which produces a change in the energy of a system.{excerpt}

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h2. Motivation for Concept

It requires effort to alter the energy of an object, as can clearly be seen when attempting to impart [kinetic energy] by pushing a car which has stalled or to impart [gravitational potential energy|gravity] by lifting a heavy barbell.  We would like to quantify what we mean by "effort".  It is clear that force alone is not enough to impart energy.  Suppose that the car or the barbell is just too heavy to move.  Then, for all the pushing or pulling that is done (a considerable _force_), no _energy_ is imparted.  For the energy of a system to change, the system must alter its position or its configuration.  In effect, the force must impart or reduce motion of the object to which it is applied.  

h2. Mathematical Definition

h4.  Work-Kinetic Energy Theorem as Postulate

Suppose that we postulate the Work-Kinetic Energy Theorem for a [point particle] as the _defining_ relationship of work.  Doing so will allow us to find a mathematical definition of work in terms of [force].

{include:Work-Kinetic Energy Theorem}

h4. Definition of Work

By comparing the derivation of the theorem to its statement, we see that in order for the theorem to be satisfied, we must make the definition:

{latex}\begin{large}\[ W_{net} = \int_{path} \vec{F}_{net}\cdot d\vec{r}\]\end{large}{latex}

which leads us to define the work done by an individual force as:

{latex}\begin{large}\[ W = \int_{path}\vec{F}\cdot d\vec{r}\]\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}\[ \int_{x_{i}}^{x_{f}} (-\mu_{k} mg)\;dx = -\mu_{k}mg(x_{f}-x_{i})\]\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 the position halfway to the final endpoint (the halfway point is _x_~m~ = ({_}x_~i{i~}+{_}x_~f{f~})/2), 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