Work in physics is defined as a path integral as seen below these paragraphs. If you are not familiar with path integrals, that’s ok. We will discuss the mathematical part in this chapter. In words, what’s important is that work is a measure of how much energy a force adds to or takes away from a system, and it depends on serveral things: The force itself, how far the object moves while being pushed, and the relative orientation of the force and motion. [br][br]What that last part means is that forces pushing in the direction something is moving tend to “help” it along and add to its energy, which I’m sure your instinct affirms. On the other hand, when the force is opposite the direction of motion the force “hinders” the motion and takes away from the object’s energy. When the force is perpendicular to the motion, it causes only the path to turn, but has no effect on the energy. All of that is neatly contained in the equation below. This is one example of why math is a much better language than English (or any other spoken language) to do physics![br][br][center][math]W=\int{\vec{F }\cdot\vec{ds}}.[/math][/center][br]Work is a means of changing the kinetic energy of a system.  It is therefore measured in units of energy.  The standard SI unit for energy is the Joule, where [math]1J\equiv 1kg\frac{m^2}{s^2}.[/math][br][br]When someone or something does positive work, as in the case of a person lifting a box, for instance, the person loses energy and the box gains it.  We are often unconcerned about the origin of that energy (at least for now), but will have occasion in the future to discuss this in more depth.  In fact, it is related to chemical energy, which is really itself related to rest energy.  One thing that should be noted about this exchange of energy is that the person's loss equals the energy gained by the box system plus any other form of energy that is created in the process, such as heat.  [br][br]The fact that energy changes forms and gets transferred between systems, but ultimately remains unchanged in total, is called [b]conservation of energy[/b], the details of which will be discussed in the next chapter.[br][br]In the context of the present chapter where we confine our discussions to single-mass systems which only contain rest energy and kinetic energy, there is a simple principle that relates work and energy. This principle is called the [b]work-kinetic energy principle[/b]. Mathematically it states[br][br][center][math]\sum W=\Delta K.[/math][/center][br]This means that if forces manage to do work on an object, the result is that the object's kinetic energy changes. If multiple forces do work as when a person lifts a box and gravity simultaneously is doing work, we must add both the work done by the person and the work done by gravity to get the total work done. This is a powerful principle, but requires that we know how to correctly calculate work done by forces, which we will focus on now.[br][br]Observe in the work expression that [math]\vec{F}[/math] is a vector field in the most general context.  This means that [math]\vec{F}[/math] might depend on location, time, velocity etc.  If it does depend on such variables, once the values of the variables are provided, a vector will be returned by the function for each set of variables which denote position, time, velocity, etc. We will only use vector fields that depend on position in this course.[br]  [br]As such, the force will often change both in magnitude and in direction as an object on which it does work traverses a path [math]s[/math]. The path is assumed to be divided into infinitesimal segments denoted [math]\vec{ds}.[/math]  [br][br]As an example of such a vector field, consider [math]\vec{F}(x,y)=x\hat{i}+y^2\hat{j}.[/math]  Given a pair of values (x and y), you can obtain a vector.  But since the function is able to be evaluated at any value of x or y, and thus the function could return a multitude of vectors, we call this object a [b]vector field[/b].  This particular vector field is plotted below in blue, along with an orange path through the field along the function [math]y=x.[/math]  The little orange arrows, each of which is a [math]\vec{ds}[/math] vector, together make the path [math]s.[/math][br]

The path vector when we do these path integrals is [math]\vec{ds} = dx\hat{i}+dy\hat{j}+dz\hat{k},[/math] when written in rectangular coordinates.  It simply means that we can move in a combination of three ways - along the [i]x[/i] direction, the [i]y[/i] direction or the [i]z [/i]direction as we take a path through the field. This vector expression replaces the standard dx found at the end of most integrals with which you're familiar.[br][br]This path vector would look different in cylindrical or spherical coordinates, but for now we will stick with rectangular coordinates.  This path is allowed to curve in any arbitrary way through space, and the specifics will sometimes need to be defined before the integral can be evaluated.  We will get to that part shortly.  [br][br]While you have probably not come across such integrals in your math classes unless you are ahead of schedule and are near the end of your third semester of calculus, I will nonetheless teach this to you.  I wish to do this not to burden you, but to enforce better understanding of this concept of work.  My goal is that you start to see the meaning of path integrals when you see them written - that you come to see a concept when you look at these path integrals rather than just an equation.[br]  [br]Let's consider an example to show how we are to evaluate a path integral.  Just before we do so, let me remind you about the dot product.  When we have two vectors [math]\vec{A}[/math] and [math]\vec{B},[/math] the following are both true:[br][br]1. [math]\vec{A}\cdot\vec{B} = A_xB_x+A_yB_y+A_zB_z[/math][br]2. [math]\vec{A}\cdot\vec{B} = AB\cos\phi\[/math] where [math]\phi[/math] is the angle between the directions the vectors point.[br][br]In the following example, we will use the first of these relations.  Just realize that the vector [math]\vec{A}[/math] will be represented by the force and the second vector [math]\vec{B}[/math] will be the displacement path vector. [br][color=#1e84cc][br]EXAMPLE: Find the work done by the vector field given above ([math]\vec{F}(x,y)=x\hat{i}+y^2\hat{j}[/math]) along the path y=x from the origin to (x,y)=(2,2). To do this we use the first form of the dot product given above and get [br][br][center][math]W=\int \vec{F}\cdot\vec{ds} \\[br]W=\int (x\hat{i}+y^2\hat{j})\cdot(dx\hat{i}+dy\hat{j}) \\[br]W=\int x\;dx + \int y^2\;dy \\[br]\text{Since x=y and dx=dy, } \\[br]W=\int (x+x^2)dx \\[br]W=\frac{x^2}{2} + \frac{x^3}{3}|_0^2 \\[br]W=4.67 J[br][/math][/center]It would have been fine to convert all terms to y instead of x or to evaluate them separately with appropriate limits on y. What you cannot do is integrate y while an 'x' still sits in the expression or vice versa.[br][br]EXAMPLE: Let's consider the work done by gravity on a mass that is allowed to fall vertically from a higher height [math]h_i[/math] to a lower height [math]h_f[/math]. Realize that this work is done by gravity on its own.  All I mean by this is that things have a tendency to fall without help.  To rise, they need help to fight against gravity.[br][br]To solve this problem, I will use the upward, vertical direction as the positive y direction.  Let's assume that the path along which the object falls is just vertically downward, in the negative y direction.  The force of gravity and the path vector are shown below. [b]Notice that I did not make the dy term negative in the path vector[/b].  Rather, the limits on the integration will decide whether I'm going upward or downward in y.  Given these vectors, the work integral becomes [br][br][center][math][br]\vec{F_g} =0\hat{i}−mg\hat{j}+0\hat{k}. \\[br]\vec{ds}=dx\hat{i}+dy\hat{j}+dz\hat{k}. \\[br]W=\int_{y_i}^{y_f}(0\hat{i}−mg\hat{j}+0\hat{k})\cdot(dx\hat{i}+dy\hat{j}+dz\hat{k})=\int_{y_i}^{y_f}-mg\,dy=−mg(y_f−y_i).[/math][/center]  [br]If we plug in values of [math]y_i[/math] and [math]y_f[/math], such as [math]y_i=20m[/math] and [math]y_f=0m,[/math][math]m=1kg,[/math] then we get [math]W=-1kg(10m/s^2)(0m-20m) = 200 J.[/math] [br][br][/color][color=#1e84cc][color=#1e84cc]Since work done is positive, according to the work-kinetic energy principle, this results tells us that the mass gains 200J of kinetic energy while falling 20m.  If the number were negative, energy would be lost - as would be the case of an object thrown upwards against gravity. It would slow as it rises, and would lose kinetic energy. [/color] [br][br]To find out how fast the object is traveling after falling that distance, we can use the work-kinetic energy principle: [math]W=\Delta K=K_f-K_i=\frac{1}{2}mv_f^2-\frac{1}{2}mv_i^2,[/math] and solve. The answer, if the object is dropped from rest, is [math]v_f=20m/s.[/math] It is merely a coincidence that the value matches the drop distance in meters. Don't assume there's a rule that forces the drop distance to match the speed. There isn't.[/color][br][br]What's worth noting about this example is that the path makes no difference!  For instance, consider what would have happened had I not used zeros for dx and dz, meaning that the objects had undergone lateral displacements while also falling due to gravity.  Since there is no horizontal component of gravity, the dot product would lead to zeros for the horizontal terms. Nothing at all would change!  This means that the work done by gravity is only affected by a change in vertical height and has no dependence on horzontal displacements. [b] In a general sense: If an object moves perpendicular to a force, no work is done by the force. [/b] Only motion parallel to a force leads to work being done. So drop a ball out of a window or roll it down a ramp to the same place and the same work is done by gravity! This path-independence is associated with what's called a [b]conservative vector field[/b]. Conservative fields like gravity are associated with conserved quantities - in this case the conservation of energy. [br][br]A related idea is that if gravity does work on an object as it falls, that the exact same work, but with opposite sign, will be done by gravity if the path is reversed. The result would be that the object would return to its initial location and no net work would have been done. [br][br]In fact, the path need not be exactly reversed. All that would need to happen is the object returns to its starting point along any path. This allows us to define mathematically a condition for a conservative vector field. A conservative vector field always satisfies [math]\oint \vec{F}\cdot \vec{ds} = 0.[/math] The little circle on the integral sign just means that the path is one that closes on itself so that the object (and the path) ends where it started.

Before moving on, I want to show mathematically how the definition of work that we used in this section leads to a change in kinetic energy.  This will only take a few steps to derive and will hopefully make it seem less like magic or just something you must memorize and more like something you understand.  It is done as follows:[br][br][center][math][br]W=\int\vec{F}\cdot\vec{ds} \\[br]W=m\int\vec{a}\cdot\vec{ds} \\[br]W=m\int\vec{a}\cdot dt(\frac{\vec{ds}}{dt}) \\[br]\text{Note that $\vec{a}\;dt = \vec{dv}$ and $\frac{\vec{ds}}{dt}=\vec{v}$, so} \\[br]W=m\int\vec{v}\cdot\vec{dv} \\[br]\text{Breaking this into magnitude and direction makes it clearer to evaluate.} \\[br]\text{Realize that the $\vec{dv}$ term can be expressed in components parallel to $\vec{v}$ or perpendicular:} \\[br]W=m\int v\hat{v} \cdot (dv \hat{v}+dv\hat{v}_{\perp}) \\[br]\text{Since $\hat{v}\cdot \hat{v}_{\perp}=0$, only the first term survives:} \\[br]W=m\int v\;dv \\[br]W=m(\frac{v_f^2}{2}-\frac{v_i^2}{2}) \\[br]W=\Delta K[br][/math][/center]