We've all done work before, whether it be for school, money, or just for fun. But in physics, the idea of work is slightly different from what we think it is in daily life. (In fact, me typing these all words out on a keyboard might sound like a lot of work, but the actual amount of (physics) work done is very little!) This lesson is designed to teach you all about work and what it means to do work in a physics context. And no, doing problems actually isn't that much "work". I'll stop with the jokes now and get into the content.
Work is defined as the dot product of force and displacement. Therefore, it has the unit of $N\cdot m$, more commonly called joules ($J$). These are both vector quantities, but work itself is a scalar quantity. However, it's still important to realize that while work doesn't have a range of directions, it can still be either positive or negative. Thus, work can "cancel out" in a way, much like vectors can, but it is unltimately always going to be a scalar quantity. Here's a mathematical description of what I just said:
$$W = \vec{F} \cdot \vec{x}$$
Now, this formula is only true if the force is constant. If we have a varying force (which we often do in more complex, more realistic scenarios) we need to write this as a differential equation:
$$dW = \vec{F} \cdot d\vec{x}$$
This will come in handy, trust me.
Figure 1: Work in daily life versus work in physics.
This formula tells us a lot of things, but most of those I've already mentioned or are pretty obvious. However, it is important to note that work can either be calculated for a single force or a net force. You can have work done by any particular force even if it is not the net force, but the net work done (by the net force, evidently) is usually more important when it comes to analysis.
Do you understand now why I said typing on your keyboard does very little work? When you press on the keys, you cause them to displace downwards a very small amount, meaning the displacement vector and therefore the work done is very small. (You also don't press down on the keys very hard, so the force is small too.)
The definition of work takes into account the dot product from earlier. In particular, it is the dot product of force and displacement (which are both vectors, but work is a scalar!). The quantity of work has the new unit of joules ($J$), which is equivalent to a newton-meter. This is a pretty simple definition, and it leads to a pretty simple formula:
$$W = \vec{F} \cdot \vec{x}$$
Figure 1: Work in daily life versus work in physics. Basically, the work is both dependent on the displacement of an object as well as the force exerted on it. This formula calculates the work done by a single force. Let's try an example, hearkening back to the beginning of this lesson, where I stated that the work done by typing on a keyboard is very little. (You should be able to understand why - the force you press on the keyboard is small, as is the displacement of the keys!)
What work is done by pressing on a keyboard key? Assume the key compresses down a distance of $x = 1.5 ~\textrm{mm}$ and the force you exert on it is $F = 0.5 ~\textrm{N}$ directly downwards?
Figure 2: An accurate representation of the scenario. This is a pretty simple calculation. Now we do have to consider that these two are vectors and the dot product is required, but since both vectors point straight downwards the angle between them is zero and thus you just multiply their magnitudes. A common mistake is to forget to convert millimeters to meters.
Technically, we have already covered the definition of work in detail. However, there is a lot of nuance to this, and there are other things that we need to cover in order to have a complete understanding of work. The first is how to find the work done on an object given a graph of force vs. displacement, almost always for an object moving in one dimension. (Calculus students, you might already know how to do this just from that!)
The key is that the work done is the area under the force vs. displacement graph! The exact explanation requires calculus, but a more basic explanation is to look at the axes. You have force and displacement as your axes, and since there's only one dimension involved you can simply multiply them. Multiplying them gives you the area under the graph! This isn't really a full definition, but it's as good as I can get without going into calculus details. You might remeber the previous work formula as a differential equation. Well, we can simply integrate this to find the work, which explains why the work done is the area under the curve for a force vs. displacement graph. This explanation is simple and elegant, but it requires knowledge of calculus. In fact, since you're calculus students, you can easily apply this principle to graphs that have irregular areas; that is, areas that can't be calculated with geometric formulas.
This might seem like a random addition to this section, and for the sake of real physics you'd probably be right. However, a large number of problems have force vs. displacement graphs and require you to find the work done by that varying force. The graphs are mostly geometric in shape and the area under the curve can be calculated exactly with geometry knowledge. I'll show one as a very basic example, just to make sure we're clear on the concept.
What is the work done on the object by the force shown in the diagram below as it moves from $x=0$ to $x=6 ~\textrm{m}$?
This is quite simple. You find the area under the graph from $x=0$ to $x=6$, which is just simple geometry. However, there are two common errors that I've seen with this kind of graph. The first is to assume that the area is measured to the bottom of the graph. However, area under a curve refers to the area between that curve and the x-axis, not the bottom of the graph. You'll notice that the x-axis is not at the very bottom of the graph, so be careful here! The next is a more elementary mistake. Area that is below the x-axis is counted as negative in calculations. Some people may neglect this, which leads to incorrect answers. But that's enough yapping, I'll get to the solution.
Using basic geometry, we can analyze the graph. Remembering that area below the x-axis is negative, we can come to the conclusion that:
$$W = \textrm{Area under the graph} = 6 + (-2) = \bbox[3px, border: 0.5px solid white]{4 ~\textrm{J}}$$
I've talked about the force part behind the work formula, so it's now time to talk about the displacement part. Actually, we're not talking about directly, but rather talking about conservative and nonconservative forces. These are the two types of forces in general when it comes to work.
The key is to focus on the different possible paths that an object can take to go from one point to another, and whether the choice of those paths affects the work done. There are technically an infinite number of paths from one point to another, but here I'll illustrate three just to show you the possible variety.
Figure 3: A work of modern art, or a representation of three different possible paths between two points? It's the latter. The conservative force is one that is path-independent. What that means is that the work it does only depends on its displacement, not the exact path it traveled on. It doesn't matter if you took any number of fancy turns or maneuvers during on your path. The work done by this force on you will be the same as someone who travelled in a straight line between the start and end points. Good examples of this are spring forces and gravitational forces.
This naturally leads to the idea of a nonconservative force. This kind of force is path-dependent, meaning the work done by it will vary on the path you took, even if the start and end points don't change. The most common of these nonconservative forces is friction. Now, let's take a moment to think about why friction is nonconservative.
One of friction's key features is that it always opposes the direction of relative motion, which means that it'll be oppositely directed to the object along its entire path. Thinking back to the differential formula for work, we realize that it's oppositely directed to the tangent to the path everywhere. This means that even though the vector differential $d\vec{x}$ is constantly changing over the path, the dot product is still the same because the direction of friction changes with this. Thus, we arrive at a formula for work done by friction: The exact reasoning here requires calculus, but just know that it has something to do with having to compute the dot product constantly because the direction of friction changes constantly. Since the direction of friction is always oppositely directed to the momentary change in position, we can write a formula for the work done by friction:
$$ W_{friction} = - F_f d \cos \theta $$
The letter $d$ here represents distance, not displacement. The cosine is there because the frictional force and the momentary displacement vector might not be exactly opposite to each other at any given moment (but they usually are). I'd say don't worry about it too much, since it doesn't show up often or even at all, since I can't think of a meaningful scenario where the cosine is actually in use.
Of course, this formula is really only applicable for kinetic friction because static friction causes no relative motion. While static friction can do work (it does work in our old penguin and sled example), most of the time we're concerned with dissipative effects due to friction, and static friction does not dissipate energy.
Speaking of energy, work and energy are actually inextricably tied. But, this is a lesson on work, so you'll have to go to the next lesson to learn about energy. So, if you're ready, hit that subscribe button next lesson button on the bottom right! The two classifications of forces depend on the idea of path independence, which is when something doesn't care about the path it took, just the start and end points. This might sound familiar, and you'd be right! The idea of displacement is path-independent, since it only considers start and end points.
Figure 3: A work of modern art, or a representation of three different possible paths between two points? It's the latter. The first type of force is a conservative force, which does the same amount of work regardless of the path taken. It doesn't matter if you took any number of fancy turns or maneuvers during on your path. This includes most field forces like gravity, as well as a few others. It's the easier of the two to deal with, since you don't have to worry about complications caused by taking a complicated path. (See what I did there?)
The other type of force is a nonconservative force that does different amounts of work depending on the path taken. These are usually forces like friction and air resistance. Now for these forces, while it is possible to calculate the work done by them, if the path is complex you either need scary calculus or just simply can't solve the problem exactly. If you want the formula for friction, you can look at it in our Algebra-based and Caclulus-based levels, but no worries if you don't feel confident. We won't be seeing these too much, so rest assured.
These two concepts conclude our discussion of work, but they lead perfectly into our discussion of energy in the next lesson. So if you're ready, just hit that button over there and let's move on!
The AntiMatter Lab 
