Video: Work

Video Transcript

In this video, we’re going to learn about work. We’ll learn its physics definition, how under that definition it can be positive or negative, and we’ll also get some practice using it in examples.

To get started, imagine at one day you and a couple of friends are out walking through the neighborhood. The three of you come across an active construction site. And you see a sign saying that help is wanted with pay given out by the net work performed. Interested in making some extra spending money, the three of you decide to sign up for a day’s work.

Once hired on, each of the three of you are assigned different tasks. Your first friend is assigned the task of using a pulley system to lift tools and supplies up to workers on the top level of the building being constructed. Once those workers are done with the tools, your friend then returns them to their original location. Your other friend is given the job of moving pallets of brick around to the back side of the building to be counted and then to bring them back to where they were. And your job throughout the day is to move all of the leftover building materials from the first floor of the building up to the second floor, where that installation will begin.

At the end of a long day, the three of you go to get paid for your work. To your friends’ great surprise, they don’t get any money for the day, but you do. To understand why that it is, we’ll want to understand a bit about work. We’ve all heard the word “work” before and probably have a sense for what it means. But in physics, there’s a precise meaning to the word “work.”

In physics, we define the work done on an object as the scalar or dot product of the force on that object and its displacement. As a symbol for work, we often use a capital 𝑊. And because work is calculated by taking a dot product, it itself is not a vector but a scalar. Another way to write out work is as the product of the magnitudes of force and displacement multiplied by the cosine of the angle between those two vectors. This cos 𝜃 term helps us understand why one of your friends didn’t get paid at the work site.

For the friend who was carrying pallets of material, the force they exerted on those objects to carry them was up to oppose gravity, but the displacement was perpendicular to that direction. The cos of 90 degrees is zero. So they did no work. Going further, if we consider the graph of the cosine function, we see that there are areas where this function goes negative, which means that it’s possible for work to be negative.

Consider this example of a book lying flat on a tabletop. Say that you pick the book up, transfer it across to the other side of the table, and then set it down. We can draw in arrows representing the book’s displacement at each of these three legs and then also draw in arrows representing the force you apply to the book as it moves.

For the first part of the book’s journey, force and displacement are aligned. So work is positive. In the second part, these two vectors are at right angles to one another. So no work is done. Then, as you set the book down, the force you exert on the book is up. But the book’s displacement is down. For this part of the book’s journey, the work you do on it is negative.

Our graph of the cosine function supports this conclusion. We see that when the angle between the force and displacement is 𝜋 radians or 180 degrees, we’ll have a negative result. This helps explain why your first friend didn’t get paid that day. This friend did positive work lifting supplies up to the top of the building and then canceled it out with negative work, lowering them back down. Just like with this book, the net work done on the supplies was zero. So the work done on an object by a force can be positive, negative, or can be zero. Let’s get some practice calculating work through a couple of examples.

A toy cart is pulled for 4.4 meters in a straight line across a floor. The force applied to the cart has a magnitude of 17 newtons and is aligned at 28 degrees above the horizontal. How much work does the applied force do?

If we draw a picture of this process, we have a cart being pulled along by an applied force, 𝐹, where the force is applied at an angle we’ve called 𝜃 above the horizontal. If we call motion to the right motion in the positive direction, we’re told that the cart is pulled 4.4 meters that way. We can call this 𝑑. We want to know the work done on the cart by this applied force, 𝐹.

To solve for this value, we can recall that work is not only equal to the dot product of force and displacement. But it’s also equal to the product of their magnitudes times the cosine of the angle between them. In our case, we know the force magnitude, 𝐹, as well as the magnitude of the displacement, 𝑑. We’re given the angle 𝜃 that separates displacement from force and so are ready to plug in and solve for 𝑊.

When we calculate this product, we find that, to two significant figures, 𝑊 is 66 newton meters or 66 joules. That’s how much work is done on the cart over this distance by the applied force.

Now let’s look at an example that will help us see work from a geometric perspective.

A crane is lifting construction materials from the ground to an elevation of 60 meters. Over the first 10 meters, the motor linearly increases the force it exerts from zero kilonewtons to 10 kilonewtons. It exerts that constant force for the next 40 meters then winds down to zero newtons again over the last 10 meters, as shown in the figure. What is the total work done on the construction materials?

Calling the net work done on these materials capital 𝑊, that work is related to the force the materials experience over the distance they experience that force. We can recall that work is equal to the dot product of force and displacement. And in our case, since the force and the displacement are in the same direction, we can write this expression as a simple product. Since work equals force times distance, looking at our graph, we can see that this will be equal to the area under our curve.

To make solving for that area a bit easier, we can divide it up into three pieces. Piece number one has an area of one-half its base, 10 meters, times its height of 10 kilonewtons. Piece number two has an area of 50 minus 10 meters, or 40 meters, multiplied by 10 kilonewtons. And the third piece has an area like the first of one-half 10 meters multiplied by a height of 10 kilonewtons. When we add the areas of these three sections together to get the total area, we find it’s equal to 500 kilojoules. That’s the total work done on the construction materials over this movement.

Let’s summarize what we’ve learned so far about work. We’ve seen that work is equal to the scalar product of force and displacement. For a given force 𝐹, the work that force does on an object is equal to the dot product of that force and the object’s displacement. We’ve also seen that work tells us a bit about the angle between the vectors of force and displacement. This cosine term means that work can be positive and it can also be negative, or of course it can be zero. And finally, we’ve seen that work, which has units of joules, is related to energy, which is expressed in the same unit.