Guys, up until now in our work problems, we've been calculating the work done by a single or just one force. In some problems, you'll be given a bunch of forces and you'll be asked to calculate the net work. I'm going to show you how to do that in this video using this super awesome flowchart that we'll use for our problems. Let's go ahead and check this out. The whole idea here, guys, is that the net or the total work that is done on an object, we've seen that word net before, is really just the sum. It's the grand total, the sum of all of the works done by all of the forces. There are actually multiple ways to calculate this net work. It really just comes down to what you're given and asked to do in a problem. So let's go ahead and take a look at our examples here, and we'll start using this flowchart.

In this first example here, we've got all the forces listed on an object. We've got an applied force. We've got friction. We have a gravitational normal. We have the displacements. In Part B, we want to calculate the work, the net work that is done. We're going from a situation here in our flowchart where we've got forces, and eventually, we want to get to net work. What we can see here from this flowchart is that there are actually two different paths to get there. You could go this left path like this or you can take this right path like this. To figure out which one, you really just need to look at what you're also asked to do in the problems. So in this first part, you're asked to calculate the works done by all of the forces. In our left and right paths, you're going to calculate works or calculate net force. So which one makes more sense? It's going to be the left path. We can always calculate work from forces by using Fd cosine θ. So, in this first part, to calculate all the works done by all the forces, you're just going to use Fd cosine θ a bunch of times.

We're going to figure out the work done by the applied force, the work done by friction, the work done by the normal force, and the work done by gravity. For our applied force, we'll notice that we have a magnitude of 15 and a displacement of 10. And both of these vectors point along the same direction. We're just going to use force times distance. So this is really just going to be 15 times 10, which equals 150 joules. Now let's move on to the friction force. The friction force, remember, is kinetic friction and it points backwards against our direction of motion. So this picks up a negative sign, and it's going to be Ff k times d. So we have negative 7 newtons times 10, and you're going to get negative 70 joules. Our last two forces, the weight and the normal force, remember, are vertical, but your direction of motion is horizontal. Therefore, the work that is done is just going to be 0 for both of them. Weight and normal do no work on this object moving horizontally.

So, what does Part B ask us now? Part B asks us to calculate the net work done on this block. This net work here, how do we do that? Well, remember, we're taking the left path to get down to net work. Once you've calculated the works by using Fd cosine θ a bunch of times, remember that we said that the net work is really just the sum of all the works done by all the forces. So one way to calculate the net work is just by adding up all of the Ws that you just calculated. So, coming down to our problem here, the net work is really just going to be the two works that ended up not zero. The 150 that's positive plus the negative 70 gives us a net work that's equal to 80 joules.

Let's take a look at the second problem now. The second problem is almost exactly identical. We have all the same forces. The second part of the problem is the same; we're going to calculate the net work. So, we're still trying to go from forces to net work, but the first part of the problem asks us to calculate the net force first. This is basically the other path that you can take. The problem is asking us to calculate the net force first, and then we'll be able to calculate the net work. To calculate the net force, we've done this a bunch of times. This is just the sum of all forces in the x and y directions. Your net force is really just going to be the sum of all forces in the x-axis because this normal force and the gravitational force (mg) are both vertical and they are both going to cancel out because this block is consistently moving horizontally. So, all we have to do here is pick a direction of positive to the right like this, then we can say that this is going to be plus 15 plus negative 7. That's our friction force. So you get a net force of 8 Newtons. Notice how it also points along the direction of our displacement.

For Part B now, how do we calculate the net work once the force is known? Now we want to figure out this net work here. How do we do that? Well, if we're taking the right path, once we figure out the net force, how do we get from force to work? We've already seen that. Work is always equal to Fd cosine θ. So, if W equals Fd cosine θ, then Wnet, the net work, is really just the net force times d cosine θ. That's what we're going to use. This Wnet here is just going to be the net force that we just calculated times the displacement times cosine θ. We have all those numbers here. Our net force is 8. Our displacement is 10, and then we have the cosine of 0 because those things are parallel. What you'll get here is 80 joules as well. We get 80 joules, and notice how we basically just got the same exact number as we did when we solved this problem using the first method, right, which kept involving all the works done, and that should be no surprise. We started off from the same forces, so we should get the same net work.

Let me kind of make sense of this here using this diagram. If you think about it, when you do these two paths when you go to the left or right, you're really just doing the same operation but kind of just in reverse order. What I mean by that is when you take the left path, you're doing Fd cosine θ a bunch of times, and then in the second step, you're going to add them all up to each other. When you take the right path, you're doing all the adding stuff first. You're adding up all the forces, and in the second step, you're using Fd cosine θ. You can see here that these two steps are basically the same, and you're just doing them in reverse order. I have one last point to mention here, which is that some problems will actually give you forces and ask you for the net work, but they won't actually guide you either to the left or the right. They won't first ask you to calculate a bunch of works or they won't ask you to calculate the net force. In those situations, all you have to do is just pick one of the paths and stick to it. Either one of those will get you the right answer. Personal preference of mine is actually doing out all the adding forces first, figure out the net force, and then you just use Fd cosine θ. That's just my preference. Either one of those will get you the right answer. So that's it for this one, guys. Hopefully, that made sense. Let me know if you have any questions.