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Calculating Net Work

Patrick Ford
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Hey guys. So up until now in our work problems, we've been calculating the work done by a single or just one force. And in some problems, you'll be given a bunch of forces and you'll be asked to calculate the network. So I'm gonna show you how to do that in this video using the super awesome flow chart that we'll use for our problems. Let's go ahead and check this out. The whole idea here guys, is that the nets 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. And the whole idea here is that there's actually multiple ways to calculate this network. It really just comes down to what you're given and asked to do in the problem. So let's go ahead and take a look at our examples here and we'll start using this flow chart. So 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're grabbing normal, we have the displacements and what we want to do in the second part part B is we want to calculate the work the network that is done. So we're going from a situation here in our flow chart where we've got forces and eventually we want to get to network and we can see here from this flow chart is that there's actually two different ways to get there. You can go this left path like this or you can take this right path like this. And to figure out which one, you're really just gonna take a look at what you're what you're also asked to do in the problems. Right? So in this first part you're asked to calculate the works done by all of the forces. So in our left and right path you're going to calculate works or you can calculate net force. So which one makes more sense? It's gonna be the left path. We can always calculate work from forces by using F. D. Cosigned data. So in this first part to calculate all the work's done by all the forces. You're just gonna use FD cosine data a bunch of times. So we're gonna figure out the work done by the applied force. The work done by friction, work done by the normal force and the work done by gravity. So here for our applied force will notice that we have a magnitude of 15 and a displacement of 10. And both of those vectors point along the same directions. We're just going to use force times distance. So this is really just gonna be 15 times 10. And this is gonna be 100 and 50 jewels. Now let's move on to the friction force. The friction force. Remember is kinetic friction and it points backwards, it points against our direction of motion. So this picks up a negative sign and it's gonna be F. K. Times D. So we have negative seven newtons times 10 and you're gonna get negative 70 jules. Now what happens is our last two forces. The weights and the normal force. Remember the normal force and the weights are vertical but your direction of motion is horizontal. So those things are perpendicular for both of those cases and therefore the work that is done is just gonna be zero for both of them waiting the normal do no work on this object moving horizontally. Alright, so that's it for part A. So what does part B ask us now? Part B asked us to calculate the network done on this block. So this network here, how do we do that? We'll remember we're taking the left path in order to get down to network. Once you've calculated the works by using F. D. Co sign data a bunch of times. Remember that we said that the network is really just the sum of all the work's done by all the forces. So one way to calculate the network is just by adding up all of the W. Is that you We just calculated. So W one W two, this is sort of like generic as many Ws as you have, you just add them all up to each other and that's the network. So to come down to our problem here we've got the network is really just gonna be R. Two works that ended up being not zero. The 150 that's positive plus the negative 70 and you get a network that's equal to 80 jewels and that's it. That's it for the first problem here. So the work is 80 jewels on this block. Let's take a look at the second problem. Now the second problem is almost exactly identical. We have all the same forces. We have, even the second part of the problem is the same. We're gonna calculate the network. So we're still trying to go from forces to network. But the first part of the problem asks us to calculate the net force first. So this is basically the other path that you can take with this problem is asking us to do is to calculate the net force first and then we'll be able to calculate the network. So let's check this out. How do we calculate net force? We've done this a bunch of times. This is just the sum of all forces in the X and Y direction. So your net force is really just gonna be the sum of all forces but it's gonna be the sum of all forces in the X axis. Because this normal force and this mgR both vertical and they're both gonna cancel out because these blocks is always moving horizontally. So really all we have to do here is point pick a direction of positive to the right like this. And then we can say that this is going to be plus 15 plus negative seven. That's our friction force. So you get a net force of eight newtons. So basically it's as if this box only had one force acting on it to the right like this. This F net here is equal to eight. Notice how it also points along the direction of our displacement. So for part B. Now, how do we calculate the network once we figure out the net force? So now we want to figure out this network here. How do we do that? Well, if we're taking the right path once we've figured out the net force, how do we get from force to work? We've already seen that work is always equal to F. D. Cosine of theta. So if F. Is if W. Equals FD cosine theta, then W Net the network is really just the net force times D. Cosine Theta. So that's what we're gonna use. This W net here is just gonna be the the net force that we just calculated times the displacement times. Cosine Theta. So we have all those numbers here are net forces. Eights are displacement is 10 and then we have the cosine of zero. Because those things are parallel. And what you'll get here is you'll get 80 jewels as Well. So we get 80 jewels. And notice how we basically just gotten the same exact number as we did when we solve this problem using the first method, right? Which is calculating all the work's done. And that should be no surprise. We started off from the same forces we should get the same network. All right. So let me kind of make sense of this here using this diagram, if you think about it, when you take when you do these these two paths, when you go to the left to right, you're really just doing the same operation, but kind of just in reverse order. What I mean by that is that when you take the left path, you're doing FD Cosine Theta a bunch of times. And then in the second step you're going to add them all up to each other. And 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 F Coast FD Cosine Theta. So you can see here that these two steps are basically like the same and you're just doing them in reverse order. Now 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 and in those situations, all you have to do is just pick one of the paths and stick to it. Right. Either one of those will get you the right answer. Personal preference of mine is actually doing all the adding forces. First figure out the Net force and then you just use FD CO. Sign data. That's just my preference. Either one of those will get you the right answer. All right. So that's it for this one, guys. Hopefully that made sense. Let me know if you have any questions.