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Table of contents

9. Work & Energy

1

concept

6m

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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.

2

Problem

You pull a 3kg box on a flat surface. The coefficient of kinetic friction is 0.6. When you pull the box horizontally through a distance of 10m, it accelerates at 2m/s2. Find the net work on the box.

A

17.6 J

B

35.3 J

C

176.4 J

D

60 J

3

Problem

To pull a 51 kg crate across a rough floor, a worker applies a force of 100 N, directed 37°above the horizontal. The coefficient of friction is 0.16. If the crate moves 3.0 m, what is the total work done on the crate?

A

29 J

B

0 J

C

240 J

D

169.6 J

4

concept

4m

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Hey guys. So in the last couple videos we've been talking a lot about work and specifically the network. Remember the definition of work is it's the quantity, it's the amount of energy transfer between objects. So for example, let's say you push a book and gain some speed against 10 jewels of kinetic energy. You've done 10 jewels of work on the spots, writes the transfer of energy. And so we're gonna talk about this video is the work energy theorem, which is basically just the mathematical relationship between work and energy. Alright, so I'm gonna show you the equation. We're gonna get straight into an example basically with the work energy theorem says, is that the network the sum of all the works done by all forces on an object actually is related to the kinetic energy. It's the change in the kinetic energy. Remember that change just means final minus initial. So that way we can write this is K, final minus K initial. Alright, that's really all there is to it. This is just the one equation work is related to kinetic energy. So I have one last point to make in this diagram here, but let's go ahead and start the problem and we'll get back to it later. So the whole idea here guys is that we have a four kg box that has an initial speed of six at point A and that later on at some point B has a speed of 10. So notice how this problem, we're actually not given any directions or told it's going to the right or up or anything like that. So what I'm gonna do is I'm gonna try to draw a sketch and I don't know if it's actually moving to the right. This is more of just like a sequence like all I know here is that at point a the velocity is six or the speed and a point b. The velocity or the speed is 10. That's all I know. I don't know if it's actually going to the right or up or down anything like that. So what I'm asked to do is I'm trying to figure out how much work is done on the box. Between these two points between A and B. There's gonna be some work done. And this makes sense because it basically picks up some speed, it goes from 6 to 10 and so it's basically gained some energy, Right? So there's gonna be some work done. All right, So how do we do this? Well, the first thing you should notice here is that they're asking us how to calculate some amount of work done. So this is w but they're not specifying by what force, in fact were not told anything about any forces in this problem. We don't know if there's an applied force or friction or anything like that. So whenever this happens, whenever you have problems where forces aren't given, but you're asked to calculate a work, it's kind of implied that this work is actually the network. You're trying to figure out this total or W. Nets. So that's what we're trying to find this problem here. So now we can actually go back to our flow chart because we know there's a couple of ways to figure out network. I can figure out network by adding up a bunch of works by using F. D. Cosign Theta. Or I can figure out if I know the net force like this. But in this problem here, we can't do either one of these things because we don't have any of the forces involved. So how do I figure out work then? And this is where the power of the kinetic energy in the work energy theorem comes into play. So we can always relate the network with the kinetic energy, remember by delta K. So really this is just a third way that we can use a third way to figure out the network. So network remember is just gonna be related to the change in the kinetic energy. It's K final which is really just K B minus K A. That's final minus initial. And remember that these kinetic energies really just are related to one half Mv squared, just mass and speed. And we actually have both of those things here. We have the speeds at point A and point B. And we also know that the mass is four. So that's all I have to do is just expand out these kinetic energy terms. So my network here is really just gonna be one half and this is going to be M V B squared minus one half, M V A squared. And so all I have to do is just go ahead and plug and chug. Right, so this is gonna be one half masses four. The velocity at point A. Is, sorry, but point B is going to be 10. So this is gonna be 10 squared minus the one half four times six squared. You just go ahead and plug all that stuff into your calculator and you're gonna get a network that's equal to 128 jewels. That's the answer. So that's what's really cool about the work energy theorem is that you can actually figure out the total work that's done on an object even though you have none of the forces involved. And it's just because it's related to the kinetic energy. All you need to know is the masses, and also the speeds at two different points. You also don't even have to know the direction that this thing is moving in. All right, so that's it for this one. Guys let me know if you have any questions.

5

Problem

A box slides across the floor with an initial speed of 3.5m/s. If the coefficient of kinetic friction is 0.15, how far will the box slide before stopping completely?

A

1.19 m

B

4.17 m

C

9 m

D

Not Enough Information

6

example

3m

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Hey guys, I have a very important problem here that's going to teach you a very important conceptual idea about the work energy theorem. That's going to allow you to sort of take a shortcut in some of your problems that you might see on homework or tests. So let's go and take a look here. We have five kg boxes on a flat surface, but it's rough, which means there's some kinetic friction. So I have this box like this, we have the masses five. And I'm going to be pushing on this box with a constant force. This is some applied force like this. Now, what happens is I'm moving at a constant eight m per second. So there has to be another forces canceling my push. And that's the force of friction. So, I've got my F. K. Here. I also have the MG and the normal force just to label all my forces. But I want to do in this. Problem is I want to figure out the network that's on the box. So remember that. To figure out network, you're just gonna have you're gonna do one of two things. You can add together all of the works done if you know all the forces right? By using some of all works. Or if you don't know all the forces you can use F net D. Cosign Theta now are applied force here is actually unknown. We've seen something like this before. So we're gonna have that the network is equal to the Net Force times D. Cosign Theta here. So all we have to do is just figure out the net force to figure out the work that's um that's the network. So we're gonna go over here and figure out the network of the net force by using F. Equals M. A. Right? So you're some of all forces equals mass times acceleration. So your forces really related to the mass times acceleration. But we actually know in this problem that we're moving in a constant eight m per second. So we know that we're going to be pushing this box, friction is gonna be opposing me but I'm going to be going at a constant speed. So because of this, I know that my acceleration is actually equal to zero. I don't have to work out. My forces actually can't because I don't know what the applied forces. Right? So basically what happens is that my net force because this is equal to zero is equal to zero. The forces have to cancel because I'm not actually accelerating. So what this means here is if I have a net force of zero, I'm just gonna plug this into my work net equation. You can probably expect what's going to happen once I multiply zero into deco Cynthia to the whole thing is going to cancel and I get a network that's equal to zero jewels. That's the answer. Here is another way you can think about this, remember that the network is equal to the change in the kinetic energy. That's the work energy theorem. So if I had zero network that's done on an object and that means that the change in the kinetic energy is also equal to zero. This brings me to a really important conceptual idea. So the work energy theorem is very useful conceptually because if you ever see that an object has constant speed in a problem and you're asked to calculate things like kinetic energies and works, you just know that they change. The kinetic energy is going to be zero and therefore the network is going to be zero. Remember that your kinetic energy is equal to one half mv squared. So if you're v never changes, your kinetic energy will never change. And that just means that there was zero work done on you net. It could just mean that there's no forces acting on you or just could mean that all the all the works are going to cancel out. So that's it for this one. Guys, let me know if you have any questions.

Additional resources for Net Work & Work-Energy Theorem

PRACTICE PROBLEMS AND ACTIVITIES (7)

- You throw a 3.00-N rock vertically into the air from ground level. You observe that when it is 15.0 m above th...
- A 1.50-kg book is sliding along a rough horizontal surface. At point A it is moving at 3.21 m/s, and at point ...
- You throw a 3.00-N rock vertically into the air from ground level. You observe that when it is 15.0 m above th...
- An 8.0 kg crate is pulled 5.0 m up a 30° incline by a rope angled 18 ° above the incline. The tension in the r...
- A 2.0 kg particle moving along the x-axis experiences the force shown in FIGURE EX9.22. The particle’s velocit...
- A red ball has a mass of 250 g. A constant force pushes the red ball horizontally and launches it at a speed o...
- A pitcher accelerates a 150 g baseball from rest to 35 m/s. How much work does the pitcher do on the ball?

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