Net Work & Work-Energy Theorem - Online Tutor, Practice Problems & Exam Prep

Understanding net work involves calculating the total work done by multiple forces acting on an object. The net work is the sum of individual works, calculated using the equation ${\frac{F}{d}}^{\cos }\left(\theta \right)$. The work-energy theorem relates net work to the change in kinetic energy, expressed as $W=\Delta K=K\_\mathrm{final}-K\_\mathrm{initial}$. This relationship allows for the calculation of work even when forces are unknown, emphasizing the connection between work and energy transfer.

Calculating Net Work

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.

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

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

The Work-Energy Theorem

Hey, guys. So in the last couple of videos, we've been talking a lot about work and specifically the network. Remember, the definition of work is that it's the quantity, it's the amount of energy transferred between objects. So for example, let's say you push a book and it gains some speed, it gains 10 joules of kinetic energy, you've done 10 joules of work on the spot. Right? It's the transfer of energy. And so what we're going to talk about in this video is the work-energy theorem, which is basically just the mathematical relationship between work and energy. Alright? So I'm going to show you the equation and we're going to get straight into an example. Basically, what the work-energy theorem says is that the net work, the sum of all the works done by all forces on an object, is related to the kinetic energy. It's the change in the kinetic energy. Remember that change just means final minus initial, so then a way we can write this is k_final - 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 4-kilogram box that has an initial speed of 6 at point A, and later on, at some point B, it has a speed of 10. So notice how in this problem, we're actually not given any directions—it's going to the right or up or anything like that. So what I'm going to do is I'm going to 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. All I know here is that at point A, the velocity is 6, and at 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 or 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. So between point A and point B, there's going to 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 going to be some work done. Alright. 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, we're 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 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 in this problem here. So now we can actually go back to our flowchart because we know that there's a couple of ways to figure out net work. I can figure out net work by adding up a bunch of works by using Fdcosine 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 and the work-energy theorem comes into play. So we can always relate the net work 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 net work. So network, remember, is just going to 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 1/2 mv², 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 4. So that's all I have to do is just expand out these kinetic energy terms. So my net work here is really just going to be 1/2, and this is going to be mV_B² minus 1/2 mV_A². And so all I have to do is just go ahead and plug and chug. Right? So this is going to be 1/2, mass is 4. The velocity at point A is oh, sorry. But point B is going to be 10. So this is going to be 10 squared minus the one-half four times 6 squared. You just go ahead and plug all that stuff into your calculator and you're going to get a net work that's equal to 128 joules. 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. Alright. So that's it for this one, guys. Let me know if you have any questions.

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

Net Work & Constant Speed

Guys, so 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 ahead and take a look here. We have 5-kilogram 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 mass is 5 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 8 meters per second so there has to be another force that's 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. What 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 going to have you're going to do 1 or 2 things. You can add together all of the works done if you know all the forces, right, by using a sum of all works or if you don't know all the forces, you can use f_netd cos(θ). Now our applied force here is actually unknown. We've seen something like this before. So we're going to have that the net work is equal to the net force times d cos(θ) here. So all we have to do is just figure out the net force, to figure out the work that's the network. So we're going to go over here and figure out the network or the net force by using F = ma. Right? So your sum of all forces equals mass times acceleration. So your forces really relate to your mass times acceleration but we actually know in this problem that we're moving at a constant speed of 8 meters per second. So we know that we're going to be pushing this box, friction is going to 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 0. I don't have to work out my forces. I actually can't because I don't know what the applied force is. Right? So basically, what happens is that my net force, because this a is equal to 0, is equal to 0. The forces have to cancel because I'm not actually accelerating. So what this means here is if I have a net force of 0, I'm just going to plug this into my work net equation. You can probably expect what's going to happen. Once I multiply 0 into d cos(θ), the whole thing is going to cancel and I get a network that's equal to 0 joules. That's the answer. Here's 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 0 net work that's done on an object, then that means that the change in the kinetic energy is also equal to 0. This brings me to a really important conceptual idea. So the work energy theorem is a 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 the change in the kinetic energy is going to be 0, and, therefore, the net work is going to be 0. Remember that your kinetic energy is equal to 1/2 m v². So if your 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 net all the works are going to cancel out. So that's it for this one, guys. Let me know if you have any questions.

Here’s what students ask on this topic:

What is the net work done on an object?

The net work done on an object is the total work resulting from all the forces acting on it. It is calculated by summing the individual works done by each force. Mathematically, it can be expressed as:

$W=\sum _{}^{i}W\_i$

where $W\_i$ is the work done by the $i$-th force. Each work can be calculated using the formula:

$W=Fdcos\theta$

where $F$ is the force, $d$ is the displacement, and $\theta$ is the angle between the force and displacement vectors.

How do you calculate net work using the work-energy theorem?

The work-energy theorem states that the net work done on an object is equal to the change in its kinetic energy. This can be expressed as:

$W=\Delta K=K\_f-K\_i$

where $K\_f$ is the final kinetic energy and $K\_i$ is the initial kinetic energy. Kinetic energy is given by:

$K=\frac{1}{2}m{v}^2$

where $m$ is the mass and $v$ is the velocity. By knowing the initial and final velocities, you can calculate the change in kinetic energy and thus the net work done.

What is the relationship between work and energy transfer?

Work is a measure of energy transfer between objects. When work is done on an object, energy is transferred to or from the object. For example, if you push a book and it gains speed, the work you do on the book results in an increase in its kinetic energy. The relationship between work and energy is given by the work-energy theorem:

$W=\Delta K$

This equation shows that the net work done on an object is equal to the change in its kinetic energy, emphasizing the direct connection between work and energy transfer.

How do you calculate work done by multiple forces?

To calculate the work done by multiple forces, you need to determine the work done by each individual force and then sum them up. The work done by a single force is given by:

$W=Fdcos\theta$

where $F$ is the force, $d$ is the displacement, and $\theta$ is the angle between the force and displacement vectors. After calculating the work for each force, sum them up to get the net work:

$W=\sum _{}^{i}W\_i$

What is the work-energy theorem?

The work-energy theorem is a fundamental principle in physics that relates the net work done on an object to its change in kinetic energy. It states that:

$W=\Delta K=K\_f-K\_i$

where $K\_f$ is the final kinetic energy and $K\_i$ is the initial kinetic energy. This theorem highlights the direct relationship between the work done on an object and the resulting change in its kinetic energy, providing a powerful tool for solving problems involving energy transfer.