Energy with Non-Conservative Forces: Study with Video Lessons, Practice Problems & Examples

Incorporating nonconservative forces like applied forces and friction into the conservation of energy equation reveals that mechanical energy is not conserved when work is done on a system. For example, when a hockey puck is pushed, the work done increases its kinetic energy. The full conservation of energy equation is expressed as ${\frac{1}{2}}^{\mathrm{mv}}+W={\frac{1}{2}}^{\mathrm{mv}}$, where W represents work done by nonconservative forces. This understanding is crucial for analyzing energy transformations in various physical scenarios.

Energy Conservation with Non-Conservative Forces

Guys, up until now, we've been using our conservation of energy equation with conservative forces. Remember that conservative forces are things like gravity and springs. We've been focusing more on gravity. So what I'm going to do in this video is I want to show you how the conservation of energy equation works when you start throwing nonconservative forces into the mix. Remember, your nonconservative forces are things like applied forces and friction. Let's check out our practice problem or example down here because it's exactly what's going on here. So, I have a hockey puck. Right? So a hockey puck is traveling with some initial speed, so the initial speed is 4. But then, what I'm going to do is I'm going to push it with my hockey stick. So my applied force is 200, and I'm going to push it through some distance here.d = 0.3 m. And then what happens is this hockey puck, because I've pushed it, we can expect that it's going to be moving faster. So because I've pushed it, it's going to have some final speed that's going to be greater than the initial 4, and that's exactly what I want to calculate here. So how am I going to calculate this puck's final speed? I'm just going to use conservation of energy. Right? So I'm just going to draw my diagram, which I've already done. Now I want to use my energy conservation equation here. So how do I do this? Well, if I write my energy conservation, my mechanical energy, I'm going to write kinitial + uinitial = kfinal + ufinal. That's how we've seen this before. However, what happens is we're going to see that there is no potential energy because we're just traveling along this horizontal surface. And what happens is we can tell, without even calculating anything, that our kinetic final is going to be greater than your kinetic initial because, again, we said that the speed is going to be greater than the initial 4 because we've been pushing through some distance. We've been doing work on it. So what ends up happening here is that we're going to find out that the mechanical energy initial is not going to be equal to your mechanical energy final here, which is really for the kinetic final. And this should make perfect sense because the force that we have acting on this puck is an applied force, and applied forces, remember, are not conservative. And remember the rule for conservation of mechanical energy. We said if nonconservative forces do the work, which means that the work done by these forces isn't 0, then your mechanical energy is not going to be conserved. So here, we have an applied force that's doing work. It's adding or removing energy to the system. So your mechanical energy is not going to be equal on the left and right sides. So how do we solve these problems? Well, it turns out that we're actually still going to use conservation of energy at our equation to solve these problems, but now we just need one more term. So we're going to have still k + u on the left and k + u on the right. But now the last term that we need is the work done by nonconservative forces. So this is known as the full conservation of energy equation. We're going to write it like this from now on every single time. Alright. So, basically, instead of using this equation now, we're going to solve this problem by using the full conservation of energy. So this is k + u initial + work done by nonconservative forces = k + u final. Now we're still going to have no gravity, you know, potential energy initial and final here. So now we're going to go ahead and start eliminating and expanding our terms. So our kinetic energy, remember, is just going to be 12 m vinitial2. So what about work done by nonconservative forces? How do we actually calculate that? Well, it turns out the work that's done by nonconservative forces is just going to be the sum of the works done by these forces, applied forces, and friction.

Box Sliding on Rough Patch

Hey, guys. Let's work this one out together here. So we have a block of unknown mass that's sliding along a floor with some initial velocity. I'm going to call this \( v_{\text{initial}} = 30 \). Well, then what happens is it's going to hit this little rough patch along the road or the surface or whatever it's sliding on, and we're going to calculate the distance that the block travels before it stops. So later, once it passes through the rough patch, it's going to come to a stop right here, and with that means that the velocity is going to be equal to 0. So we want to calculate basically what is the distance that thing travels across that rough patch. I'm going to call that \( d \) here. Alright. So what happens is we're going to use conservation of energy for this. We've got our diagram. Let's go ahead and write out our energy conservation equation. There's really only 2 points, an initial and a final here. So this is my initial and this is my final. So I've got \( K_{\text{initial}} + U_{\text{initial}} + \text{any work done by non-conservative forces} = K_{\text{final}} + U_{\text{final}} \). Alright. So we have some initial velocity here. So we have some initial kinetic energy. There's no, gravitational potential or, you know, there's no spring energy or anything like that here. So there's actually no potential in either case, initial or final. What about work done by non-conservative forces? Remember, work done by non-conservative forces is work done by you and also friction. What happens is you're not pushing it, you're just watching this thing as it's sliding along, so there's nothing here. But there is going to be work that's done by friction because as this block is sliding through this rough patch, there is a force of friction that's going to oppose that direction of motion here. So that's where our work is going to come into. What about \( K_{\text{final}} \) here? Is there any kinetic energy final? We just said here that the final velocity is going to be 0 once it comes to a stop, so there is no kinetic energy. So what happens here is you might be thinking, whoa, I thought mechanical energy has to be conserved. It's only conserved if you have only conservative forces. We have a non-conservative force that's acting here. So basically, there's some work that is done by that non-conservative force. So the point I want to make here is that whenever mechanical energy isn't going to be conserved, the work is always going to basically make up the difference. Let's check out how this works here. So I've got my \( K_{\text{initial}} \) which is going to be \( \frac{1}{2} m v_{\text{initial}}^2 \), and then the work that's done by friction. So the work done by friction, remember, is going to be \(- f_k d\), right? So got negative there very important that you add that negative sign there because friction is going to remove energy from the system. Alright, and this is going to equal 0 on the right side, right? Both of the terms cancel out here. So what happens is I can go ahead and rearrange, and actually sort of expand out this friction equation, so I'm going to do this over here. So remember that friction is just equal to \( k \) times the normal force, and what happens is if you're just sliding along a normal, a flat surface, then you have an \( mg \) downwards like this and you also have a normal force and these two things have to balance each other because there are only 2 forces in the vertical direction. So basically what happens is we're just going to get \( \times mg \). So now what I can do here is I can say that \( \frac{1}{2} m v_{\text{initial}}^2 + \text{negative}\mu g d = 0 \), and I can actually just move the whole term over to the other side. So what I end up getting here, is I get \( \frac{1}{2} m v_{\text{initial}}^2 = \mu k \times mgd \). Notice how the masses actually cancel, which is great because we didn't actually know what the initial mass was. And now we just want to calculate the distance here. So basically, what happens is when you calculate the distance, I'm just going to go ahead and write an expression for this. You're going to get \( v_{\text{initial}}^2 \), you're going to get \( \frac{1}{2} v_{\text{initial}}^2 \), and then you're going to divide over the \( \mu k \) and also gravity. So \( \mu k \times g \). So this is just going to be, let's see here. This is going to be the initial speed of 30 squared divided by 2 times \( \mu k \). Right? The coefficient here is 0.6 and then times 9.8. When you go ahead and work this out, what you're going to get is you're going to get 76.5 meters, and that's the answer. Right? So what happens here is that you've had all this initial kinetic energy, right, this \( \frac{1}{2} mv_{\text{initial}}^2 \), but it gets removed by this non-conservative force. Basically, friction removes that energy until the box is left with nothing. So what happens is that this work done by non-conservative forces always is going to make up the difference between the left and the right side of your equation here. And what happens is if you end up getting, you know, something that's non-zero like we did here, then that work done is always energy that's either added or removed from the system. Alright? So that's it for this one, guys. Let me know if you have any questions.

An 800kg car skids to a stop from 30m/s through a distance of 90m. Calculate the coefficient of friction between the car and the road.

• A. 0.51
• B. 1.23
• C. 0.73
• D. 0.02

Energy Conservation with Air Resistance

Hey, guys. Let's put this one out together here. So we have a 2-kilogram object. It's dropped from a height of 80 meters and it reaches the floor with 30. But in this case, there's actually some air resistance here. We're going to calculate the work that's done. Let's check out this problem here. Let's go ahead and draw our diagram. So we've got the floor like this, this is y equals 0. I've got my 2-kilogram objects and it's going to be dropped, which means that the initial speed is 0. But once it falls to the floor, what happens is that it's going to reach the floor with some final speed. This is v final and this equals 30. It doesn't matter if it's positive or negative because their kinetic energy is always going to be final. Right? Or it's always going to be positive. Now what happens is throughout the motion, this object, it's being pulled down by gravity, there's mg, but there's also air resistance. So then I'm going to call this f air. Usually, we kind of neglect air resistance, but in this problem, we want to calculate the work that is done by this force. So let's go ahead and check out our energy conservation equation. Right? We're going to use conservation of energy. We've got our diagram, and now we're going to go ahead and do our conservation of energy. So this is going to be k initial, plus u initial, plus work done by nonconservative equals k final plus u final. So I forgot one last thing here. You're actually falling from your initial height. This is y initial of 80. So this is going to be y initial. Okay. So we have no initial kinetic energy because the initial speed is 0. We do have some gravitational potential because we're at 80 meters and that's above our zero point. So we're going to set the 0 point of gravitational potential here. There is work done by nonconservative forces because we do have air resistance, and that's a nonconservative force. Remember that work nonconservative is either work that's done by you, which there's none of in this problem, plus the work that's done by friction. Basically, friction and air resistance are kind of the same thing. Air resistance is really just friction through the air. So this is really just going to be f air, the work that's done by this force here. And this is going to be equal to k final plus u final. So there is some kinetic energy final, but there's no gravitational potential energy because you're at the ground. So what happens is our equation sort of simplifies to this is going to be mgyinitialplustheworkthat'sdonebyfairandthenequals the kinetic energy final is going to be 12mvfinal2. So let's see here. I know m, I know g, I know the initial heights. I also know m and I know v final squared. So all I have to do is just go ahead and move everything over. So this w f air here, w f air is just going to be, this is going to be, let's see. You actually can't cancel out the masses because it doesn't exist in every single one of these terms. Right? This w f air doesn't have an m in it, so we can't cancel that out. So what ends up happening is you're going to get, let's see. You're going to get 12 of 2 and then times 80. So that's the first term and then when you subtract it, you're going to subtract it from 2 times 9.8, and then this is going to be oh, I'm sorry. This is not 80. This is v final. So this is actually 30 squared. There we go. Sorry about that. So we've got 12, 2, and then we've got, 30 squared minus 2 times 9.8 times 80. Alright? So that's the initial height. So then you go ahead and work this out. What you're going to get is you're going to get negative 670 joules. So why do we get a negative sign? It's because work is actually removing energy from the system. That's exactly what we should what we should expect. So we got a negative number here because we have energy that's being taken out by air resistance. Alright? So we can use energy conservation equations to solve problems with resistive forces, like when we have air water resistance, basically because they just act like friction. So we can just sort of calculate them as a nonconservative work. Alright. So let me go ahead and quickly solve part b now. Part b is now asking for the average force of air resistance. Alright? So we want to figure out basically what is this f air. Whoops. What is this f air? Well, what happens is you can think of this f air as being a constant force that's acting on this object as it falls. So it's a constant force that's being exerted over some distance d, which is really just my delta y. So what happens is I can use the work energy or sorry, the work done by constant force equation. Work is really just going to be f air times d, times the cosine of the angle between those two things. So your displacement's down, your force is up, therefore, this is just going to simplify. This turns into a negative one. And basically, what happens is you're going to get that the w, f air is equal to f air times d. So now we want to figure_out what's this force. We actually know what the distance is. It's just going to be my delta y. So I can go ahead and calculate this. So my f air is really just going to be the work that's done, which I know is negative 670 divided by my delta y. And my delta y is actually just 80 or it's actually rather just negative 80 because technically, it's going to go downwards like this. Also, I just want the negative signs to cancel so that I get the magnitude of the force, and the force is equal to what I get is 8.38 newtons. So that's the average force of air resistance. Alright? 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 difference between conservative and nonconservative forces?

Conservative forces, such as gravity and spring forces, are path-independent and do not dissipate mechanical energy. The work done by these forces depends only on the initial and final positions, not the path taken. Nonconservative forces, like friction and applied forces, are path-dependent and can dissipate mechanical energy as heat or other forms. The work done by nonconservative forces can add or remove energy from a system, making the total mechanical energy not conserved.

How do you incorporate nonconservative forces into the conservation of energy equation?

To incorporate nonconservative forces into the conservation of energy equation, you add a term for the work done by these forces. The full conservation of energy equation is:

$\frac{1}{2}m{v}_{\mathrm{initial}}+W=\frac{1}{2}m{v}_{\mathrm{final}}$

where $W$ represents the work done by nonconservative forces. This term accounts for any energy added or removed from the system by forces like friction or applied forces.

How do you calculate the work done by nonconservative forces?

The work done by nonconservative forces is calculated using the formula:

$W=Fdcos\theta$

where $F$ is the force applied, $d$ is the distance over which the force is applied, and $\theta$ is the angle between the force and the direction of motion. This formula accounts for the component of the force that does work in the direction of the displacement.

Why is mechanical energy not conserved when nonconservative forces are present?

Mechanical energy is not conserved when nonconservative forces are present because these forces can add or remove energy from the system. For example, friction converts mechanical energy into heat, and applied forces can increase or decrease the system's energy. This results in a change in the total mechanical energy, making it different from the initial mechanical energy. The work done by nonconservative forces accounts for this energy change.

Can you provide an example of a problem involving nonconservative forces and how to solve it?

Consider a hockey puck of mass 0.5 kg initially moving at 4 m/s. An applied force of 200 N pushes it over a distance of 0.3 m. To find the final speed, use the full conservation of energy equation:

$\frac{1}{2}m{v}_{\mathrm{initial}}+W=\frac{1}{2}m{v}_{\mathrm{final}}$

Calculate the work done by the applied force:

$W=Fdcos0=200\times 0.3\times 1=60J$

Substitute values into the energy equation:

$\frac{1}{2}\left(0.5\right)\left(4{)}^2\right)+60=\frac{1}{2}\left(0.5\right)v_{\mathrm{final}}{)}^2$

Solve for $v_{\mathrm{final}}$ to get 16 m/s.