Kinetic Friction - Online Tutor, Practice Problems & Exam Prep

Kinetic Friction Problems

Guys, in this video, we're going to talk a little bit more closely about a force that we've seen before in our free body diagrams, the force of friction. And specifically, we're going to talk about kinetic friction. In a later video, we'll talk about the other kind of friction. So let's get started. Kinetic friction is written with the symbol \( f_k \), \( k \) for kinetic. And basically, what it is, it's a resisting force that happens whenever you have two rough surfaces that are rubbing or slipping or sliding past each other. So you'll see a few words for this: rub, slide, slip. The easiest example is if you take your hands and rub them against each other, and it gets warm, that's because of friction. So, up until now, we've been assuming that all of our surfaces have been frictionless. But we know that's not the case in everyday life. If you were to push a book across the table, eventually, it's going to come to a stop, and that's because of friction, this resisting force. So this kinetic friction tries to stop all motion that happens between the surfaces. For example, you have your book that's sliding to the right with some velocity, and so kinetic friction tries to stop it by acting to the left. If you were to put a book on an incline or a ramp, and it starts sliding down the ramp like that, then kinetic friction tries to stop that by going up the ramp like this. So, basically, we can see that the kinetic friction is always going to be opposite of the velocity. They're always going to be opposite of \( v \). So that's the direction. But what about this magnitude? The magnitude in the equation is very straightforward. It's just this letter \( \mu_k \) times the normal. Right? So the normal just means that we know that we have two surfaces in contact. Right? We have a normal force here and here. This friction force is actually proportional to this normal. And this other letter here, which is this Greek letter, what it is, it's called the coefficients; it's the coefficients of kinetic friction. Basically, it's just a measure of how rough these two surfaces are. It's just a property of the two surfaces that are in contact with each other, and it's basically just a unitless number between 0 and 1. So if we in our problems, what we've seen is when we have perfectly smooth surfaces, what that means is that the coefficient of kinetic friction is actually 0. There is no roughness. Right? These two things are actually perfectly smooth, and they just slide with no friction. But if you were to grab two, like, ice chunks or something like that and rub them against each other, there's not a whole lot of resistance. There's not a whole lot of friction. And this coefficient is actually pretty low. It's closer to 0 than it is to 1. And if you grab two cinder blocks or two bricks. Right? Just imagine grabbing two bricks and rubbing them against each other. There's going to be a lot of resistance, a lot of friction, and so that \( \mu_k \) is going to be somewhere high. It's going to be basically closer to 1 than it is to 0. But that's really all there is to it, guys. So let's go get to this problem here. Alright. So we've got this 10 kilogram box that moves on this flat surface at 2 meters per second. So I've got this box like this, and I've got it, the 10 kilograms. I know it's going to be moving to the right with \( v = 2 \). I'm told what the kinetic the so coefficient of friction is. It's 0.4 and I want to calculate the kinetic friction force and then the acceleration. Let's get started. So, in part a, what I want to do is I want to figure out the kinetic friction force, which is \( f_k \). The first thing I want to do is just draw a free body diagram for what's going on here. Right? I know this block is being drawn free body diagram for what's going on here. I know this block is being moved across this flat surface, but I want the free body diagram. So let's just draw it really quickly. Remember, we look for the weight force. This is going to be my \( mg \). Then we look for any applied forces or tensions. We don't have any applied forces. Now, if the box is moving to the right with some velocity, friction wants to stop that by acting to the left. So our friction force actually acts this way. This is our \( f_k \), and this is what we want to find here. So that's the free body diagram. So the next thing we want to do is we want to write \( f = ma \), but we actually don't have to do that in this case because we're not trying to find an acceleration. Remember, this \( f_k \) here has an equation that we just saw before, that we just saw. It's basically just \( \mu_k \) times the normal force. So we know this is \( \mu_k \) times the normal. So basically, if we want to figure out \( f_k \), we know this \(\mu_k \) is 0.4. Now we just have to figure out the normal force. So what happens is, what is this \( n \)? Well, if this box is only going to be sliding across the surface like this, and these

Pushing a 10-kg toolbox across the floor, you find that the box moves at a constant speed when you push horizontally with a force of 39 N. What is the coefficient of kinetic friction between the floor and the toolbox?

• A. 0.2
• B. 0.4
• C. 2.5

You push on a 3-kg box to give it an initial speed of 5 m/s across a floor. If μ_k = 0.3, how far does the box travel before coming to a stop?

• A. 8.6m
• B. 2.9m
• C. 7.7m
• D. 4.3m

Pushing Down on a Block with Friction

Guys, let's check out this problem here. We've got a 20 kilogram box that's moving along the floor, and we've got a downward force on it. So, let me go ahead and sketch this out. We have a 20 kilogram box that's on the floor, and we're pushing down on this box. I'm going to call this f down, and that's 30 newtons. We know that the box is going to keep its velocity constant, v equals 2. We want to figure out how hard we have to push the box horizontally so that we can keep this box at a constant speed. So, basically, there's another force right here, which I'll just call regular f, and that's basically what we're trying to figure out. We have the coefficient of friction. So what we want to do first is draw a proper free-body diagram. Let's go ahead and do that. So, basically, our free-body diagram is going to look like this. We have a downwards mg, and we look for any applied forces. We know there's 2. We have one that acts downwards. That's f down. We know what that is. And then we have our horizontal force, which is our f that we're trying to look for. Now remember there are 2 other forces. We have a normal force because it's on the floor. And then because these two surfaces are in contact and they're rough, we have some coefficient of friction. There's going to be some friction. Now, if this box is moving to the right, then remember, kinetic friction always has to oppose that motion. It always is in the opposite direction of velocity. So your f_k points to the left like this. That's your free-body diagram.

Now, we want to figure out this force here. So what we want to do is write our F = ma. But first, I'm going to pick a direction as positive. So I'm just usually going to choose up and to the right to be positive. That's what we'll do here. So your sum of all forces in the x-axis equals mass times acceleration. We're going to start with the x-axis because that's where that force pops up. Alright. So we got our forces. When we expand our sum forces, we got f is positive and then we've got f_k is to the left. What about this acceleration here? What about the right side of this equation? Well, remember, what we're trying to find is how hard we need to push this so that the box is moving at a constant 2 meters per second. If the velocity is constant, then that means the acceleration is equal to 0. So really, this is an equilibrium problem. So we've got 0 here. So basically, our applied force, our mystery f, has to balance out with the kinetic friction. So, basically, when you move this to the other side, it's equal to f_k. So that means your force is equal to μkN. Remember that is the equation for kinetic friction. We have μ_k = 0.3. What about this normal force here? You might be tempted to write mg in place of the normal force, but I want to warn you against that because you should never assume that normal is equal to mg. When you look at your free-body diagram, you have to look at all the forces that are acting in the vertical axis and then basically use F = ma to solve for that normal. So, we're going to go over to the y-axis here and solve for normal. This is the sum of all forces in the y-axis equals ma_y. Similar to the x-axis, an acceleration is going to be 0 in the y-axis because that would mean, basically, the block isn't going to go flying into the air or go crashing into the ground. That doesn't make any sense. So now we expand our forces. We've got normal, then you've got mg, and then you've got this f down. This basically this additional force that's pushing the block down, and that's equal to 0. So, when you solve for this, you're going to get n is equal to when you move both of these over to the other side, you're going to get 20 times 9.8 plus 30. If you go ahead and work this out in your calculator, you're going to get 226. So this is the number that you plug back into this equation here. So, basically, your force is equal to 0.3 times 226. So if you go ahead and work this out, what you're going to get is 67.8, and that's your answer. So you get 67.8 Newtons is how hard you need to push this thing.