1

concept

## Kinetic Friction Problems

6m

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Hey guys in this video, we're gonna 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 and a later video, we'll talk about the other kind of friction. So let's get started. Kinetic friction is written with the simple F K K for kinetic. And basic what it is is it's a resisting force that happens whenever you have to 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 you 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 services 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 gonna 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. So, 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, always gonna be opposite of v. So that's the direction. But what about this magnitude? The magnitude of the equation is very straightforward. It's just this letter mu K times the normal. Right? So the normal just means that we know we have two services 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 mu sub K. This greek letter what it is, it's called the coefficient the coefficient 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 unit list number between zero and one. So, in our problems we've seen is when we have perfectly smooth surfaces, what that means is that the coefficient of kinetic friction is actually zero. There is no roughness. These two things are actually perfectly smooth and they just slide with no friction. But if you were to grab to, 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 zero than it is to one if you grab two cinder blocks or two bricks, right? Just managing, grabbing two bricks and rubbing them against each other. There's gonna be a lot of resistance, a lot of friction. And so that mu K. Is gonna be somewhere high, it's gonna be basically closer to one than it is to zero. But that's really all there is to it guys. So let's go ahead and get to this problem here. Alright, so we've got this 10 kg box that moves on this flat surface at two m per second. So I've got this box like this and I've got the 10 kg. I know it's gonna be moving to the right with the equals two, I'm told with the kinetic coefficient of friction, is it 0.4? And I want to calculate the kinetic friction force and then the acceleration. So let's get started. So, in part a what I wanna do is I want to figure out the friction force which is F of 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 blocks 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 gonna be my MG. And we look for any applied forces or tensions. We don't have any applied forces or tensions, but we know we do have a normal force because these two surfaces are in contact. And then we also have a friction force. 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 R. F. K. And this is what we want to find here. So that's the free body diagram. So the next thing we wanna do is we want to write F. Equals M. A. But we actually don't have to do that in this case because we're not trying to find an acceleration. Remember this F. F. K. Here has an equation that we've just seen before that we just saw. It's basically just mu K. Times the normal force. So you 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 is what is this end? Well, if this box is only gonna be sliding across the surface like this and these are the only two forces that act in the in the vertical direction, then they have to cancel each other because the box wouldn't be flying up into the air or crashing through the surface. So this normal is equal to this MG. If those are the only two forces. So we can do here is we can just say that this normal is actually just equal to MG. And so your friction force is just 0.4 times 10 times 9.8. So if you work this out you're gonna get 39.2 newtons. Now remember when we sulfur forces were always gonna get positive numbers and then we just look back into the diagram and see if we can make sense of the sign. We have this F. K. Here is gonna be 39.2 and it points to the left. Alright so let's get now let's uh let's go ahead and start with part B. Which is now we have to find the acceleration of the box. So now we're trying to figure out the acceleration and we have forces. Now we're going to write our F. Equals M. A. So we're only just gonna right F. Equals M. A. In the X axis. Because it only is just gonna accelerate along the flat surface. This is M. A. X. We only have one force. It's R. F. K. But remember we have to choose a direction of positive so we can just choose the up and right direction to be positive like this. And so therefore our friction is actually gonna be negative. This is gonna be negative M. K. F. K. Equals M. A. X. So now this is gonna be now negative 39.2. And then we're gonna divide over the mass when we bring this to the other side. This is gonna be 10. So this is equal to A. X. So basically we get an acceleration of negative 3.92 m per second square. This makes sense that we get a negative sign because if the velocity was positive V equals to this way, then that means the acceleration is going to be negative to bring it to a stop. Alright, so that's it for this one. Guys let me know if you have any question.

2

Problem

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

3

Problem

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

4

example

## Pushing Down on a Block with Friction

3m

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All right, guys, let's check out this problem here. We've got a 20 kg 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've got 20 kg boxes on the floor and we're pushing down on this box. I'm gonna call this F down and it's 30 Newton's. We know that the box is going to be keeping its velocity constants. V equals two. We want to figure out how hard do we have to push the box horizontally so that we can keep this box at 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 diagrams gonna look like this. We have a downwards mg and we look for any applied forces. We know there's two. We have one that acts downwards. That's f down. You know what that is? And then we have our horizontal force, which is our F that's we're trying to look for. Remember, there's two other forces. We have a normal force because it's on the floor. And then because these two surfaces are in contact and the rough we have some coefficient of friction, there's gonna 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. So that's your free body diagram now. So now we want to figure out this force here. So what we wanna do is write r f equals Emma. But first, I'm gonna pick a direction of positive. So I'm just usually going to choose up into the right to be positive. That's what we'll do here. So you're some of all forces in the X axis equals mass times acceleration. We're gonna start with the X axis because that's where that force pops up. All right, so we got our forces. When we expand our some forces, we got f. It's positive. And then we've got f k is to the left. So what about this acceleration here. What about the right side of this equation? Well, remember, we're trying to find is how hard we need to push this so that the box is moving at a constant 2 m per second if the velocity is constant and that means the acceleration is equal to zero. So, really, this is an equilibrium problem. So we've got zero here, so basically are 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 mu k times the normal. Remember, that is the equation for kinetic friction. We have the UK it's 03 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're never going to assume that normal is equal to mg when you look at your free body diagram, what you have to do is you have to look at all the forces that are acting in the vertical axis and then basically use f equals m A to solve for that Normal. So we're gonna go over to the Y axis here and solve for normal. So this is the sum of all forces in the Y Axis equals M A. Y, similar to the x axis. And acceleration is going to be zero in the Y axis because that would mean basically, the block isn't gonna go flying into the air or go crashing into the ground. That doesn't make any sense. So now you expand our forces, we've got normal. Then you've got mg, and then you've got this f down. That's basically this additional force that's pushing the block down, and that's equal to zero. So when you saw for this, you're gonna get N is equal to when you move both of these over to the other side, you're gonna get 20 times 9.8 plus 30 if you go out and work this out in your calculator, going to 2 to 26. So this is the number that you plug back into this equation here, so basically your force is equal to 0. times to 26. So if you go ahead and work this out, what you're gonna get uh, you're gonna get 67.8, and that's your answer. So you get 67.8 Newtons, is how hard you need to push this thing.

Additional resources for Kinetic Friction

PRACTICE PROBLEMS AND ACTIVITIES (8)

- In a laboratory experiment on friction, a 135-N block resting on a rough horizontal table is pulled by a horiz...
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- A box of bananas weighing 40.0 N rests on a horizontal surface. The coefficient of static friction between the...
- A 45.0-kg crate of tools rests on a horizontal floor. You exert a gradually increasing horizontal push on it, ...
- A 45.0-kg crate of tools rests on a horizontal floor. You exert a gradually increasing horizontal push on it, ...
- You throw a baseball straight upward. The drag force is proportional to υ2. In terms of g, what is the y-compo...
- You throw a baseball straight upward. The drag force is proportional to υ2. In terms of g, what is the y-compo...
- (b) If the skydiver’s daughter, whose mass is 45 kg, is falling through the air and has the same D (0.25 kg/m)...