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

- 0. Math Review31m
- 1. Intro to Physics Units1h 23m
- 2. 1D Motion / Kinematics3h 56m
- Vectors, Scalars, & Displacement13m
- Average Velocity32m
- Intro to Acceleration7m
- Position-Time Graphs & Velocity26m
- Conceptual Problems with Position-Time Graphs22m
- Velocity-Time Graphs & Acceleration5m
- Calculating Displacement from Velocity-Time Graphs15m
- Conceptual Problems with Velocity-Time Graphs10m
- Calculating Change in Velocity from Acceleration-Time Graphs10m
- Graphing Position, Velocity, and Acceleration Graphs11m
- Kinematics Equations37m
- Vertical Motion and Free Fall19m
- Catch/Overtake Problems23m

- 3. Vectors2h 43m
- Review of Vectors vs. Scalars1m
- Introduction to Vectors7m
- Adding Vectors Graphically22m
- Vector Composition & Decomposition11m
- Adding Vectors by Components13m
- Trig Review24m
- Unit Vectors15m
- Introduction to Dot Product (Scalar Product)12m
- Calculating Dot Product Using Components12m
- Intro to Cross Product (Vector Product)23m
- Calculating Cross Product Using Components17m

- 4. 2D Kinematics1h 42m
- 5. Projectile Motion3h 6m
- 6. Intro to Forces (Dynamics)3h 22m
- 7. Friction, Inclines, Systems2h 44m
- 8. Centripetal Forces & Gravitation7h 26m
- Uniform Circular Motion7m
- Period and Frequency in Uniform Circular Motion20m
- Centripetal Forces15m
- Vertical Centripetal Forces10m
- Flat Curves9m
- Banked Curves10m
- Newton's Law of Gravity30m
- Gravitational Forces in 2D25m
- Acceleration Due to Gravity13m
- Satellite Motion: Intro5m
- Satellite Motion: Speed & Period35m
- Geosynchronous Orbits15m
- Overview of Kepler's Laws5m
- Kepler's First Law11m
- Kepler's Third Law16m
- Kepler's Third Law for Elliptical Orbits15m
- Gravitational Potential Energy21m
- Gravitational Potential Energy for Systems of Masses17m
- Escape Velocity21m
- Energy of Circular Orbits23m
- Energy of Elliptical Orbits36m
- Black Holes16m
- Gravitational Force Inside the Earth13m
- Mass Distribution with Calculus45m

- 9. Work & Energy1h 59m
- 10. Conservation of Energy2h 51m
- Intro to Energy Types3m
- Gravitational Potential Energy10m
- Intro to Conservation of Energy29m
- Energy with Non-Conservative Forces20m
- Springs & Elastic Potential Energy19m
- Solving Projectile Motion Using Energy13m
- Motion Along Curved Paths4m
- Rollercoaster Problems13m
- Pendulum Problems13m
- Energy in Connected Objects (Systems)24m
- Force & Potential Energy18m

- 11. Momentum & Impulse3h 40m
- Intro to Momentum11m
- Intro to Impulse14m
- Impulse with Variable Forces12m
- Intro to Conservation of Momentum17m
- Push-Away Problems19m
- Types of Collisions4m
- Completely Inelastic Collisions28m
- Adding Mass to a Moving System8m
- Collisions & Motion (Momentum & Energy)26m
- Ballistic Pendulum14m
- Collisions with Springs13m
- Elastic Collisions24m
- How to Identify the Type of Collision9m
- Intro to Center of Mass15m

- 12. Rotational Kinematics2h 59m
- 13. Rotational Inertia & Energy7h 4m
- More Conservation of Energy Problems54m
- Conservation of Energy in Rolling Motion45m
- Parallel Axis Theorem13m
- Intro to Moment of Inertia28m
- Moment of Inertia via Integration18m
- Moment of Inertia of Systems23m
- Moment of Inertia & Mass Distribution10m
- Intro to Rotational Kinetic Energy16m
- Energy of Rolling Motion18m
- Types of Motion & Energy24m
- Conservation of Energy with Rotation35m
- Torque with Kinematic Equations56m
- Rotational Dynamics with Two Motions50m
- Rotational Dynamics of Rolling Motion27m

- 14. Torque & Rotational Dynamics2h 5m
- 15. Rotational Equilibrium3h 39m
- 16. Angular Momentum3h 6m
- Opening/Closing Arms on Rotating Stool18m
- Conservation of Angular Momentum46m
- Angular Momentum & Newton's Second Law10m
- Intro to Angular Collisions15m
- Jumping Into/Out of Moving Disc23m
- Spinning on String of Variable Length20m
- Angular Collisions with Linear Motion8m
- Intro to Angular Momentum15m
- Angular Momentum of a Point Mass21m
- Angular Momentum of Objects in Linear Motion7m

- 17. Periodic Motion2h 9m
- 18. Waves & Sound3h 40m
- Intro to Waves11m
- Velocity of Transverse Waves21m
- Velocity of Longitudinal Waves11m
- Wave Functions31m
- Phase Constant14m
- Average Power of Waves on Strings10m
- Wave Intensity19m
- Sound Intensity13m
- Wave Interference8m
- Superposition of Wave Functions3m
- Standing Waves30m
- Standing Wave Functions14m
- Standing Sound Waves12m
- Beats8m
- The Doppler Effect7m

- 19. Fluid Mechanics2h 27m
- 20. Heat and Temperature3h 7m
- Temperature16m
- Linear Thermal Expansion14m
- Volume Thermal Expansion14m
- Moles and Avogadro's Number14m
- Specific Heat & Temperature Changes12m
- Latent Heat & Phase Changes16m
- Intro to Calorimetry21m
- Calorimetry with Temperature and Phase Changes15m
- Advanced Calorimetry: Equilibrium Temperature with Phase Changes9m
- Phase Diagrams, Triple Points and Critical Points6m
- Heat Transfer44m

- 21. Kinetic Theory of Ideal Gases1h 50m
- 22. The First Law of Thermodynamics1h 26m
- 23. The Second Law of Thermodynamics3h 11m
- 24. Electric Force & Field; Gauss' Law3h 42m
- 25. Electric Potential1h 51m
- 26. Capacitors & Dielectrics2h 2m
- 27. Resistors & DC Circuits2h 7m
- 28. Magnetic Fields and Forces2h 23m
- 29. Sources of Magnetic Field2h 30m
- Magnetic Field Produced by Moving Charges10m
- Magnetic Field Produced by Straight Currents27m
- Magnetic Force Between Parallel Currents12m
- Magnetic Force Between Two Moving Charges9m
- Magnetic Field Produced by Loops and Solenoids42m
- Toroidal Solenoids aka Toroids12m
- Biot-Savart Law (Calculus)18m
- Ampere's Law (Calculus)17m

- 30. Induction and Inductance3h 37m
- 31. Alternating Current2h 37m
- Alternating Voltages and Currents18m
- RMS Current and Voltage9m
- Phasors20m
- Resistors in AC Circuits9m
- Phasors for Resistors7m
- Capacitors in AC Circuits16m
- Phasors for Capacitors8m
- Inductors in AC Circuits13m
- Phasors for Inductors7m
- Impedance in AC Circuits18m
- Series LRC Circuits11m
- Resonance in Series LRC Circuits10m
- Power in AC Circuits5m

- 32. Electromagnetic Waves2h 14m
- 33. Geometric Optics2h 57m
- 34. Wave Optics1h 15m
- 35. Special Relativity2h 10m

# Kinetic Friction - Online Tutor, Practice Problems & Exam Prep

Kinetic friction, denoted as *f*_{k}, is a resisting force that occurs when two rough surfaces slide against each other. It acts opposite to the direction of motion, with its magnitude calculated using the equation f_{k }= μ_{k }× N, where *μ*_{k} is the coefficient of kinetic friction and *N* is the normal force. The coefficient varies between 0 (smooth surfaces) and 1 (rough surfaces), influencing the frictional force experienced during motion.

### Kinetic Friction Problems

#### Video transcript

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?

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?

### Pushing Down on a Block with Friction

#### Video transcript

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*. So that means your force is equal to μkN. Remember that is the equation for kinetic friction. We have

_{k}*μ*= 0.3. What about this normal force here? You might be tempted to write

_{k}*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*. 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

_{y}*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.

## Do you want more practice?

More sets### Your Physics tutor

- In a laboratory experiment on friction, a 135-N block resting on a rough horizontal table is pulled by a horiz...
- A box of bananas weighing 40.0 N rests on a horizontal surface. The coefficient of static friction between the...
- 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)...
- A 45.0-kg crate of tools rests on a horizontal floor. You exert a gradually increasing horizontal push on it, ...
- (II) Police investigators, examining the scene of an accident involving a car and an old truck, measure 72-m-l...
- So-called volcanic 'ash' is actually finely pulverized rock blown high into the atmosphere. A typical ash part...
- What is the magnitude of the acceleration of a skydiver at the instant she is falling at one-half her terminal...
- An E. coli bacterium can be modeled as a sphere that has the density of water. Rotating flagella propel a bact...
- An object with cross section A is shot horizontally across frictionless ice. Its initial velocity is v₀ₓ at t₀...
- A 1.0-cm-diameter, 2.0 g marble is shot horizontally into a tank of 20°C olive oil at 10 cm/s. How far in cm w...
- b. A 4.0-cm-diameter, 55 g ball is shot horizontally into a tank of 40°C honey. How long will it take for the ...
- a. A spherical particle of mass m is shot horizontally with initial speed v₀ into a viscous fluid. Use Stokes'...
- A medium-sized jet has a 3.8-m-diameter fuselage and a loaded mass of 85,000 kg. The drag on an airplane is pr...
- So-called volcanic 'ash' is actually finely pulverized rock blown high into the atmosphere. A typical ash part...
- (b) Below what speed does a 3.0-mm-diameter ball bearing in 20°C air experience linear drag?
- (a) Above what speed does a 3.0-mm-diameter ball bearing in 20°C water experience quadratic drag?
- A ball is shot from a compressed-air gun at twice its terminal speed. a. What is the ball's initial accelerat...
- A 10 kg crate is placed on a horizontal conveyor belt. The materials are such that mu(s) = 0.5 and mu(k) = 0.3...
- A 250 g ball is launched with a speed of 35 m/s at a 30° angle. A strong headwind exerts a constant horizontal...
- A 68-kg water skier is being accelerated by a ski boat on a flat ('glassy') lake. The coefficient of kinetic f...
- A 68-kg water skier is being accelerated by a ski boat on a flat ('glassy') lake. The coefficient of kinetic f...
- (II) A package of mass m is dropped vertically onto a horizontal conveyor belt whose speed is v=1.5m/s , and t...
- (II) A package of mass m is dropped vertically onto a horizontal conveyor belt whose speed is v=1.5m/s , and t...
- A 68-kg water skier is being accelerated by a ski boat on a flat ('glassy') lake. The coefficient of kinetic f...