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Work/energy problem with friction

A conservation of energy problem where all of the energy is not conserved. Created by Sal Khan.

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  • blobby green style avatar for user RShevde
    does friction always act opposite to the motion?
    (60 votes)
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    • blobby green style avatar for user Shreyan Sama
      No. Friction always opposes relative slipping between two objects. If one body is placed over another and both the bodies move with the same velocity, there is no slipping between them and friction does not act. Now if the lower body is pulled with some force and there is no friction between the ground and the lower body, the frictional force between the 2 bodies will actually push the upper body forward so that there is no relative slipping between the 2 bodies. Here friction is acting along the direction of motion.
      However if there is friction between the ground and the lower body, it will depend whether the force applied is less than the max. frictional force between the ground and the body. If the applied force is more that the max. frictional force, the frictional force between the upper body and the lower body comes into play else there is no friction between the upper and lower body as there is no relative slipping between them.
      (52 votes)
  • mr pink red style avatar for user Abijeeth
    At ,How Is Initial Energy + Work done by friction = Final energy
    (20 votes)
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    • orange juice squid orange style avatar for user Connor Hallett
      The system is losing energy to friction, and work is energy entering or exiting the system. Therefore, we will be adding all the work done on the system to the initial energy to achieve the final energy. In this problem, work done by friction will be negative because the force is applied in the opposite direction as the motion, and will decrease the initial energy, giving us a Final Energy result that is less than the initial energy because energy is lost to friction.
      (32 votes)
  • blobby green style avatar for user docbritt12
    When we are trying to find the work due to friction why are we using W=FD and not W=FDcos5? When is it appropriate to use either formula?
    (22 votes)
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    • blobby green style avatar for user anmol.panchal
      It's not W = FDcos5, it's W = FDcos(180). This is a common misconception, remember that the angle theta in W = FDcos(theta), is equal to the angle BETWEEN the force vector and the distance vector. Since the distance vector is facing the complete opposite direction of the force vector, then the angle between them will be 180 degrees. COS(180) evaluates to "-1", which is why Sal always says that the work done by friction is NEGATIVE.
      (3 votes)
  • leaf green style avatar for user doodle duck
    Is work negative when the direction of force is opposite to the direction of motion? If yes, then why do we always have a positive value for work done against resistance(which is opposite to the direction of motion)?
    (5 votes)
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  • piceratops tree style avatar for user Art
    What is a conservative force? How is it different from a non conservative force? Mathematically too?
    (6 votes)
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    • leaf grey style avatar for user Qeeko
      What Andrew has written is correct, but I thought I'd just add something.

      Let F be a force field (vector field) in three dimensions. One can say that F is a conservative force if and only if there exists a potential function ƒ for which grad(ƒ) = ∇ƒ = F. We can also say that F is a conservative force if and only if curl(F) = ∇ x F = 0.

      If the force is conservative, as Andrew has pointed out, the path of the object is irrelevant. When dealing with conservative forces, what is more interesting is the initial and the final position of the object in question. This is useful when one for instance want to calculate the work done on an object moving from point A to point B through the force field. W = ∫ F · d r along C, where C is the path of the object from A to B. Instead of calculating this line integral, one can for instance find the function ƒ and evaluate ƒ(B) - ƒ(A).
      (3 votes)
  • blobby green style avatar for user isaakzwaan
    What about the length of the ramp? Why only the height? The length can easily be much longer? And the height still the same. Doesn't that affect the potential energy?
    (3 votes)
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  • leaf blue style avatar for user Maryam
    i only wanted the whole topic on friction
    but i cant find the topic friction can anybody help me
    i want the definition of friction
    (2 votes)
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  • blobby green style avatar for user anmolpatel09
    Why do you use W=Fd to calculate the final energy and not use W=Fcos theta x d when it's on an inclined ramp? I thought you include the angle if there is one and in this case it was 5 degrees.
    (2 votes)
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  • male robot hal style avatar for user Dakshina Sarma
    what is coefficient of friction/
    (1 vote)
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    • male robot johnny style avatar for user sam fisher
      The coefficient of friction is a number that has been found through experimentation that can be put into the friction formula so that it matches the materials you are using.
      it's a way of saying weather you are pushing against rubber versus a smooth table top.
      (4 votes)
  • blobby green style avatar for user Haydee Fernandez
    You summed initial energy + work, but your units were 38.5 KJ + 8455 J, so wouldn't it actually be (3000 J/45)^1/2 in order to find the final velocity? Which would be 25.8 m/s instead of 13.7m/s? My professor says to always pay attention to units so I'm just wondering if you made a mistake or if that's okay to do.
    (2 votes)
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    • leaf red style avatar for user nachristopher98
      No, in the video to make things a little bit simpler he declares that 38455 J is roughly equivalent to 38.5 KJ because 38455/1000 is roughly equivalent to 38.5.
      then you take the work due to friction so 60N*500m and get 30000J. Since friction is always an opposing force you subtract this from the 38.5KJ and get the 8455J mentioned. This is the kinetic energy so 1/2mv^2 and you then multiply both sides by 2 and get 16910 = mv^2. The mass is 90kg so divide both sides by 90 and get v^2=187.8889. Square root this and you end up with 13.7m/s. He just consolidated the 38455J for the sake of brevity in the video. Think of the law of conservation of energy when non-conservative forces are at work as the PEi+KEi+NC(Non=conservative Force)=PEf+KEf. Hope this helps you. I am an AP Physics 1 Student and this is how we learned how to deal with non-conservative forces.
      (2 votes)

Video transcript

Welcome back. Welcome back. Welcome back. I'll now do another conservation of energy problem, and this time I'll add another twist. So far, everything we've been doing, energy was conserved by the law of conservation. But that's because all of the forces that were acting in these systems were conservative forces. And now I'll introduce you to a problem that has a little bit of friction, and we'll see that some of that energy gets lost to friction. And we can think about it a little bit. Well where does that energy go? And I'm getting this problem from the University of Oregon's zebu.uoregon.edu. And they seem to have some nice physics problems, so I'll use theirs. And I just want to make sure they get credit. So let's see. They say a 90 kilogram bike and rider. So the bike and rider combined are 90 kilograms. So let's just say the mass is 90 kilograms. Start at rest from the top of a 500 meter long hill. OK, so I think they mean that the hill is something like this. So if this is the hill, that the hypotenuse here is 500 hundred meters long. So the length of that, this is 500 meters. A 500 meter long hill with a 5 degree incline. So this is 5 degrees. And we can kind of just view it like a wedge, like we've done in other problems. There you go. That's pretty straight. OK. Assuming an average friction force of 60 newtons. OK, so they're not telling us the coefficient of friction and then we have to figure out the normal force and all of that. They're just telling us, what is the drag of friction? Or how much is actually friction acting against this rider's motion? We could think a little bit about where that friction is coming from. So the force of friction is equal to 60 newtons And of course, this is going to be going against his motion or her motion. And the question asks us, find the speed of the biker at the bottom of the hill. So the biker starts up here, stationary. That's the biker. My very artful rendition of the biker. And we need to figure out the velocity at the bottom. This to some degree is a potential energy problem. It's definitely a conservation of mechanical energy problem. So let's figure out what the energy of the system is when the rider starts off. So the rider starts off at the top of this hill. So definitely some potential energy. And is stationary, so there's no kinetic energy. So all of the energy is potential, and what is the potential energy? Well potential energy is equal to mass times the acceleration of gravity times height, right? Well that's equal to, if the mass is 90, the acceleration of gravity is 9.8 meters per second squared. And then what's the height? Well here we're going to have to break out a little trigonometry. We need to figure out this side of this triangle, if you consider this whole thing a triangle. Let's see. We want to figure out the opposite. We know the hypotenuse and we know this angle here. So the sine of this angle is equal to opposite over hypotenuse. So, SOH. Sine is opposite over hypotenuse. So we know that the height-- so let me do a little work here-- we know that sine of 5 degrees is equal to the height over 500. Or that the height is equal to 500 sine of 5 degrees. And I calculated the sine of 5 degrees ahead of time. Let me make sure I still have it. That's cause I didn't have my calculator with me today. But you could do this on your own. So this is equal to 500, and the sine of 5 degrees is 0.087. So when you multiply these out, what do I get? I'm using the calculator on Google actually. 500 times sine. You get 43.6. So this is equal to 43.6. So the height of the hill is 43.6 meters. So going back to the potential energy, we have the mass times the acceleration of gravity times the height. Times 43.6. And this is equal to, and then I can use just my regular calculator since I don't have to figure out trig functions anymore. So 90-- so you can see the whole thing-- times 9.8 times 43.6 is equal to, let's see, roughly 38,455. So this is equal to 38,455 joules or newton meters. And that's a lot of potential energy. So what happens? At the bottom of the hill-- sorry, I have to readjust my chair-- at the bottom of the hill, all of this gets converted to, or maybe I should pose that as a question. Does all of it get converted to kinetic energy? Almost. We have a force of friction here. And friction, you can kind of view friction as something that eats up mechanical energy. These are also called nonconservative forces because when you have these forces at play, all of the force is not conserved. So a way to think about it is, is that the energy, let's just call it total energy. So let's say total energy initial, well let me just write initial energy is equal to the energy wasted in friction-- I should have written just letters-- from friction plus final energy. So we know what the initial energy is in this system. That's the potential energy of this bicyclist and this roughly 38 and 1/2 kilojoules or 38,500 joules, roughly. And now let's figure out the energy wasted from friction, and the energy wasted from friction is the negative work that friction does. And what does negative work mean? Well the bicyclist is moving 500 meters in this direction. So distance is 500 meters. But friction isn't acting along the same direction as distance. The whole time, friction is acting against the distance. So when the force is going in the opposite direction as the distance, your work is negative. So another way of thinking of this problem is energy initial is equal to, or you could say the energy initial plus the negative work of friction, right? If we say that this is a negative quantity, then this is equal to the final energy. And here, I took the friction and put it on the other side because I said this is going to be a negative quantity in the system. And so you should always just make sure that if you have friction in the system, just as a reality check, that your final energy is less than your initial energy. Our initial energy is, let's just say 38.5 kilojoules. What is the negative work that friction is doing? Well it's 500 meters. And the entire 500 meters, it's always pushing back on the rider with a force of 60 newtons. So force times distance. So it's minus 60 newtons, cause it's going in the opposite direction of the motion, times 500. And this is going to equal the ending, oh, no. This is going to equal the final energy, right? And what is this? 60 times 500, that's 3,000. No, 30,000, right. So let's subtract 30,000 from 38,500. So let's see. Minus 30. I didn't have to do that. I could have done that in my head. So that gives us 8,455 joules is equal to the final energy. And what is all the final energy? Well by this time, the rider's gotten back to, I guess we could call the sea level. So all of the energy is now going to be kinetic energy, right? What's the formula for kinetic energy? It's 1/2 mv squared. And we know what m is. The mass of the rider is 90. So we have this is 90. So if we divide both sides. So the 1/2 times 90. That's 45. So 8,455 divided by 45. So we get v squared is equal to 187.9. And let's take the square root of that and we get the velocity, 13.7. So if we take the square root of both sides of this, so the final velocity is 13.7. I know you can't read that. 13.7 meters per second. And this was a slightly more interesting problem because here we had the energy wasn't completely conserved. Some of the energy, you could say, was eaten by friction. And actually that energy just didn't disappear into a vacuum. It was actually generated into heat. And it makes sense. If you slid down a slide of sandpaper, your pants would feel very warm by the time you got to the bottom of that. But the friction of this, they weren't specific on where the friction came from, but it could have come from the gearing within the bike. It could have come from the wind. Maybe the bike actually skidded a little bit on the way down. I don't know. But hopefully you found that a little bit interesting. And now you can not only work with conservation of mechanical energy, but you can work problems where there's a little bit of friction involved as well. Anyway, I'll see you in the next video.