https://youtu.be/YjWrlxDPYmM
PHYS 1101: Lecture Sixteen, Part Seven
In this example, we have an automobile that approaches a barrier at a speed. Again, let me flag that word for you. Right away, I think energy conservation might be the right way to go. It’s a speed of 20 meters per second along a level road. The driver locks the brakes at a distance of 50 meters from the barrier. Conservation of energy looking even better because I’ve got some position change information. What’s the minimum coefficient of kinetic friction that’s required to stop the automobile before it hits the barrier?
On one hand, the mention of kinetic friction here, the frictional force, might suggest that you should start out applying Newton’s Second Law to the problem. Turns out, you could apply Newton’s Second Law, but it would take you many more steps to solve it. You’ll see that with using energy arguments, you’ll just get there quicker. And again, the flag that tells you that you may be able to use energy arguments is this fact that I only care about speeds and distances. I am going to have to think about the frictional force. That’s going to play a role, in that that’s going to do negative work on this car. It’s going to take energy away from the car, which is what brings it to a stop. But, I don’t have to consider that force in terms of Newton’s Second Law. I can just consider the work that that force does, and you’ll see that this is going to lead me to an equation that has that variable Mu sub K, which is ultimately what you have to solve for.
Okay. So, I’ve recognized speed and distance. I’m going to attempt to solve it using my energy conservation ideas. My basic equation is: work due to non-conservative forces equals the change in kinetic plus the change in potential. Let me draw a sketch of it, of the picture, so I can have the right scope in mind, the initial and final instant, what the forces are, etc. Here’s my car. That’s excellent. Here’s my stretch of road. Here’s the wall that the car needs to stop in front of. This car hits the brakes, and it’s moving over this distance s. What I want is the frictional force to be large enough that it brings this car to a stop, in a distance of… It can’t be any bigger than 50 meters.
So, I’m going to look at the requirement that s is just equal to 50 meters, which will be the minimum friction that I need so that the car stops just at the wall, before it makes contact. Okay. What forces do I have on the car? I have the weight. I have a normal force. Then, I know I’ve got a friction force. I’ll draw it over here. Let me go label these forces. Normal force never does any work. Don’t have to consider it. Force due to gravity. The work that it does is going to be captured in my potential energy terms. I don’t need to worry about it on the left side.
I have a third force here, though. That’s the kinetic friction force. This force does have a component along the direction of motion. In fact, friction, as it always is, is completely opposite to the motion. It definitely does work. It does a large amount of negative work. That’s why this car slows down.
There’s no other forces directly on the car. The only thing touching the car is the road. The road is providing a normal force and kinetic friction. That’s all I’ve got.
Okay. So, here’s my object. I got my picture sketched. Let’s think about the left side of our starting equation. I need to represent the work done by everything, all forces except gravity. We said this doesn’t do any work. I don’t count gravity. I do have to count the work that friction does. So, my work due to non-conservative forces is going to include one term. That’s going to become, or it’s equal to, the work that friction does. It’s the only force. From this, then, we conclude that work due to non-conservative forces, the left side, is not 0.
Quiz question 20: Is mechanic energy conserved for this problem? For the right side, I need to think about an initial speed, a final speed, an initial altitude, and a final altitude. Okay. Initial speed. That was a given in the problem. The car starts out going 20 meters per second. What’s my initial altitude value? Well, I need to set my 0. That would be my lowest altitude; I’m going to call the 0 height. Well, this is a problem and example where the height doesn’t change. I’m just going to call that height 0. Both at the start and at the end of this problem, those two instances, my altitude is the same. I’m going to call it 0. Both of these are 0.
Both of these potential energy terms are just going to drop out. Only when I have a height change or an altitude change is gravitational work going to come into play. Therefore, do I need, and I’m going to have these potential energy terms to capture that. If ever a problem involves work and speeds and distance change, where the motion is all horizontal, potential energy will always just drop out. You can set that to 0. Okay. That’s enough thinking to figure out, think about the right side of the equation.
Our list of knowns. Again, I have most of them listed there, h0 and hf, those are both 0. What other info do I have? I only have the distance of 50 meters, s is 50. Again, I don’t know the mass. Let’s hope it drops out, as it has before. The thing that I want, what variable represents that? I need to determine the minimum coefficient of kinetic friction. That means I want the smallest frictional force. The variable that’s going to be in that term is Muk. So, the variable I want is Muk. Alright.
Now, we’ve got our picture drawn. We’ve oriented ourselves on what we’re going to need to plug into the left side and plug into the right side here. Now, let’s get to work. Let’s write our main equation. Right away, before I start doing any substitutions or anything, I know I can kill off a lot of these terms. The motion is purely horizontal for this problem. That’s my lowest altitude. Both of my potential energy terms just drops out. I realize I had a typo in that equation. Let me scan up and fix that typo. My apologies. It’s always the final minus the initial.
Okay. I have another 0 here to get rid of. The final velocity is 0. In fact, I forgot to write that in my list of knowns and thinking about the right side. I know that because I’m picturing the scope where the car comes to a stop just before it hits the wall. So, the final speed will be 0. Let me add that here to my list. Final speed 0. That means my final kinetic energy is going to be 0. Okay. Simplifies quite a bit. Let’s rewrite it. See what we have.
Okay. Let’s read that. This says that whatever work gets done that’s not gravity, that that energy is equal to the negative of the initial kinetic energy. In essence, friction is the only force that’s doing work here, and it has to take away or remove all of the kinetic energy from this car. Let’s go in and keep customizing, and see how that evolves or emerges. The only work due to non-conservative forces, as we argued above, is the work due to friction. That, I’m just plugging it in for the left. What is the work due to friction? If you go back previously in our lecture, we know that that’s minus the size of the friction force times the distance.
Okay. Before I plug in even much more, I’m going to do a little simplifying here and cancel out those sign differences. The negative there, remember, meant that friction takes energy away from this object. That’s why it’s equal to the negative of the kinetic energy. So, let me rewrite it. For kinetic energy initial, I’m going to go ahead and make that substitution or break that down. It’s equal to 1/2 mv0 squared. I don’t have a direct number for the friction force right now. I have numbers for everything else. Oops. I don’t have a number for m. Let me go break this down.
What is the friction force? How large is it? Well, it’s always the coefficient times the normal force. Okay. Again now, I need to break it down further. What’s my normal force? To do that, let me go back and do a quick visual glance here at my forces. I only have these three. I’m in equilibrium vertically. So, my normal force equals mg. It’s the only vertical force that’s countering mg, balancing it. So, for this, I can substitute in mg. So, I end up with Muk times mg times s is equal to 1/2 mv0 squared. We see again that the mass cancels out. Very common in these problems.
So, it turns out that this coefficient of friction that we’re trying to solve for doesn’t depend on the mass of the car. I need the same kind of tires and roughness between the tire and the road for a small car as I do for a big truck, in order to ensure that it’s going to stop within that 50 meters. It’s kind of counter-intuitive. So, I’m trying to solve for the coefficient of kinetic friction. I need to divide both sides then by gs. Then, I have that my coefficient is going to be given by 1/2 v0 squared divided by g times s. 1/2 then times my initial speed was 20 meters per second. I have to square that. Then, I have to divide it by 9.8 meters per second squared times my distance. That was 50 meters. That was the distance I needed to stop over to ensure I didn’t hit the wall.
When I multiply those numbers out, I end up with 0.41. Notice that the units all cancel, which is good. These coefficients should always be unit-less, dimensionless. I have meters squared, seconds squared in the numerator, and I have meters squared, seconds squared in the denominator. So, I feel pretty good about that value. It’s a number between 0 and 1. It’s physically reasonable. No units. That’s the best checking that I know how to do.