You are driving your 1800 kg car at 25 m/s over a circular hill that has a radius of 150 m. A deer running across the road causes you to hit the brakes hard while right at the summit of the hill, and you start to skid. The coefficient of kinetic friction between your tires and the road is 0.75. What is the magnitude of your acceleration as you begin to slow?

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Welcome back, everybody. We are making observations about a toy car that is rolling on a circular vertical loop here. I'm gonna represent our car with a little rectangle here and I'm going to set up this, this little cross hair if you will, which we will use to represent our forces acting on our car. Now, on this car, you're going to have the force due to gravity and the normal force and you are also going to have a friction force asking our acting to the right. Now. This left arrow is not gonna be a force. I just wanted it to be there to denote that we are having the car moved to the left with a given velocity. Now, we are told a couple of these numbers are, well, we're told that the diameter of the loop is one m and we are told that the mass of the car is 250 g or 2500.25 kg, told that the velocity is 0.25 m per second and that there is a coefficient of friction between the car and the loop Of .45. And we are tasked with finding what the tangential acceleration is at the top of the loop. Luckily, we have a formula for this, we know that this is equal to the negative of the friction force divided by mass. But what is our friction force? Well, our friction force is just equal to our coefficient of friction times our normal force, but we don't know what our normal force is. So here's what I'm gonna do. I'm gonna use Newton's second law, but I'm gonna use it in the centripetal or circular direction. Meaning that this direction towards the center of the circle is going to be positive. We have the sum of all centripetal forces is equal to mass times our centripetal acceleration here. So let's see, we have MG minus N is equal to our mass times are centripetal acceleration. But I'm gonna sub in our centripetal acceleration with just our velocity squared over R. Now let me add end to both sides and then subtract this value from both sides. And you'll see that this cancels out and this cancels out. And what we are left with is that our normal force is equal to our mass times our gravity minus velocity squared over R. So let's sub this into this formula and then sub this into this formula here. So we have, let me change colors here real quick. That are tangential acceleration is equal to negative one over M times our coefficient of friction times our normal force, which we just found to be mass times gravity minus philosophy squared over R its mass on top. And this mass on bottom, we're going to cancel out. Meaning our tangential acceleration is just equal to the negative of our coefficient of friction times acceleration due to gravity minus tangential velocity squared all divided by our radius. Okay. Perfect. So now let's go ahead and plug in all of our values here. So we have tangential acceleration is equal to negative 0. times 9.81 minus our velocity of 0.25 squared divided by our radius or radius is just going to be half of our diameter. So one divided by two is 20.5. When you plug all of this into our calculator, we get a tangential acceleration of negative 4.4 m per second squared corresponding to our final answer. Choice of a. Thank you all so much for watching. Hope this video helped. We will see you all in the next one.

Verified Solution

Key Concepts

Centripetal Acceleration

Friction and Kinetic Friction Coefficient

Newton's Second Law of Motion