# How Do Forces, Friction, and Slopes Affect Automobile Movement?

• robdude
In summary, the speaker is a Comp Sci major who hasn't touched physics since high school. They are trying to finish their sick girlfriend's physics homework due in less than 30 minutes. They have two questions left and are asking for help. The first question asks for the total retarding force needed to drive a 1900-kg car at 60.0 km/h on a level road, which is 671 N. The second question asks for the power needed to drive the car up a 10.0% grade, which is 24.9 HP, and down a 1.00% grade, which is 13.8 HP. The final question asks for the grade at which the car would coast at
robdude
Hi all,

Let me first apologize for the nature of my post. If this sort of thing isn't welcome here, just say the word. The truth is, I haven't touched physics since my second year of high school (that was many years ago, I went on to college a Comp Sci major, graduated). And yet, here I am, doing homework. :)

Long story short, my girlfriend has a homework assignment that is due at midnight (less than 30 minutes). She is sick and only finished about half of the assignment before going to bed. For the last three hours I've been trying to do her homework for her. It's been going good, but very, very slowly. I have two questions remaining and so far, I'm averaging one question every hour or so.

If someone could help me finish this problem, I would be eternally grateful. I'm not sure what level of traffic this site gets, if this post is more than 30 minutes old when you see, no sense in wasting your time. But if you see this right away and can do these sorts of things...

If 15.0 HP are required to drive a 1900-kg automobile at 60.0 km/h on a level road, what is the total retarding force due to friction, air resistance, and so on?

671 N (This is correct and was as far as I've gotten so far...)

What power is necessary to drive the car at 60.0 km/h up a 10.0% grade (a hill rising 10.0m vertically in 100.0m horizontally)?

What power is necessary to drive the car at 60.0 km/h down a 1.00% grade?

Down what percent grade would the car coast at 60.0 km/h?

Any help would be much appreciated. Thanks!The power to drive the car up a 10.0% grade is 24.9 HP.The power necessary to drive the car down a 1.00% grade is 13.8 HP.The car will coast at 60.0 km/h on a -3.28% grade.

Hi there,

First of all, it is not appropriate to ask for help with homework on this forum. However, since you have already put in effort and have a specific question related to physics, I will provide a brief response.

The total retarding force on the car can be calculated using the equation F = ma, where F is the force, m is the mass of the car, and a is the acceleration. In this case, the acceleration is 0 since the car is moving at a constant speed. Therefore, the total retarding force is equal to the force required to overcome friction and air resistance. In this case, it is 671 N.

To calculate the power required to drive the car up a 10.0% grade, we can use the equation P = Fv, where P is power, F is force, and v is velocity. The force required to overcome the slope can be calculated using the equation F = mg(sinθ), where m is the mass of the car, g is the acceleration due to gravity, and θ is the angle of the slope. In this case, the angle is 10.0% or 0.1. Plugging in the values, we get F = 1900 kg x 9.8 m/s^2 x sin(0.1) = 186.2 N. Multiplying this by the velocity of 60.0 km/h (16.67 m/s), we get a power of 3107 Watts.

To calculate the power required to drive the car down a 1.00% grade, we can use the same equation as above but with a negative force since the car is going downhill. The force required can be calculated using the same equation as above, but with a negative angle of -0.01. This results in a power of -33.3 Watts.

The grade at which the car would coast at 60.0 km/h can be calculated using the equation F = mg(sinθ) = 0, since the car is coasting and there is no net force acting on it. Solving for θ, we get θ = 0. Therefore, the car would coast at 0% grade or on a level road at 60.0 km/h.

I hope this helps you understand the concepts of forces, friction, and slopes better. Remember, it is important to put in effort and try to understand the material rather than

## 1. What is the difference between a force and a friction?

A force is a push or pull on an object, while friction is a force that opposes the motion of an object. In other words, force is what causes an object to move, while friction is what causes an object to slow down or stop.

## 2. How does the slope of a surface affect the amount of friction?

The steeper the slope, the greater the amount of friction. This is because a steeper slope means there is a greater force pulling the object downhill, which results in a higher amount of friction opposing that force.

## 3. What factors affect the amount of friction between two surfaces?

The two main factors that affect friction are the types of materials involved and the force pressing the surfaces together. Rougher surfaces and heavier weights generally result in more friction.

## 4. How can friction be reduced?

Friction can be reduced by using a lubricant, such as oil or grease, between two surfaces. Another way to reduce friction is by using smoother surfaces or reducing the force pressing the surfaces together.

## 5. How is the force of friction calculated?

The force of friction can be calculated using the equation F = μN, where F is the force of friction, μ is the coefficient of friction, and N is the normal force (the force pressing the surfaces together). The coefficient of friction is a measure of how rough or smooth the surfaces are in contact with each other.

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