Solving Force Due to Friction on a Car

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Homework Help Overview

The discussion revolves around a physics problem involving a car experiencing friction while moving at a constant velocity. The original poster is attempting to analyze the forces acting on a 1500 kg car traveling at 25 m/s on a dry asphalt surface, specifically focusing on the force of friction and its relationship to other forces acting on the car.

Discussion Character

  • Exploratory, Assumption checking

Approaches and Questions Raised

  • The original poster calculates the normal force and frictional force but questions the implications of constant velocity and the absence of an applied force. Some participants discuss the relationship between applied force and friction, particularly in the context of static friction.

Discussion Status

Participants are exploring the implications of the problem's conditions, including the role of static friction and the relationship between forces. There is an ongoing inquiry into the original poster's reasoning about the applied force being equal to the weight of the car, with some seeking clarification on this point.

Contextual Notes

The problem specifies to neglect air resistance, which may influence the discussion about the forces acting on the car.

xxharulover
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I am having a problem with this force equation since it does not state an acceleration but there is a constant velocity. Any help?
Also it does not state any force applied. All i can figure out is normal force and gravitational force.

Here the equation:
A 1500 kg car driving over a dry asphalt surface at a constant speed of 25m/s. There is static friction because you have your anti lock breaks on.

All I know is that the applied force due to constant velocity equals the weight of the car.

From what I know Normal Force Gravitational force is equal so
(9.81 m/s^2)(1500kg)= 14,715N

Then used The equation Ff = u Fn = (0.85)(14,715N) = 12507.75 N which is the force of friction.

Since the force of the applied force and the force of friction are going in opposite directions I subtracted 14,715- 12507.5 and used the Fnet=ma equation getting 1.465 m/s^2.

Would this make sense?

The question:
What is the force of the car?
 
Last edited:
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If the applied force were greater than the force of friction your wheels would begin to slip, the problem indicates static friction. So the wheels can't be slipping and the force applied must be equal to or less than the static friction, I believe. Does it give you any info for air-resistance?
 
It says to neglect all air-resistance.
 
All I know is that the applied force due to constant velocity equals the weight of the car.

What do you mean by this exactly?
 

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