Finding coefficient of friction

[CaptainSFS]

Homework Statement

On a level road with its brakes on, the shortest distance in which a car traveling with 80 km/hr can stop is 95 m. This shortest distance occurs when the driver uses anti-lock brakes which means that that the car brakes without skidding.

I found the deceleration to be 2.599 m/s²

I need to find the coefficient of friction between the tires and the pavement.

Homework Equations

I'm not completely sure.
F=ma?

The Attempt at a Solution

I don't know how to solve this problem without mass. My only guess would be to find two equations and set them equal to mass in order to cancel it. I can't seem to find two equations, so perhaps there's another way? Any help would be great. Thanks again.

 

[hage567]

Yes, you use F=ma. What is the equation for the frictional force? You don't need to worry about the mass, it will cancel out.

 

[CaptainSFS]

okay, so the frictional force is described by F_f = u * F_n where the normal force is mass * gravity. So... F_f = u * m * g.

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Okay, So I solved my question. The F_f equals the F in F=ma. The m's did cancel and I found u to equal 0.265. I'm not entirely sure why I can substitute that F_f in for the other F, but my guess would be because the car comes to a complete stop?

 

[hage567]

I'm not entirely sure why I can substitute that Ff in for the other F

I'm not sure what you mean. The F in F=ma is really the net force of all the forces that act on an object (in a specified direction). It is not a specific force in and of itself. So in your problem, the only force acting on the car (in the direction of motion) is the force of friction, which is slowing it down. So in this case F = F_f. Does that help?

 

[CaptainSFS]

yeah, that does make sense. I guess I didn't think about that. It is the only force acting against it. okay cool. Thanks!