Coefficient of Friction of brakes (1 Viewer)

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You are driving a 2500.0-kg car at a constant speed of 14.0 m/s along an icy, but straight and level road. While approaching a traffic light, it turns red. You slam on the brakes. Your wheels lock, the tires begin skidding, and the car slides to a halt in a distance of 25.0 m. What is the coefficient ([tex]\mu[/tex])of sliding friction between your tires and the icy roadbed?

W=mg=2500.0(9.8) = 24 500 N
Normal force=Fn=W=24 500 N
Exerted force = Fa = ma = 2500.0 (-3.92) = -9800 N
How do I go on from here?
 
I'm confused by your last line:

Exerted force = Fa = ma = 2500.0 (-3.92) = -9800 N

I'm not sure what you're doing there.

You can solve this problem using conservation of energy, you know the initial and final energies of the car and any difference between the two must be due to work exerted by the frictional force.
 
We haven't learned about conservation of energy yet. By the last line, I was trying to calculate what force I am exerting on the car to make it stop.
 
Oh I see, the "Fa" is "force which causes acceleration" and not "force times acceleration." Sorry :smile:

OK, so you've calculated the acceleration that the car underwent in order to come to a stop and used that to determine the force which must have been applied to the car in order to stop it, now you need to relate that to your frictional force.

So I'll ask you a question: How do you calculate a frictional force, and what do you need to know in order to do so?
 
Is the frictional force the opposite of the force that was applied to the car in order for it to stop?
 
The frictional force is the force that stopped the car. Don't worry about the minus sign though, it's just an issue of which way you call positive. Really the equation for the frictional force should be something like:

[tex]|F| = \mu N[/tex]

... but I can't recall ever having seen it written that way.
 
Frictional Force shldn be the opposite to the direction of the force..

Frictional forces shld occur opposite to the direction of motion. In this case, friction is still acting on the back of the car. Not opposite to the force applied to the car in order for it to stop.
 
I tot [tex] \mu [/tex] was what the qn was asking?
 
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So, since [tex]F=\mu N, \mu = F/N = 9800 N/24500 N = 0.40[/tex]?
 
If the qn assumes that the icy road bed is frictionless.. then yes, the ans is 0.4..
Becoz, the only artificial friction left is the braking force.. and that will be your frictional force.
 
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sweet877 said:
So, since [tex]F=\mu N, \mu = F/N = 9800 N/24500 N = 0.40[/tex]?
Yup, that's what I got.
 
Thanks for your help!
 
My pleasure :smile:
 

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