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Finding the coefficient of frictional force

  • #1
You are driving 2500 Kg car at a constant speed of 14 m/s along an icy, but stright and lvel road. While approaching a traffic light, it turns red. You slam on the brakes. You wheels lock, the tires begin skidding and the car slides to a halt in a distance of 25 m. What is the coefficient of sliding friction between the tires and the icy road?

-------------------------------------
Calculations:

m = 2500 Kg
Vi = 14 m/s
Vf = 0 m/s
deltaD = 25 m
a = (vf^2 - vi^2)/(2 x deltaD)
= -14 m/s / 50
= 2.8 m/s^2

W = (2500) (9.81)
= 24525 N

Fnet = (2500)(-0.28)
= -700 N

I`m not sure what to do after this.
 

Answers and Replies

  • #2
Nabeshin
Science Advisor
2,205
16
Looks good so far.

You know that friction is the force producing this acceleration (and thus net force), so what is the equation for the force of friction?
 
  • #3
Looks good so far.

You know that friction is the force producing this acceleration (and thus net force), so what is the equation for the force of friction?
F = uFn

700 = u(24525)

Is that what I am supposed to be doing?
 
  • #4
Nabeshin
Science Advisor
2,205
16
F = uFn

700 = u(24525)

Is that what I am supposed to be doing?
Yup. Have a little confidence :)
 
  • #5
Yup. Have a little confidence :)
I have an answer sheet and it says the answer is supposed to be 0.4 so I'm a little confused.
 
  • #6
Nabeshin
Science Advisor
2,205
16
Sorry I wasn't checking your arithmetic, the approach was correct though.

The acceleration should be 3.92m/s^2:
[tex]14^2/(50)=3.92[/tex]

Also, for some reason in your f=ma you used your a value divided by ten instead of the a you derived...

But yeah, see if you can get the .4 now.
 
  • #7
Sorry I wasn't checking your arithmetic, the approach was correct though.

The acceleration should be 3.92m/s^2:
[tex]14^2/(50)=3.92[/tex]

Also, for some reason in your f=ma you used your a value divided by ten instead of the a you derived...

But yeah, see if you can get the .4 now.
Thank you so much :)
 

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