Finding the coefficient of frictional force

[drinkingstraw]

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?

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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.

 

[Nabeshin]

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?

 

[drinkingstraw]

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?

 

[Nabeshin]

F = uFn

700 = u(24525)

Is that what I am supposed to be doing?

Yup. Have a little confidence :)

 

[drinkingstraw]

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.

 

[Nabeshin]

Sorry I wasn't checking your arithmetic, the approach was correct though.

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

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.

 

[drinkingstraw]

Sorry I wasn't checking your arithmetic, the approach was correct though.

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

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 :)