Stopping Distance with Vi and Coefficient of friction.

[kitenyos]

Homework Statement

A car is traveling at 66.9 mi/h on a horizontal highway.
The acceleration of gravity is 9.8 m/s2 .
If the coefficient of friction between road and tires on a rainy day is 0.13, what is the minimum distance in which the car will stop? (1 mi = 1.609)
Answer in units of m.
Vi=29.9m/s
mu=.13
Vf=0m/s

Homework Equations

Ff=(Fn)(mu)

The Attempt at a Solution

There is no mass so i cannot calculate the normal force and i cannot calculate the Ff. At first I thought this problem was impossible because of lack of information but i did some research and all i could find is that "the mass cancels out". I do not understand this.

 

[Quinzio]

Write down the complete formulas without plugging in the numbers.
Then see you don't need the mass.

If you still can't get it, decide yourself a mass (eg. 1000kg) then try another mass to see if the result changes.

 

[kitenyos]

Which formulas? The Ff=(FN)(mu) ?

 

[Quinzio]

The complete formula/s !
Then what do you do with Ff ?

 

[kitenyos]

Well i plugged in 1000kg for mass and got 9800N for normal force so Ff=1274N. Not i need to figure out what to do with Ff. If the mass is 1000kg then it is raveling with a force of 29900N (F=MV). So there is 29900N forcer going forward and a 1275.3N Force of friction. I do not know what i do from here.

 

[Quinzio]

F=MV ?
Really ? Force = mass x velocity ?

 

[kitenyos]

No it doesn't i was wrong. F=mass x acceleration. Acceleration is 0 because it is a constant velocity. So wouldn't that mean that force=0 because 1000kg x 0= 0. So then wouldn't there only be force of friction?

 

[Quinzio]

Yes, and friction is slowly stopping the car. How long does it take ?

 

[kitenyos]

What is the formula for that. This is just a guess. T=A/Ff? So T=29900/1274? T=23.47s? And then i used the kinematics equation DeltaX= vf+vi/2 x t and got 350.88 m

 

[kitenyos]

Yes this is correct. Thank you very much!