Stopping Distance (Friction and Kinematics)

[Chandasouk]

Homework Statement

If the coefficient of kinetic friction between tires and dry pavement is 0.800, what is the shortest distance in which an automobile can be stopped by locking the brakes when traveling at 27.9 m/s?
Take the free fall acceleration to be g = 9.80 m/s^2.


On wet pavement the coefficient of kinetic friction may be only 0.250. How fast should you drive on wet pavement in order to be able to stop in the same distance as in part (a)? (Note: Locking the brakes is not the safest way to stop.)

For the first part, I did

Weight = mg -> m(-9.80) = w

And Fn = 9.80m

I found the frictional force = \muk*Fn

Frictional Force = (.800)*9.80m = 7.84m

From here what should I do? Fnet = ma so therefore

ma=7.84m

a = 7.84m/s^2

Then using Vfinal^2=Vinitial^2+2a\Deltax

\Deltax= 49.6 m ?

 

[Dick]

Sure. That's fine. If you are writing it up to turn in I'd suggest putting units on numbers that have units. Like '9.8' should be '9.8*m/s^2'. To avoid confusion with m=mass and m=meters you might want to use a different symbol for mass.

 

[Chandasouk]

For part B, is the answer 15.6m/s?

Frictional Force = (.250)(9.80m) = 2.45m

ma=2.45m

a=2.45m/s^2

Vf = 0

and plugging in the values

Vfinal^2=Vinitial^2+2a/\ x

Vintial = 15.6m/s ?

The thing is, if I solved it using that equation, I ended up having to square root a negative number which is not possible to do. I had to square root -243.04 but instead just squared 243.04 to get my answer. Is it even correct? If so, where did I go wrong with ending up with a negative sign?

 

[Dick]

Your answer is right again. Vfinal should be 0. And in both cases 'a' should correctly speaking, be negative. You are decelerating. You should have run into the same problem in the first case, if you were really paying close attention to signs. I wasn't.