- #1
Chandasouk
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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 = [tex]\mu[/tex]k*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[tex]\Delta[/tex]x
[tex]\Delta[/tex]x= 49.6 m ?