Coefficient of Friction vs Braking?

Question

Okay, with a coeffeicient of friction of 1 the braking time from 50-0 is 0.71 seconds in a 1000 kg car. To get the braking time with a coefficient of 0.5 do I divide the 0.71 by 0.5? please help

thanks jeff =) kinda figured that, but its a take home exam so i dont wanna make a mistake

Answer 1

This about what a change in the Coefficient of friction effects.
With a higher coefficient of friction, the frictional force acting on the object increases,
F_f = mu * F_n
where F_f is the frictional force, F_n is the normal force, and mu is the coefficient of friction.
If we hold F_n constant, an increase in mu results in an increase in F_f.

What does the frictional force do? The frictional force is the force which acts to slow the object down.
F = ma
A force causes an acceleration,
in this case the force is the frictional force which causes an acceleration (slows the car down).
With a higher frictional force, the resulting acceleration (holding m constant) is higher.
Therefore a higher coefficient of friction translates into a higher frictional force which means a higher acceleration, which means a quicker stopping time.

So now take the opposite case, a lower coefficient of friction.
With a lower value of mu, the frictional force is reduced, and additionally, so is the accereration acting to slow/stop the car.
If the coefficient of friction is 1/2 its original value, the the frictional force is 1/2 its original value, which means the acceleration is 1/2 its original value. But if something is accerating at only 1/2 its original value, it will take twice as long to experience the same change in velocity.

So in other words, the car will take twice as long to stop. You need to divide the original stopping time (.71 seconds) by the factor that the coefficient of friction changes (by 1/2).

old mu * x = new mu
1 * x = .5,
x = .5

Old time / x = new time,
.71 s / .5 = 1.42 s

.71 seconds / 1/2 = 1.42 seconds would be the new stopping time.