Sliding Friction Question?

Question

You are driving a 2330 kg car at a constant speed of 20.5 m/s along an icy, but straight and level road. While approaching a traffic light, it turns red. You slam on the breaks. Your wheels lock, the tires begin skidding, and the car slites to a halt in a distance of 65.0 m. What is the coefficeint of sliding friction between your tires and the road?

Thank you for any insight into this problem.

Answer 1

First, let's find the rate of acceleration (or decceleration):

V^2 = Vo^2 + 2adx
0 = 420 + 2a(65.0)
-130a = 420
a = -3.23 m/s^2 (negative because you're slowing down)

Now, let's sum up the forces in the x-direction:

Fx = max
-Ff = max
-umg = ma (Fn = mg when ay = 0) (u = coeff. of friction)
a = -ug
u = -a/g
u = -(-3.23 / 9.80)
u = 0.33

Answer 2

Hi. The best way to approach this question is to realize that the friction force slows down the car, causing it to stop in 65 m. Because the friction force is = coefficient * Normal force, you know it is constant, since the normal force does not change. Thus, there is a constant deceleration ( = Ff/m ).
Draw a velocity vs. time curve. The line intersects the velocity axis at 20.5 m/s and intersects the time axis at "t"

slope = deceleration = V/t.
area of triangle = distance so
0.5 * t * V = d... so t = 2d/V
so slope = V^2/2d

so Ff = ma = mV^2/(2d)
coef = Ff/N = Ff/(mg) = mV^2/(mg2d) = V^2/(2dg)
= (20.5m/s)^2/(2*65.0m*9.8m/s^2) = 0.33

Answer 3

The only force acting on the car is the friction of the ice. Force of friction is equal to the coefficient of friction * the normal force of the car (in this case, its weight in Newtons).
To find the force of friction, use the formula
vf^2 = vi ^2 + 2ad
where vf is the final velocity, vi is the initial velocity, d is the distance, and a (the only unknown you have) is the acceleration. Use the acceleration times the mass to find the force of friction.