physics question

Question

Suppose a car is driving along a street with a velocity of of 54.0 mi/h. The street is horizontal and is not banked.

(a) If the coefficient of static friction between street and tires on a rainy day is 0.102, what is the minimum distance in which the car will stop?

(b) What is the stopping distance when the surface is dry and 𝜇_s = 0.597?

Answer 1

Let

m (unknown) be the mass of the car,

g = 9.81 m/s² be gravitational acceleration,

v = 54.0 mi/h be the initial speed of the car,

μ = 0.102 or 0.597 be the coefficient of static friction,

x (to be computed) be the distance travelled.

When a car is breaking there is a force F acting against the direction of motion.

That force F is partially due to the breaks, air resistance, friction, but it does not matter its source.

When the force is small it is not sufficient to overcome static friction, and the wheels turn.

But if the force gets large, it overcomes static friction, the wheels lock,

and the breaks stop dissipating the kinetic energy;

instead the kinetic energy is dissipated through kinetic friction.

In general, the amount of force needed to overcome static friction is the

product of the normal force and the coefficient of static friction.

In this example that is

mgμ

Thus the maximal force F that does not cause the wheels to lock is

F = mgμ.

If the wheels lock then the force breaking the car is the force of kinetic friction, which

is smaller than static friction.

Therefore the minimal braking distance will be achieved with maximal force F, which is mgμ.

All the initial kinetic energy (which is mv²/2) needs to be dissipated by the work of the force F (which is Fx):

mv²/2 = Fx = mgμx

x = v²/2gμ

Before substituting actual numbers be careful to convert the initial speed v to m/s.