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

If all the energy the car has while in motion is dissipated as work then:

E_K = W where E_K is kinetic energy and W is work (due to friction)

1/2mv² = F_Fd where "m" is the mass of the car, "v" is the velocity of the car, FF is the force of friction, and "d" is the distance (stopping distance)

But F_F = μF_N where μ is the coefficient of friction and F_N is the normal force.

Summing the forces in the y-direction, we see that F_N = F_G (F_G is the force of gravity) and we know that F_G = mg therefore, substituting:

F_F = μF_N

F_F = μF_G

F_F = μmg

1/2mv² = μmgd

1/2v² = μgd

d = v²/(2μg)

Change 54 mi/hr into m/s:

54 mi/hr = 24.14 m/s

d = v²/(2μg) so just plug in the numbers