Question

A 940-kg race car can drive around an unbanked turn at a maximum speed of 43 m/s without slipping. The turn has a radius of 180 m. Air flowing over the car's wing exerts a downward-pointing force (called the downforce) of 13000 N on the car. (a) What is the coefficient of static friction between the track and the car's tires

Answer 1

0.74

Step-by-step explanation:

Ft = 13000N

M = 940kg

V = 43 m/s

r = 180m

Force acting (downward) = Force acting on friction

F = mv² / r

F = (940 * 43²) / 180

F = 9655.88N

Coefficient of static friction (μ) = force acting downwards / force acting due to friction

μ = 9655.88 / 13000

μ = 0.74