Question

To test the performance of its tires, a car
travels along a perfectly flat (no banking) circular track of radius 333 m. The car increases
its speed at uniform rate of

(USE IMAGE)

until the tires start to skid.
If the tires start to skid when the car reaches
a speed of 31 m/s, what is the coefficient of
static friction between the tires and the road?
The acceleration of gravity is 9.8 m/s^2

Answer 1

Answer:

0.45

Explanation:

Sum of forces in the y direction:

∑F = ma

N − mg = 0

N = mg

There are friction forces in two directions: centripetal and tangential.  The centripetal acceleration is:

ac = v² / r

ac = (31 m/s)² / 333 m

ac = 2.89 m/s²

The total acceleration is:

a = √(ac² + at²)

a = √((2.89 m/s²)² + (3.32 m/s²)²)

a = 4.40 m/s²

Sum of forces:

∑F = ma

Nμ = ma

mgμ = ma

μ = a / g

μ = 4.40 m/s² / 9.8 m/s²

μ = 0.45