An 805-kg race car can drive around an unbanked turn at a maximum speed of 54 m/s without slipping. the turn has a radius of curvature of 155 m. air flowing over the car's wing exerts a downward-pointing force (called the downforce) of 10600 n on the car. (a) what is the coefficient of static friction between the track and the car's tires? (b) what would be the maximum speed if no downforce acted on the car?

This is about relationship between centripetal force and frictional force.

A) Coefficient of static friction = 0.312

B) Max speed with no downward force = 14.16 m/s

Mass of car; m = 80 kg

Max speed; v = 54 m/s

Radius of curvature; R = 155 m

Downward force; D = 10600 N

F = mv²/R

Where;. v is velocity

m is mass

R is radius of curvature

Now, since there is a downward force, the normal force will be;

N = mg + D

Thus, frictional force is given by the formula; F = μN = μ(mg + D)

Where μ is coefficient of static friction

This frictional force will be equal to the centripetal force.

Thus;

μ(mg + D) = mv²/R

Making μ the subject, we have;

μ = mv²/(R(mg + D))

Plugging in the relevant values, we have;

μ = (80 × 54²)/(155((80 × 9.8) + 10600))

μ = 0.132

Thus;

N = mg + 0

N = mg

Thus;

μmg = mv²/R

m will cancel out to give;

μg = v²/R

v² = μgR

v² = 0.132 × 9.8 × 155

v² = 200.508

v = √200.508

v = 14.16 m/s

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