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Suppose a NASCAR race car rounds one end of the Martinsville Speedway. This end of the track is a turn with a radius of approximately 57.0 m . If the track is completely flat and the race car is traveling at a constant 27.5 m/s (about 62 mph ) around the turn, what is the race car's centripetal (radial) acceleration? What is the Coefficient of friction?

User B K
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Answer:

Step-by-step explanation:

The centripetal acceleration of the race car is given by the formula:

a = v^2 / r

where v is the speed of the race car and r is the radius of the turn.

Substituting the given values, we get:

a = (27.5 m/s)^2 / 57.0 m = 13.3 m/s^2

So the centripetal acceleration of the race car is 13.3 m/s^2.

To find the coefficient of friction, we need to use the formula:

f = μN

where f is the force of friction, μ is the coefficient of friction, and N is the normal force.

The normal force is equal to the weight of the car, which we can calculate as:

N = mg

where m is the mass of the car and g is the acceleration due to gravity (9.81 m/s^2).

Assuming the mass of the car is 1500 kg, we get:

N = 1500 kg × 9.81 m/s^2 = 14,715 N

The force of friction is equal to the centripetal force required to keep the car moving in a circle:

f = ma = (1500 kg)(13.3 m/s^2) = 19,950 N

Substituting the values of N and f into the formula for friction, we get:

19,950 N = μ(14,715 N)

Solving for μ, we get:

μ = 1.35

So the coefficient of friction is 1.35.

User Aldin Bradaric
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