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If a curve with a radius of 80 m is perfectly banked for a car traveling 75 km per hour, what must the coefficient of static friction for the car not to skid when traveling at 95 km per hour?

a. Insufficient information.
b. 0.21.
c. 0.14.
d. 0.07.

1 Answer

4 votes

Final answer:

The coefficient of static friction necessary for a car that travels at 95 km/h on a curve banked for 75 km/h with a radius of 80 m cannot be determined without additional information about the bank angle or the forces at the ideal speed. Therefore, the correct answer is (a) Insufficient information.

Step-by-step explanation:

To calculate this, we can use the relationship between the force of static friction (Fs), centripetal force (Fc), and the component of gravitational force along the banked curve. At the ideal speed, no friction is needed because the bank angle provides the necessary centripetal force. However, when speed increases, static friction is required to provide the excess centripetal force. The centripetal force needed for the car to turn is given by ω=((v^2)/r), where ω is the centripetal force, v is velocity, and r is the radius. If the centripetal force exceeds the force provided by the banking of the curve alone, the difference must be provided by static friction, Fs=ω-(Force provided by banking). From this, we can find the coefficient of static friction (μ) using μ=Fs/N, where N is the normal force (equal to the component of gravitational force perpendicular to the banked curve). The coefficient of friction (mu) when the car travels faster than the ideal speed can be found using these equations and the specified speed of 95 km/h (which converts to 26.39 m/s). For this question, the relevant information to calculate the coefficient of friction is not provided, which likely leads to answer choice (a): Insufficient information.

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