A 1500-kg car rounds an unbanked curve with a radius of 52 m at a speed of 14 m/s. What minimum coefficient of friction must exist between the road and tires to prevent the car from slipping? (g = 9.8 m/s2)Select one:a.0.38b.0.28c.0.30d.0.18
Question
A 1500-kg car rounds an unbanked curve with a radius of 52 m at a speed of 14 m/s. What minimum coefficient of friction must exist between the road and tires to prevent the car from slipping? (g = 9.8 m/s2)Select one:a.0.38b.0.28c.0.30d.0.18
Solution
To solve this problem, we need to use the formula for the force of static friction, which is what keeps the car from slipping. The force of static friction is equal to the coefficient of static friction times the normal force. In this case, the normal force is equal to the weight of the car, which is its mass times the acceleration due to gravity.
The force of static friction also provides the centripetal force that keeps the car moving in a circle. The formula for centripetal force is mass times velocity squared divided by the radius of the circle.
Setting these two expressions for the force of static friction equal to each other gives us:
μN = mv²/r
where μ is the coefficient of static friction, N is the normal force, m is the mass of the car, v is the velocity of the car, and r is the radius of the curve.
We can substitute mg for N to get:
μmg = mv²/r
Solving for μ gives us:
μ = v²/(rg)
Substituting the given values into this equation gives us:
μ = (14 m/s)² / (52 m * 9.8 m/s²) = 0.38
So, the minimum coefficient of friction that must exist between the road and the tires to prevent the car from slipping is 0.38. Therefore, the answer is a. 0.38.
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