A 2000-kg car rounds an unbanked curve with a radius of 45 m at a speed of 15 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)

µ = v^2 / Rg

11.5 m/s

sike that's wrong my bad ^^^

To find the minimum coefficient of friction required to prevent the car from slipping, we can use the concept of centripetal force.

The centripetal force acting on the car is given by the equation:
Fc = mv^2 / r

where Fc is the centripetal force, m is the mass of the car, v is the velocity, and r is the radius of the curve.

In this case, Fc is provided by the frictional force between the tires and the road. The maximum frictional force that can be provided by the tires is given by the equation:
Ff = μN

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

In equilibrium, the maximum frictional force (Ff) is equal to the centripetal force (Fc). Therefore, we can equate these two equations:
Ff = Fc

Substituting the values:
μN = mv^2 / r

The normal force (N) acting on the car is equal to the weight of the car (mg), where g is the acceleration due to gravity:
μmg = mv^2 / r

Simplifying the equation gives:
μ = (v^2) / (rg)

Now let's plug in the given values:
μ = (15 m/s)^2 / (45 m * 9.8 m/s^2)

Calculating the value gives:
μ ≈ 0.323

Therefore, the minimum coefficient of friction that must exist between the road and tires to prevent the car from slipping is approximately 0.323.