What is the maximum speed with which a 1150-kg car can round a turn of radius 71 m on a flat road if the coefficient of static friction between tires and road is 0.80?

May someone walk me through this problem step by step? Thank you

*The equation friction coefficient= v^2/Rg does not make any sense. Because, I do not have a velocity.

mv²/R= F(fr)

F(fr)=μN=μmg
mv²/R= μmg
μ= v²/Rg
Follow up
v= sqrt(μ•R•g) =
=sqrt(0.8•71•9.8) =23.6 m/s

To find the maximum speed at which a car can round a turn without slipping, you can use the equation for the maximum static friction force:

Maximum static friction force (fs) = coefficient of static friction (μs) * normal force (N)

In this case, the normal force is equal to the weight of the car, which is given by the formula:

Normal force (N) = mass (m) * gravitational acceleration (g)

Given:
Mass of the car (m) = 1150 kg
Radius of the turn (R) = 71 m
Coefficient of static friction (μs) = 0.80
Gravitational acceleration (g) = 9.8 m/s^2 (approximate value)

1. Calculate the normal force:
N = m * g = 1150 kg * 9.8 m/s^2 = 11270 N

2. Determine the maximum static friction force:
fs = μs * N = 0.80 * 11270 N = 9016 N

3. The maximum static friction force provides the centripetal force required for the car to round the turn without slipping. The centripetal force is given by the formula:

Centripetal force (Fc) = mass (m) * velocity (v)^2 / radius (R)

Setting the maximum static friction force equal to the centripetal force, we get:

fs = Fc
9016 N = (1150 kg * v^2) / 71 m

4. Rearrange the equation to solve for velocity (v):
v^2 = (fs * R) / m
v^2 = (9016 N * 71 m) / 1150 kg
v^2 = 558256 / 1150
v^2 ≈ 485.22

5. Take the square root of both sides to find the velocity:
v ≈ √(485.22) ≈ 22.03 m/s

Therefore, the maximum speed with which the 1150-kg car can round the turn of radius 71 m is approximately 22.03 m/s.