An unbanked curve has a radius of 80.0 m. What is the naximum speed at which a car can make the turn if the coefficient of static friction is 0.8?

Ff=Fx

mg*mu=mv^2/r solve for v

Thank you

Vmax=√musrg

To find the maximum speed at which a car can make the turn on an unbanked curve, we need to consider the forces acting on the car during the turn.

In this case, the primary force at play is the friction force between the car's tires and the road. The maximum friction force that can exist between two surfaces is given by the equation:

Friction force (max) = coefficient of friction × normal force

Here, the coefficient of friction is given as 0.8. To calculate the normal force, we can consider the vertical forces acting on the car during the turn. Since the curve is unbanked, the vertical forces are the weight (mg) and the normal force (N) acting in the opposite direction.

The normal force can be calculated as:

Normal force (N) = weight (mg)

Now, let's calculate the normal force:

Weight (mg) = mass × acceleration due to gravity

Let's assume the mass of the car is 1000 kg and the acceleration due to gravity is 9.8 m/s^2:

Weight (mg) = 1000 kg × 9.8 m/s^2 = 9800 N

Therefore, the normal force (N) is 9800 N.

Now, let's calculate the maximum friction force:

Friction force (max) = coefficient of friction × normal force
= 0.8 × 9800 N
= 7840 N

The maximum friction force (Fmax) is 7840 N.

During the turn, the centripetal force acting on the car is provided by the friction force:

Centripetal force = Friction force (max)

Also, the centripetal force is given by the equation:

Centripetal force = mass × (velocity^2 / radius)

Here, the radius of the curve is given as 80.0 m. We can rearrange the equation to solve for velocity:

Velocity = √(Centripetal force × radius / mass)

Plugging in the values, we get:

Velocity = √(7840 N × 80.0 m / 1000 kg)
= √627,200 m^2/s^2
= 790.6 m/s

Therefore, the maximum speed at which a car can make the turn on an unbanked curve with a radius of 80.0 m and a coefficient of static friction of 0.8 is approximately 790.6 m/s.