A race car is making a U-turn at constant speed. The coefficient of friction between the tires and the track is μs = 1.0. If the radius of the curve is 40 m, what is the maximum speed at which the car can turn without sliding? Assume that the car is performing uniform circular motion.

v=sqrt (m*g*r)

To find the maximum speed at which the car can turn without sliding, we can use the centripetal force equation:

Fc = mv²/r

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

In this case, the maximum speed occurs when the centripetal force is equal to the maximum static friction force. The maximum static friction force can be calculated using the equation:

fs,max = μs * N

Where μs is the coefficient of friction between the tires and the track and N is the normal force.

Since the car is on a flat track and making a U-turn, the normal force is equal to the weight of the car:

N = mg

Where m is the mass of the car and g is the acceleration due to gravity.

Putting it all together and substituting for N, we have:

fs,max = μs * m * g

Setting the maximum static friction force equal to the centripetal force:

fs,max = Fc

μs * m * g = mv²/r

Simplifying:

μs * g = v²/r

Solving for v:

v = sqrt(μs * g * r)

Plugging in the given values:

v = sqrt(1.0 * 9.8 m/s² * 40 m)

v = sqrt(392) m/s

v ≈ 19.8 m/s

Therefore, the maximum speed at which the car can turn without sliding is approximately 19.8 m/s.

To find the maximum speed at which the car can turn without sliding, we can use the concept of centripetal force and the maximum friction force.

We know that the net force acting on the car in uniform circular motion is the centripetal force given by:

Fc = (mv^2) / r

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

At the maximum speed, the friction force should be equal to the maximum static friction force, which is given by:

Ffmax = μs * N

Where μs is the coefficient of friction and N is the normal force acting on the car.

We can express the normal force N as N = mg, where g is the acceleration due to gravity.

Setting the friction force equal to the centripetal force, we have:

μs * N = (mv^2) / r

Substituting N = mg, we get:

μs * mg = (mv^2) / r

Canceling out the mass, we have:

μs * g = (v^2) / r

Solving for v, we get:

v^2 = μs * g * r

Taking the square root of both sides:

v = √(μs * g * r)

Now, plugging in the given values:

μs = 1.0
g = 9.8 m/s²
r = 40 m

v = √(1.0 * 9.8 * 40)

Calculating this expression, we find:

v ≈ 19.8 m/s

Therefore, the maximum speed the car can turn without sliding is approximately 19.8 m/s.