A 2050 kg car rounds a circular turn of radius 18.5 m. If the road is flat and the coefficient of friction between tires and road is 0.680, how fast can the car go without skidding?

forcefriction=centripetal force

mu*mg=m*v^2/r

solve for v.

To determine the maximum speed at which the car can go without skidding, we need to consider the centripetal force acting on the car as it rounds the turn. The centripetal force is provided by the friction between the car's tires and the road.

The maximum frictional force that can be exerted between the tires and the road is given by the equation:

Frictional force (Ff) = coefficient of friction (μ) * normal force (Fn)

The normal force (Fn) is the force acting perpendicular to the surface of the road, and in this case, it is equal to the weight of the car (mg).

So, Fn = m * g

Where:
m = mass of the car = 2050 kg
g = acceleration due to gravity = 9.8 m/s^2

Substituting the values, we find:

Fn = 2050 kg * 9.8 m/s^2

Next, we calculate the maximum frictional force (Ff) using the given coefficient of friction (μ):

Ff = 0.680 * Fn

Now, we know that the centripetal force required for the car to go around the turn is given by:

Centripetal force (Fcp) = Ff

But the centripetal force is also given by the equation:

Fcp = (mass of the car) * (velocity^2) / (radius of the turn)

Where:
velocity = speed of the car

Substituting the values, we have:

Fcp = (2050 kg) * (velocity^2) / (18.5 m)

Setting the two equations equal to each other, we get:

(2050 kg) * (velocity^2) / (18.5 m) = 0.680 * (2050 kg * 9.8 m/s^2)

Now, we can solve this equation to find the maximum speed (velocity) at which the car can go without skidding:

velocity^2 = (0.680 * 2050 kg * 9.8 m/s^2 * 18.5 m) / 2050 kg

Finally, taking the square root of both sides of the equation, we can find the maximum speed (velocity):

velocity = √(0.680 * 9.8 m/s^2 * 18.5 m)

Evaluating this equation, we find:

velocity ≈ 16.8 m/s

Therefore, the car can go at a maximum speed of approximately 16.8 m/s without skidding.