what is the maximum speed with which a car can round a turn of radius of 80.0m on a flat road if the coefficient of friction between tires and the road is .700?
My teacher said two substitutions and I don't have a mass. Thanks (:
Centipetal force = frictional force mv2/r=Rmu m will be canceled and thus v will be equal to under root of ugr putting the values 80*700*10=560000 taking the under root answer will be 748.33
To find the maximum speed with which a car can round a turn, we can use the concept of centripetal force. The maximum speed occurs when the frictional force between the tires and the road is equal to the maximum static friction that can act to provide the necessary centripetal force.
First, let's set up the equations. The centripetal force is given by:
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 turn.
The maximum static frictional force is given by:
Ffriction max = μ * N
where Ffriction max is the maximum static frictional force, μ is the coefficient of friction, and N is the normal force.
Notice that the normal force cancels out when we equate these two equations, so we don't actually need the mass of the car.
Let's substitute the expressions for Fc and Ffriction max:
(mv^2) / r = μ * N
Now, we can replace N with the weight of the car, which is equal to mg, where g is the acceleration due to gravity:
(mv^2) / r = μ * mg
Simplifying, we can cancel out the masses:
v^2 = μ * g * r
Finally, we can solve for the maximum speed, v:
v = sqrt(μ * g * r)
Plugging in the given values:
v = sqrt(0.700 * 9.8 * 80.0)
v ≈ 29.6 m/s
Therefore, the maximum speed with which the car can round the turn is approximately 29.6 m/s.