What is the maximum speed with which a 1500-{\rm kg} car can round a turn of radius 67 m on a flat road if the coefficient of static friction between tires and road is 0.65?

20.65

when car moves on a straight road the limiting force of friction should be greater than or equal to centripetal force
i.e v=sq.root of meu r g

To find the maximum speed at which a car can round a turn, we can use the concept of centripetal force.

The centripetal force is the force that keeps an object moving in a circular path, and in this case, it is provided by the static friction between the car's tires and the road.

The formula for centripetal force is given by:

F = m * v^2 / r

Where:
F is the centripetal force
m is the mass of the car (1500 kg)
v is the velocity of the car
r is the radius of the turn (67 m)

In this case, the maximum centripetal force is given by the static friction force, which can be calculated using the formula:

F_friction = μ * N

Where:
F_friction is the static friction force
μ is the coefficient of static friction (0.65 in this case)
N is the normal force, which is the weight of the car (mg)

The normal force, N, is equal to the weight of the car since the car is on a flat road. The weight of the car can be calculated using:

N = mg

Finally, combining these equations, we can solve for the maximum velocity.

F_friction = F
μ * N = m * v^2 / r
μ * (mg) = m * v^2 / r
μ * g = v^2 / r
v^2 = μ * g * r
v = sqrt(μ * g * r)

Now we can substitute the values given in the question into this equation to find the maximum speed with which the car can round the turn on a flat road.