What is the maximum speed with which a 1050- car can round a turn of radius 68 on a flat road if the coefficient of static friction between tires and road is 0.60?
centripetal force= friction force
m v^2/r=mu*mg
Your problem statement did not include units. Units are a student's best friend.
To determine the maximum speed at which a 1050-kg car can round a turn of radius 68 m on a flat road, we can use the formula for centripetal force. The centripetal force required to keep the car in a circular path is provided by the friction between the tires and the road. The maximum frictional force can be calculated using the coefficient of static friction.
The formula for centripetal force is:
F = m*v^2 / r
where:
F is the centripetal force,
m is the mass of the car (1050 kg),
v is the velocity of the car, and
r is the radius of the turn (68 m).
The maximum frictional force can be calculated using the formula:
f_max = μ * N
where:
f_max is the maximum frictional force,
μ is the coefficient of static friction (0.60), and
N is the normal force.
The normal force N is equal to the weight of the car, which can be calculated using:
N = m * g
where:
g is the acceleration due to gravity (9.8 m/s^2).
Now, substituting the values into the equations:
f_max = μ * N = μ * m * g
Set the maximum frictional force equal to the centripetal force:
f_max = F
μ * m * g = m*v^2 / r
Simplifying the equation, we can solve for the velocity v:
v^2 = μ * r * g
v = sqrt(μ * r * g)
Now we can substitute the given values:
v = sqrt(0.60 * 68 * 9.8)
Calculating this gives us:
v ≈ 12.45 m/s
Therefore, the maximum speed at which a 1050-kg car can round a turn of radius 68 m on a flat road, with a coefficient of static friction of 0.60, is approximately 12.45 m/s.