To maintain circular motion, there must be a centripetal force acting towards the center of the circle. In this case, the force needed to maintain the car's circular motion is provided by the frictional force between the car's tires and the pavement.
a. To find the force needed to maintain the car's circular motion, we can use the equation for centripetal force:
F = m * (v^2 / r)
where:
F 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 circular path.
Plugging in the values:
F = (1250 kg) * ((48 km/h)^2 / 35 m)
First, let's convert the velocity from km/h to m/s:
v = (48 km/h) * (1000 m/1 km) * (1 h/3600 s) ≈ 13.33 m/s
Now, let's calculate the force:
F = (1250 kg) * ((13.33 m/s)^2 / 35 m)
F ≈ 6317.5 N
Therefore, the force needed to maintain the car's circular motion is approximately 6317.5 N.
b. The available frictional force is given by the product of the coefficient of kinetic friction (μk) and the normal force (N) between the car and the pavement.
The normal force (N) is equal to the weight of the car (mg), where g is the acceleration due to gravity (approximately 9.8 m/s^2).
N = mg
Plugging in the values:
N = (1250 kg) * (9.8 m/s^2)
N ≈ 12250 N
Now, let's calculate the available frictional force:
Frictional force = μk * N
Frictional force = (0.5) * (12250 N)
Frictional force ≈ 6125 N
Therefore, the available frictional force is approximately 6125 N.
c. To determine if the available frictional force is large enough to maintain the automobile's circular motion, we compare it to the force needed to maintain circular motion.
The available frictional force (6125 N) is less than the force needed to maintain circular motion (6317.5 N). Since the available frictional force is smaller, it is not large enough to maintain the automobile's circular motion. As a result, the car would not be able to follow the circular path and would skid or slide off the road.