Final answer:
The normal force exerted by the car on the driver as it goes over the top of the hill is 9555 N.
Step-by-step explanation:
To calculate the normal force exerted by the car on the driver as it goes over the top of the hill, we can start by determining the net force acting on the car at the highest point of the hill.
At the highest point, the car is still moving forward, so there is a centripetal force directed towards the center of the circular path. This force is provided by the normal force and the force of gravity acting on the car.
The net force can be calculated using the equation:
Net Force = Centripetal Force = (mass x velocity^2) / radius
Plugging in the given values:
Mass = 975 kg (car + driver)
Velocity = 18.0 m/s
Radius = 88.0 m (rounded top of the hill)
Net Force = (975 kg x (18.0 m/s)^2) / 88.0 m = 3367.05 N
Now, we can consider the forces acting on the car at the highest point. The weight of the car (mg) is exerted downwards, and the normal force (N) is exerted upwards. At the highest point, the normal force must be larger than the weight of the car to provide the centripetal force needed to keep the car moving in a circular path.
Weight of car = mass x gravity
Weight of car = 975 kg x 9.8 m/s^2 = 9555 N
Therefore, the normal force exerted by the car on the driver as the car goes over the top of the hill is 9555 N.