A roller-coaster car has a mass of 1170 kg when fully loaded with passengers. As the car passes over the top of a circular hill of radius 16 m, its speed is not changing.
(a) What are the magnitude and direction of the force of the track on the car at the top of the hill if the car's speed is 11 m/s?
______N
upward
downward
no force
(b) What are the magnitude and direction of the force of the track on the car at the top of the hill if the car's speed is 17 m/s?
_________N
upward
downward
no force
force on track=mg-mv^2/r
To find the magnitude and direction of the force of the track on the car at the top of the hill, we can start by analyzing the forces acting on the car at that point.
(a) When the car's speed is 11 m/s:
At the top of the hill, the car experiences two main forces: gravitational force (mg) and a normal force (N) exerted by the track.
The normal force acts perpendicular to the track's surface and opposes the gravitational force.
In this case, the car's speed is not changing, indicating that the forces are in equilibrium. Therefore, the magnitude of the force of the track (N) must be equal to the magnitude of the gravitational force (mg).
To find the magnitude, we can use the equation:
N = mg
The mass of the car when fully loaded with passengers is given as 1170 kg.
So, N = 1170 kg * 9.8 m/s^2 = 11,466 N
Therefore, the magnitude of the force of the track on the car at the top of the hill is 11,466 N in the downward direction.
(b) When the car's speed is 17 m/s:
Similarly, at the top of the hill, the forces acting on the car are still gravitational force (mg) and the normal force (N).
Since the car's speed is higher, it indicates that there is an additional force acting on the car, causing the car to move in a circular path. This force is called the centripetal force.
The centripetal force is given by the equation:
F_c = (mv^2) / r
Where:
m = mass of the car
v = velocity of the car
r = radius of the circular path
In this case, the radius of the circular hill is given as 16 m.
Using the above equation, we can calculate the centripetal force:
F_c = (1170 kg * (17 m/s)^2) / 16 m = 12,153.75 N
Since the force of gravity (mg) is always acting downward, and the centripetal force acts towards the center of the circular path (which is downward), the normal force (N) will have to be the vector sum of these two forces.
N = mg + F_c = 1170 kg * 9.8 m/s^2 + 12,153.75 N
N ≈ 23,724 N
Therefore, the magnitude of the force of the track on the car at the top of the hill is approximately 23,724 N in the downward direction.
In summary:
(a) The magnitude and direction of the force of the track on the car at the top of the hill when the car's speed is 11 m/s is approximately 11,466 N downward.
(b) The magnitude and direction of the force of the track on the car at the top of the hill when the car's speed is 17 m/s is approximately 23,724 N downward.