What is the force of the track on the car?

In summary, the problem involves a roller-coaster with a mass of 500 kg and a speed of 8.0 m/s at the top of a hill with a radius of 15 m. The question asks for the force of the track on the car at the top of the hill. Using the equations F=ma and N-mg=0, the solution is 2.8 kN up.
  • #1
ScullyX51
36
0

Homework Statement


A roller-coaster has a mass of 500 kg when fully loaded with passengers. The car passes over a hill of radius 15 m. When at the top of the hill, the car has a speed of 8.0 m/s. WHat is the force of the track on the car at the top of the hill?


Homework Equations


F=ma


The Attempt at a Solution


I drew an FBD: wtih mg pointing down, N pointing up and static friction pointing to the left. ( I think this may be wrong, but I figured it may need static friction to stay on the track).
Then when broken into i and j components I got:
i= -MsN=-MV2/r ...(because the acceleration is centripical)
then for j:
I have N-mg=0
N=mg
and since it was asking for the force of the track on the car, I thought this was the normal force, so I just plugged in the values for m ang, and got 4.9 kN up. This is wrong, it is saying the correct answer is 2.8 kN up. Any suggestions on where I went wrong?
 
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  • #2
The car is at the top of the hill; you seem to have your i and j mixed up. The centripetal accelertion is vertically down. Ignore friction.
 
  • #3


Your approach is correct, but you made a mistake in your calculation for the i component. The correct equation for the i component should be F=mv^2/r, where m is the mass of the car and v is its velocity. Plugging in the given values, we get:

F=(500 kg)(8.0 m/s)^2 / 15 m = 2133.33 N = 2.13 kN

This is the force of the track on the car at the top of the hill. To find the normal force, we need to subtract the weight of the car from this force, since the normal force is the force that balances the weight. The weight of the car is mg, which is (500 kg)(9.8 m/s^2) = 4900 N. So the normal force is:

N = F - mg = 2133.33 N - 4900 N = -2766.67 N = -2.77 kN

Since the normal force is pointing upwards, we take the absolute value to get the magnitude, which is 2.77 kN. This is very close to the given answer of 2.8 kN.

Therefore, the force of the track on the car at the top of the hill is 2.77 kN upwards.
 

Related to What is the force of the track on the car?

1. What is the force of the track on the car?

The force of the track on the car is known as the normal force. It is a reaction force that is exerted by the track on the car's tires in the direction perpendicular to the track's surface.

2. How is the force of the track on the car calculated?

The force of the track on the car can be calculated using the formula F = m * a, where F is the force, m is the mass of the car, and a is the acceleration of the car. The normal force is equal in magnitude to the weight of the car, which is also equal to m * g, where g is the acceleration due to gravity. Therefore, the normal force can also be calculated by multiplying the mass of the car by the gravitational acceleration.

3. Does the force of the track on the car change during a race?

Yes, the force of the track on the car can change during a race. This is because the normal force is dependent on the weight of the car, which can change if the car's fuel load decreases or if parts of the car break off during the race. Additionally, the acceleration of the car can also change, which can affect the normal force.

4. How does the force of the track on the car affect the car's performance?

The force of the track on the car is crucial for the car's performance. It helps to keep the car on the track and allows it to accelerate, brake, and turn effectively. The normal force also affects the car's grip on the track, which is essential for maintaining control and achieving high speeds.

5. Can the force of the track on the car be manipulated by the driver?

Yes, the force of the track on the car can be indirectly manipulated by the driver through their actions, such as braking and accelerating. However, the actual magnitude of the normal force is determined by the car's weight and the track's surface, which cannot be directly controlled by the driver.

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