Conservation of Energy roller coaster car

In summary: The normal force and the gravitational force act in opposite directions. Thus, you would in fact subtract them to find the net force. The force due to gravity is stronger than the normal force, so the net force points towards the center of curvature, as it should. However, I am getting 80 m for my answer, just like you. I cannot figure out why 30 m is recorded as the correct answer.
  • #1
kiwikahuna
61
0

Homework Statement


A roller coaster car starts from rest and descends 40 m before rising 20 m to the top of a hill. The passenger of mass 75 kg at the top of the hill feels a normal force of 368 N. The radius of curvature of the hill is...?


Homework Equations



F = m * a
a = v^2/r

The Attempt at a Solution



Because we are given the force and the mass the person, I found the acceleration and set acceleration = v^2/r. Velocity isn't given so I used the kinematics equation: Vf^2 = Vi^2 + 2a(distance)
where Vi =0 and distance = 60. Once I found the velocity I went back and plugged that number into a =v^2/ r to find the radius but I'm not getting the right answer at all. Any help would be great.
 
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  • #2
The title of your post is "conservation of energy." It doesn't appear that you've applied this concept to your attempt at the solution.

I suggest you find the velocity at the top of the hill by simply considering the change in gravitational potential energy associated with the roller coaster's descent and subsequent ascent.

Please also note the kinematics equation you provided cannot be used because that equation is only valid under constant acceleration.
 
  • #3
Thank you for the suggestions.
I used the conservation of energy formula:

mg(h1) = 1/2 mv^2 + mg(h2)
where h1 = 40 m and h2 =20 m
I find the velocity to be 19.8 m/s at the top of the 20 m hill.

I plugged the velocity back into a = v^2/r and I get r to be 80 m but it's still not the right answer (which is actually 30m )
 
  • #4
kiwikahuna said:
Thank you for the suggestions.
I used the conservation of energy formula:

mg(h1) = 1/2 mv^2 + mg(h2)
where h1 = 40 m and h2 =20 m
I find the velocity to be 19.8 m/s at the top of the 20 m hill.

I plugged the velocity back into a = v^2/r and I get r to be 80 m but it's still not the right answer (which is actually 30m )

Draw a free-body diagram of the car, noting that it as at the TOP of the hill. Also note that the net force must be equal to the quantity (mv^2)/r. How do the normal force and force due to gravity relate? What does this say about the radius of curvature?
 
  • #5
there's a gravitational force down (with the mass of both the cart and the person)

There's a normal force going up. But this is the normal force that the person feels. To find the net force, you would have to add these two forces together? Any more explanation on this topic would be very helpful. Conceptually I don't understand why you would have to add these two forces together as opposed to subtracting them.
 
  • #6
The normal force and the gravitational force act in opposite directions. Thus, you would in fact subtract them to find the net force. The force due to gravity is stronger than the normal force, so the net force points towards the center of curvature, as it should. However, I am getting 80 m for my answer, just like you. I cannot figure out why 30 m is recorded as the correct answer.

I've done countless problems like this in the past, so I do not know where I'm going wrong.

It'd be great if anyone else could chime in.


kiwikahuna said:
there's a gravitational force down (with the mass of both the cart and the person)

There's a normal force going up. But this is the normal force that the person feels. To find the net force, you would have to add these two forces together? Any more explanation on this topic would be very helpful. Conceptually I don't understand why you would have to add these two forces together as opposed to subtracting them.
 

1. What is the Conservation of Energy principle in a roller coaster car?

The Conservation of Energy principle states that energy cannot be created or destroyed, but rather it can only be transferred or transformed. In a roller coaster car, this means that the total energy at any point in the ride (kinetic energy + potential energy) will remain constant.

2. How does the roller coaster car conserve energy?

The roller coaster car conserves energy by using various mechanisms such as ramps, loops, and friction to transfer or transform energy from one form to another. For example, when the car goes up a hill, its potential energy increases while its kinetic energy decreases. Then, as it goes down the hill, its potential energy decreases while its kinetic energy increases.

3. What factors affect the conservation of energy in a roller coaster car?

The conservation of energy in a roller coaster car is affected by factors such as the height and shape of the track, the mass and speed of the car, and the presence of friction. These factors determine the amount of potential and kinetic energy the car has at any given point in the ride.

4. Can the conservation of energy be violated in a roller coaster car?

No, the conservation of energy cannot be violated in a roller coaster car. This principle is a fundamental law of physics and applies to all objects, including roller coaster cars. Any perceived violation of energy conservation in a roller coaster ride is due to the transfer or transformation of energy, not its destruction.

5. How does the conservation of energy affect the safety of a roller coaster ride?

The conservation of energy is crucial for the safety of a roller coaster ride. By carefully designing and maintaining the track and car, engineers ensure that the total energy of the car remains constant throughout the ride. This prevents dangerous situations where the car may lose too much energy and not be able to complete the ride, or gain too much energy and become unsafe for passengers.

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