Work & Energy for a Roller Coaster

In summary, the conversation discusses a problem involving a roller coaster and its speed at the bottom of a hill. The participants discuss the use of kinetic and potential energy equations, breaking forces into components, and the effects of friction on the object's motion. They also explore different approaches to solving the problem, including calculating net work and using the energy method.
  • #1
peaches1221
15
0
[SOLVED] Work and Energy

Hello, I'm a little new to this forum but I would really appreciate some help with this particular problem.

A roller coaster reaches the top of the steepest hill with a speed of 5.8 km/h. It then descends the hill, which is at an average angle of 25 degrees and is 50m long. What will the speed be when it reaches the bottom. assume the coefficient of friction (uk) = 0.12

So the first thing I did was convert the 5.8 km/h to 1.61 m/s to get consistent units. I know that KE initial + PE initial = KE final + PE final. I also know I have to break things up into components. I know Fx=mgcos25. But I think I'm screwing up the Fy component. After that I'm lost and I'm not sure where to go from there. Any suggestions and help will be greatly appreciated.
 
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  • #2
peaches1221 said:
I know Fx=mgcos25.

What do you mean by Fx here?
 
  • #3
peaches1221 said:
So the first thing I did was convert the 5.8 km/h to 1.61 m/s to get consistent units.
OK.
I know that KE initial + PE initial = KE final + PE final.
What about the work done by friction?
I also know I have to break things up into components. I know Fx=mgcos25.
That's the component of the weight normal to the surface. What's the component of the weight parallel to the surface?

What's the friction force? The work done by friction?
 
  • #4
Calculate the forces first

Hi peaches! Welcome to PF! :smile:

Hint: Be systematic.

1. How many forces are there?

2. What is the value of each force?

Then you can start deciding what to do. :smile:

(btw, if you type alt=m, it prints µ for you!)
 
  • #5
thank you all for your responses. I looked back at this problem and looked again at the forces affecting the object, the normal force, the weight, and friction. Please tell me if I'm going in the right track. If I find the net work, that would be delta KE. Then I would just have to solve for the final velocity. Does that make sense?

I know that force of friction is µkFN.
X: mgsin(theta)= ma
Y: FN - mgcos(theta) = 0
thus, FN = mgcos(theta)

the work of friction when simplified is equal to µkmgcos(theta)dcos(180)
there is no work by the normal force cos(90)= 0

is there work done by the weight??
 
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  • #6
Go back to the start, when you had written initial KE+PE = final KE+PE. That would have been correct if there were no frictional force.

The object has to to do work against the frictional force when coming down the incline. If F is the frictional force, then work done against friction is F*distance.

So, initial (KE+PE) = (final KE+PE) + F*d.

You know F=µmgcosθ. Now you should be able to solve directly.
 
  • #7
peaches1221 said:
I looked back at this problem and looked again at the forces affecting the object, the normal force, the weight, and friction. Please tell me if I'm going in the right track. If I find the net work, that would be delta KE. Then I would just have to solve for the final velocity. Does that make sense?
That's a perfectly valid approach as long as you make sure you deal with the work done by every force.

I know that force of friction is µkFN.
Good.
X: mgsin(theta)= ma
The net force equals ma (you forgot the friction).
Y: FN - mgcos(theta) = 0
thus, FN = mgcos(theta)
Good.
the work of friction when simplified is equal to µkmgcos(theta)dcos(180)
there is no work by the normal force cos(90)= 0
Good. Note that the work done by friction is negative.

is there work done by the weight??
Most definitely. What's the component of the weight along the direction of motion?

Note: While this is a perfectly valid approach, it requires that you account for the work done by gravity. If you use the "energy method" that you started with (and as explained by Shooting Star in post #6), that work is automatically taken care of as a PE term.

It might be easier to understand the energy method if you write it as:
initial (KE + PE) + Work done by friction = final (KE + PE)

Since the work done by friction is negative (= -F*d), that becomes:
initial (KE + PE) -F*d = final (KE + PE)

(which is equivalent to what SS wrote, of course)
 
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  • #8
thank you all for your help!
 

Related to Work & Energy for a Roller Coaster

1. How does the height of a roller coaster affect its potential energy?

The potential energy of a roller coaster is directly proportional to its height. This means that the higher the roller coaster is, the more potential energy it will have. When the roller coaster reaches the top of a hill, it has the maximum potential energy, which is then converted into kinetic energy as it moves down the hill.

2. What is the role of friction in a roller coaster's energy?

Friction plays a crucial role in a roller coaster's energy. It acts as a resistance to the motion of the roller coaster, converting some of its kinetic energy into heat. This is why roller coasters need to be regularly maintained and lubricated to reduce friction and maintain their speed and energy.

3. How does the speed of a roller coaster impact its kinetic energy?

The kinetic energy of a roller coaster is directly proportional to its speed. This means that the faster the roller coaster is moving, the more kinetic energy it will have. As the roller coaster goes through its twists, turns, and loops, its speed and kinetic energy will constantly change.

4. How is the conservation of energy applied to a roller coaster?

The law of conservation of energy states that energy cannot be created or destroyed, only transferred from one form to another. This applies to roller coasters as well. As the roller coaster moves, its potential energy is converted into kinetic energy and back again, while the total energy of the system remains constant.

5. How does gravity affect a roller coaster's energy?

Gravity is a fundamental force that plays a crucial role in the energy of a roller coaster. As the roller coaster moves through its track, it is constantly fighting against the force of gravity, which is pulling it down towards the ground. This causes the roller coaster to gain and lose energy, depending on the height and angle of the track.

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