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Conservation of Mechanical Energy on a Roller Coaster

  • Thread starter Foopyblue
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  • #1
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Homework Statement


A roller coaster at an amusement park is at rest on top of a 30 m hill (point A). The car starts to roll down the hill and reaches point B which is 10 m above the ground, and then rolls up the track to point C, which is 20 m above the ground.
(A) A student assumes no energy is lost, and solves for how fast the car is moving at point C using energy arguments. What answer does he get?
(B) If the final speed at C is actually measured to be 2 m/s, what percentage of energy was "lost" and where did it go?
2u8xq2v.jpg

Homework Equations


Kf + Uf = Ki + Uf
K = 1/2mv2
Ug = mgh

The Attempt at a Solution


I believe there is a typo in the book for Part A, I think they got the height wrong, their answer is 20 m/s. For Part B, I think I did something wrong because the height error has been corrected.

My Work for Part A:
I set potential energy at A equal to kinetic energy at C. My reference point is considering the height of C as ground level.

mgh = 1/2mv2
gh = 1/2v2
10*10 = 1/2v2
100 = 1/2v2
200 = v2
14 = v

Their Work for Part A
2ldkawy.jpg


My Work for Part B

Find ratio of Final Energy to Initial Energy and subtract from 100%
[itex]\frac{\frac{1}{2}mv^2}{mgh}[/itex]

[itex]\frac{\frac{1}{2}v^2}{gh}[/itex]

[itex]\frac{\frac{1}{2}*2^2}{10*10}[/itex]

[itex]\frac{2}{100}[/itex]

2%

So 98% is lost to heat. This is radically different from their answer:

2qsxlpl.jpg

For this part I again used a reference height of the height at C, but am I not allowed to do that?
 
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Answers and Replies

  • #2
Nathanael
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They seem to have messed up; they solved for the speed at point B. I agree with your answer for the speed at point C

You still messed up for part B, though. You seem to be saying the total energy of the system is m*g*10
 
  • #3
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If I set my reference height of C as ground level, is that not the total energy of the system? It has no Kinetic at point A, all Potential, right?
 
  • #4
Nathanael
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If I set my reference height of C as ground level, is that not the total energy of the system? It has no Kinetic at point A, all Potential, right?
Ah, good point; the question is ambiguous because the answer depends on where you choose the zero for your potential. The reason the answer differs so much is because they chose their PE=0 at the actual ground.
 
  • #5
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view the attached file please for your answer
 

Attachments

  • #6
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view the attached file please for your answer
I hate to bother you Arjun but I'm having trouble reading it because of the quality, do you have another picture you could upload?
 
  • #7
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Please do check this. I have attached a clearer image. Please do reply if the information was good. Thank you. Always happy to help. :) the variable of gravity g changes with height so the h should not be changed
DSCN0180.JPG
 
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  • #8
BvU
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My calculator has something else for 202/300
 
  • #9
BvU
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Ah, good point; the question is ambiguous because the answer depends on where you choose the zero for your potential. The reason the answer differs so much is because they chose their PE=0 at the actual ground.
The answer does not depend on the choice for zero potential. That's the fun of potentials in general: the only thing that matters is the difference in potential. It's about ##\Delta h##, not about ##h## itself.
 
  • #10
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The textbook and I both agree that there was a change in height of 10 m, but where we differ is that they are calculating the potential energy from the height above the ground(30m) while I am calculating the potential energy from the height above point C(10m). That's why their ratio is different from mine, at least I think.
 
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  • #11
Nathanael
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The answer does not depend on the choice for zero potential. That's the fun of potentials in general: the only thing that matters is the difference in potential. It's about ##\Delta h##, not about ##h## itself.
Yes, but in this case it's a bit different, isn't it? Because we're calculating the percentage change in energy... which is ("final energy") / ("initial energy"), but this is ambiguous because as you said, only changes in potential energy make sense, so neither final energy nor initial energy are clearly defined (and the ambiguity does not cancel out).

The answer would be, (KE+C) / (10*mg+C) Where "C" represents the arbitrary potential energy due to the choice of where the zero potential is. This depends on C; the bigger C is, the closer the expression will be to 1. Their solution chose C=20*mg and the OP chose C=0.
 
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  • #12
BvU
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I humbly agree. If there was any percentage worth mentioning here it is indeed 98%.
Kinetic energy ideally 10 m x mg turns out to be only 0.5 x (2m/s)2.

It's not the book answer that is wrong, it's the book question that is disastrously bad. Kudos for Foopy for not going with that nonsense -- and idem for Nathaniel
 

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