How to find height of original hill (roller coaster loop problem)

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SUMMARY

The discussion focuses on calculating the height of the original hill in a roller coaster loop problem, given a loop radius of 20 m, a cart mass of 225 kg, and gravitational acceleration of 10 m/s². Key findings include a velocity of 14.14 m/s at the top of the loop, gravitational potential energy (GPE) at the top calculated as 90,000 N, and kinetic energy (KE) at the bottom determined to be 180,000 N. The conversation highlights the importance of correctly applying energy conservation principles and clarifying terminology, particularly distinguishing between energy and heat.

PREREQUISITES
  • Understanding of gravitational potential energy (GPE) and kinetic energy (KE) equations
  • Familiarity with the conservation of energy principle in physics
  • Knowledge of basic mechanics, including force and mass calculations
  • Ability to differentiate between units of energy (Joules) and force (Newtons)
NEXT STEPS
  • Study the conservation of energy in mechanical systems
  • Learn about the relationship between potential energy and height in gravitational fields
  • Explore the implications of energy losses in real-world scenarios
  • Review the definitions and applications of energy units, specifically Joules and Newtons
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Students studying physics, particularly those focusing on mechanics and energy conservation, as well as educators seeking to clarify concepts related to energy and motion in roller coaster dynamics.

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Homework Statement


Radius of loop: 20 m
Mass of cart: 225 kg
Gravity: 10 m/s/s (once I figure out how to do this I'll use 9.8 in the final project)

Found so far:

B) Velocity (top of loop) = 14.14 m/s
C) GPE at the top of the loop = 90000 N
C) KE top of loop = 22500 N

D) KE (bot) = 180000 N

Homework Equations



A) Fg = mg = 225 kg * 10 m/s/s = 2250 N
B) Fg = mv^2/r
C) GPE (top of loop) = KE (top of loop) - heat (lost going through loop)


The Attempt at a Solution



B) Using these equations I found that the velocity at the top of the loop is 14.14 m/s.

C) 90000 N (GPE = mgh) = 22500 (KE =1/2mv^2) - heat.
That means heat would have to be (-)67500 N. Because of conservation of energy, those three energies added together should be the energy at the bottom (before) the loop. Where all energy would be allocated to KE because height at that point is 0. So

D) KE (bot) = 180000 N


I'm confused as to where to go next. The KE (bot) should be equal to the GPE at the top - whatever heat was lost going down the hill. But how would I find the heat lost there? Would it be the same as before (67500 N)?
 
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I think you mean "energy" where you are using the word "heat". Heat is a very specific form of energy that does not really apply here. Also you say 'GPE= 9000 N'. N is standard for "Newton" which is a unit of force, not energy. The MKS unit of energy is the "Joule".
 
Mikhowl said:
C) GPE (top of loop) = KE (top of loop) - heat (lost going through loop)
Why should that be true? In particular, GPE depends on the arbitrary definition of "zero height", the other two expressions do not.
Energy conservation gives GPE(starting point) = GPE(top of loop) + KE (top of loop) + losses
 

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