- #31
crpcrpcrp
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is the initial velocity=0ms-1
Kinetic and potential energy and speed
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SUMMARY
The discussion focuses on the principles of kinetic and potential energy in physics, specifically addressing two problems involving energy conservation. The first problem involves determining the height at which kinetic energy equals potential energy for an object dropped from 10 meters. The second problem calculates the speeds of a 100 kg roller coaster car at different heights on a track. Key equations discussed include the conservation of energy, where total energy remains constant, and the relationship between potential energy (PE = mgh) and kinetic energy (KE = 1/2 mv²).
PREREQUISITES- Understanding of basic physics concepts, specifically kinetic and potential energy.
- Familiarity with the conservation of energy principle.
- Knowledge of gravitational force and its impact on potential energy calculations.
- Ability to manipulate algebraic equations to solve for unknowns.
- Study the conservation of mechanical energy in closed systems.
- Learn how to derive the equations for kinetic and potential energy.
- Explore real-world applications of energy conservation in roller coasters and other mechanical systems.
- Practice solving problems involving energy transformations and speed calculations in physics.
Students studying physics, educators teaching energy concepts, and anyone interested in understanding the mechanics of energy conservation in physical systems.
- #32
Hootenanny
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Correct! So since the object is at rest before it is dropped, all the energy is in the form of potential energy. So the constant in our equation must be...crpcrpcrp said:
- #33
crpcrpcrp
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0?
- #34
Hootenanny
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If the total energy of the object must be constant then the total energy must be the same as the total energy at the start. Does that make sense?crpcrpcrp said:0?
- #35
crpcrpcrp
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I understand
- #36
Hootenanny
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Good, so now we have
2*potential energy = constant
2mgh = 20mh = 100m
20h = 100
Can you now go from here?
2*potential energy = constant
2mgh = 20mh = 100m
20h = 100
Can you now go from here?
- #37
crpcrpcrp
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i'm still a little confused. so what would be the constant?
- #38
crpcrpcrp
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Hootenanny said:
what do i do form here?
- #39
crpcrpcrp
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wait...
where did u get the 100?
where did u get the 100?
- #40
Hootenanny
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The constant would be the 100m as you stated yourself.crpcrpcrp said:
You should solve for h, the height of the ball when the potential energy is equal to the kinetic energycrpcrpcrp said:what do i do form here?
You derived it yourself;crpcrpcrp said:
crpcrpcrp said:
- #41
wisienkas
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at starting height you know that potential energy = constant
you know the constant is potential energy + kinetic energy.
therefore the potential energy and the kinetic energy is the same if 2*potential energy=constant
you can therefore say constant/2 = potential energy or constant/2= kinetic energy
since you get the potential energy from gravity*mass*height you can simply just isolate the height since you have te energy from the constant.
its more math than physics however. hope it helps for any looking in here
you know the constant is potential energy + kinetic energy.
therefore the potential energy and the kinetic energy is the same if 2*potential energy=constant
you can therefore say constant/2 = potential energy or constant/2= kinetic energy
since you get the potential energy from gravity*mass*height you can simply just isolate the height since you have te energy from the constant.
its more math than physics however. hope it helps for any looking in here