Applying Kinetic Energy and Power Formulas on Inclined Surfaces

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In summary, questions were asked about using kinetic energy and the power formula to calculate velocity on an incline or slope, as well as the applicability of GPE and KE on a spiral or helix. The expert suggests using KE + PE = constant and clarifies that the power formula is only useful in certain situations. The expert also explains the calculation process for finding the final velocity on an incline and addresses confusion about finding centripetal acceleration on a spiral or helix.
  • #1
Procrastinate
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I just have some questions about kinetic energy.

1. If you are going up an incline on a roller coaster and you are trying to calculate the KE lost; is it possible to use this to find the velocity at the top, or does this only apply to downwards motion?

2. Can you apply the power formula to a roller coaster going down an incline, or is this only useful for when a motor is in charge of the energy outlay rather than gravity?

3. Can you use GPE and KE, even if you are on an incline/slope?
 
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  • #2
Hi Procrastinate! :smile:
Procrastinate said:
1. If you are going up an incline on a roller coaster and you are trying to calculate the KE lost; is it possible to use this to find the velocity at the top, or does this only apply to downwards motion?

3. Can you use GPE and KE, even if you are on an incline/slope?

Yes, KE + PE = constant works in all circumstances (assuming, of course, no energy loss to friction etc). :smile:

Incidentally, potential energy (PE) is just another name for (minus) work done by a conservative force (and gravity is conservative). :wink:
2. Can you apply the power formula to a roller coaster going down an incline, or is this only useful for when a motor is in charge of the energy outlay rather than gravity?

Sorry, what's the "power formula" ? :confused:
 
  • #3
tiny-tim said:
Hi Procrastinate! :smile:

Sorry, what's the "power formula" ? :confused:

P = E/t - is this only used when a motor is involved, rather than being controlled by gravity.


So, back on Kinetic energy. If say for example, a 20kg mass was moving up an incline from a height of 3m to a height of 10m and I wanted to find the final velocity at the top? I would do:

GPE_top - GPE_bottom = KE lost=1372

1/2xmv2 = KE lost
1372=.5mv2
11.7=velocity
 
  • #4
well in case of gravity power isn't dissipated but changed into another form
but i think u can stil apply that formula(actually more proper would be to apply f.v though it is very much derived from e/t) and u will know the ke that converted into potential

well not sure about ur calculation but the method for the example is right
 
  • #5
Procrastinate said:
P = E/t - is this only used when a motor is involved, rather than being controlled by gravity.

ah, that's the definition of power … power = energy (or work) per time.

you can only really use power when for some reason you know exactly what it is … eg your example of a motor, with a given power.

(and I personally have never seen an exam question involving power, so you can forget it for solving rollercoasters etc :wink:)
So, back on Kinetic energy. If say for example, a 20kg mass was moving up an incline from a height of 3m to a height of 10m and I wanted to find the final velocity at the top? I would do:

GPE_top - GPE_bottom = KE lost=1372

1/2xmv2 = KE lost
1372=.5mv2
11.7=velocity

That's fine, except the last two lines …

1372 is the difference in KE …

to find the final velocity, start with the initial velocity, and subtract 1372 from its KE. :smile:
 
  • #6
tiny-tim said:
ah, that's the definition of power … power = energy (or work) per time.

you can only really use power when for some reason you know exactly what it is … eg your example of a motor, with a given power.

(and I personally have never seen an exam question involving power, so you can forget it for solving rollercoasters etc :wink:)


That's fine, except the last two lines …

1372 is the difference in KE …

to find the final velocity, start with the initial velocity, and subtract 1372 from its KE. :smile:

Ok, thank you.

Just to check, can you use GPE and KE on a spiral that is going down? Or is it better to use circular motion formulas? When i am finding acceleration centripetal, I get confused as to whether to use initial velocity first going around the curve, or final velocity going around the curve.
 
  • #7
Hi Procrastinate! :smile:

(just got up :zzz: …)
Procrastinate said:
Just to check, can you use GPE and KE on a spiral that is going down? Or is it better to use circular motion formulas?

GPE and KE are almost always better, because they involve v instead of a …

in other words, they automatically do the job for you of performing a possibly difficult integration! :biggrin:
When i am finding acceleration centripetal, I get confused as to whether to use initial velocity first going around the curve, or final velocity going around the curve.

I'm not sure what you mean :confused:

centripetal acceleration is always v2/r, and on (for example) a rollercoaster, you can always find v using KE + PE = constant.
 
  • #8
tiny-tim said:
Hi Procrastinate! :smile:

(just got up :zzz: …)

I'm not sure what you mean :confused:

centripetal acceleration is always v2/r, and on (for example) a rollercoaster, you can always find v using KE + PE = constant.


When I am substituting v in the equation to find centripetal acceleration, and this is on a spiral that is going down, the velocity at the top of the curve will be different from the velocity at the bottom of the curve. What do I do then?
 
  • #9
Procrastinate said:
When I am substituting v in the equation to find centripetal acceleration, and this is on a spiral that is going down, the velocity at the top of the curve will be different from the velocity at the bottom of the curve. What do I do then?

(btw, do you mean a helix?)

The centripetal acceleration will change … is that what's confusng you? :confused:

it will always be v2/r.

(but why are you trying to find the centripetal acceleration anyway?)
 
  • #10
tiny-tim said:
(btw, do you mean a helix?)

The centripetal acceleration will change … is that what's confusng you? :confused:

it will always be v2/r.

(but why are you trying to find the centripetal acceleration anyway?)

It looks a bit like this: I posted it here (I wanted to delete it but I can't): https://www.physicsforums.com/showthread.php?p=2599095#post2599095 [Broken]



I need to compare the centripetal acceleration on my X and Z acceleration graphs from the accelerometer for the primary data that was collected.
 
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  • #11
hmm … it doesn't look like there's any way to calculate r (the radius of curvature of the track) … is there?

(also, it looks to me as if the centripetal acceleration will not be horizontal)

If not, you'll have to get the centripetal acceleration from the accelerometer (and make allowance for g, if it's not horizontal)
 
  • #12
tiny-tim said:
hmm … it doesn't look like there's any way to calculate r (the radius of curvature of the track) … is there?

(also, it looks to me as if the centripetal acceleration will not be horizontal)

If not, you'll have to get the centripetal acceleration from the accelerometer (and make allowance for g, if it's not horizontal)

If I do use my acceleration graphs, which axis should I use? The Z acceleration (I got it mixed up with X before, which actually wasn't provided) or the Y acceleration?

I found the radius of the track from Google Earth, even if it is not 100% accurate.
 
  • #13
Sorry, I've no idea what you're talking about. :redface:

What are these acceleration graphs? :confused:
 
  • #14
tiny-tim said:
Sorry, I've no idea what you're talking about. :redface:

What are these acceleration graphs? :confused:

Actually, I get it now, so it doesn't matter.

However, is it better to leave calculating centripetal motion out at the curves I mentioned in the diagram since they are not completely horizontal?
 
  • #15
You can always use centripetal acceleration, but if the bend is on a slope, the centripetal acceleration will along that slope also. :smile:
 
  • #16
tiny-tim said:
You can always use centripetal acceleration, but if the bend is on a slope, the centripetal acceleration will along that slope also. :smile:

Do you mean change along the slope?


Thank you for your help by the way.
 
  • #17
oops!

oops! i left out the word "be" :rolleyes:

… the centripetal acceleration will be along that slope (and not horizontal or vertical) :smile:
 
  • #18


tiny-tim said:
oops! i left out the word "be" :rolleyes:

… the centripetal acceleration will be along that slope (and not horizontal or vertical) :smile:

Thank you very much.
 

1. What is kinetic energy/power?

Kinetic energy is the energy an object possesses due to its motion, while power is the rate at which energy is transferred or converted.

2. How is kinetic energy/power calculated?

Kinetic energy is calculated using the formula KE = 1/2 * m * v^2, where m is the mass of the object and v is its velocity. Power is calculated as P = W/t, where W is the work done on the object and t is the time taken to do the work.

3. What are some examples of kinetic energy/power?

Examples of kinetic energy include a moving car, a rolling ball, and a swinging pendulum. Power can be seen in actions such as pushing a heavy object, running, or lifting weights.

4. How does kinetic energy/power relate to potential energy?

Kinetic energy and potential energy are the two main forms of mechanical energy. Kinetic energy is the energy of motion, while potential energy is the energy an object has due to its position or state, such as gravitational potential energy. The total energy of a system is the sum of its kinetic and potential energies.

5. How can kinetic energy/power be converted or transferred?

Kinetic energy can be converted or transferred through work done by a force, such as friction or a collision. Power can be transferred through various forms of energy, such as heat, electricity, or mechanical work.

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