Need an aerodynamics guru to help me out! Please!

[niccolo98]

Hey guys,
I take a physics AP course. We wrote a test yesterday, and there was one question that did not seem right. The question was:
A body with the mass of 3.6 kg gets dropped from a height of 30m. The body hits the floor at 20 meters per second. Calculate the mechanic energy that gets turned into heat energy DURING the fall, not the impact.

To my understanding, at those speeds, the heat that is created due to friction is so little, that it barely has an effect on the loss of energy. So, thats what I wrote. He wrote a big fat wrong next to that. This is how he said it is calculated.

E potential= m (mass) times g (gravitational pull. about 10) times h (height)
If you put all those numbers in there, you get

1080N for the potential Energy.

Do this for the kinetic energy aswell:

E kinetic= m (mass) divided by 2 times v squared (speed squared)
Put the numbers in, and get 720N.

He then says, you subtract 720N from the 1080N.
Get 360N. He says, that this then equals the energy that is converted to heat.

That is one third of the total potential energy. I am almost certain, that at those speeds, not one third of the total energy gets converted to heat. I am also sure, that for one to calculate the heat created, that the frontal area of the object is needed. Again, something that is not given in the question. Please help me. I have been searching the internet for hours.

One more question is, what factors acctually take (convert) energy in a falling object? The 360N might be right, but I dont think that all the energy gets converted to heat.

Please help before I go insane looking through physic forums.
Thanks in advance,
Nick

[niccolo98]

nobody?

[airriley]

I don't think the problem was intended to be taken literally. I think your professor just wanted to see if you would catch on to the fact that "heat" energy must have been the difference between total potential and kinetic. I agree with you that frontal area or overall shape would affect the amount of lost or heat energy, but in this case the impact velocity was given, so all that doesn't matter anymore.

[shannah]

	Quote: Originally Posted by niccolo98
	Hey guys, I take a physics AP course. We wrote a test yesterday, and there was one question that did not seem right. The question was: A body with the mass of 3.6 kg gets dropped from a height of 30m. The body hits the floor at 20 meters per second. Calculate the mechanic energy that gets turned into heat energy DURING the fall, not the impact. To my understanding, at those speeds, the heat that is created due to friction is so little, that it barely has an effect on the loss of energy. So, thats what I wrote. He wrote a big fat wrong next to that. This is how he said it is calculated. E potential= m (mass) times g (gravitational pull. about 10) times h (height) If you put all those numbers in there, you get 1080N for the potential Energy. Do this for the kinetic energy aswell: E kinetic= m (mass) divided by 2 times v squared (speed squared) Put the numbers in, and get 720N. He then says, you subtract 720N from the 1080N. Get 360N. He says, that this then equals the energy that is converted to heat. That is one third of the total potential energy. I am almost certain, that at those speeds, not one third of the total energy gets converted to heat. I am also sure, that for one to calculate the heat created, that the frontal area of the object is needed. Again, something that is not given in the question. Please help me. I have been searching the internet for hours. One more question is, what factors acctually take (convert) energy in a falling object? The 360N might be right, but I dont think that all the energy gets converted to heat. Please help before I go insane looking through physic forums. Thanks in advance, Nick

Hi Nick,
This isn't an aerodynamics issue. It is just straight physics. All the energy gets converted to heat. Don't try to think of it in a real world or rational sense. Just do the math. The example you are working with isn't real world, so don't expect the answer to reflect anything that might be rationalized by observation or logic.

[niccolo98]

Ok, saying that is what he meant, even though he said this is as close to reality as it gets, I would really still like to know. Thanks,
nick

[todhog]

I could never graduate from high school now.

Todd

[rede2fly]

WHHHHHAAAAATTTTTTT yes it will 3-d that is the answere.

[airbike]

Since this freefall takes place in some sort of fluid medium (such as air), wouldn't the drag of the body during the fall be the primary mechanic of energy lost? I would say that the heating of the adjacent molecules would be a much smaler component of the energy used.

and if the freefall was in a vacuum...there should be no energy lost

[FireFighterFSFD]

Anything that produces energy creates heat. yes friction from just the air will produce a effect on the object but there is not enough information to answer how much heat is produced by friction. if the object is at the temperature of 500* the objest will not heat more it will cool from the air flow like our motors do on our planes. Like I said, find out how much energy is released and what the formula is for heat to energy release ratio and there is your answer. BUT on the other hand I could be wrong. that is just how I see you can produce a answer with the information given

oh and GOOD LUCK dude it took me 2 times to pass honors calculus II and 2 times to pass statistics. I was ready to just shoot my self in the head by the end of it all!!

dont give up dude

[CRG]

A good aero guy can probably tell you if that's correct. It's a case of fluid friction (drag), and some kinetic energy is converted to heat. Perhaps not all of it, there's going to be some noise also. The drag is what's stopping the object from accelerating as it would if in free-fall (gravity only, no atmosphere), and depends on the nature of the fluid, shape, material and speed of the object etc.

Wish I could help more.

Craig.

[Vic3D]

Hey Nicc,

It's been a couple years since I took intro physics my freshman year of engineering at college so I'll try to help as best as I can. You're prolly not too thrilled to hear this, but your professor is "right". I put the quote marks on right because the result is right, but I have no clue why he would use the approach he used towards you to "teach" it. It's just grounds for confusion, so I'll try to explain what he's doing..

"mechanical energy" = Kinetic energy

"heat energy" = Thermal energy


ΔE = ΔK + ΔV + ΔU Meaning that the change in total energy is the addition of all 3 changes in the 3 different types of energy (thermal, kinetic, potential). All energy is conserved. Energy can not be created or destroyed, it can only be changed from one form to another, right? ( Law of conservation of energy) This means that the CHANGE in energy within a closed system is 0.

Using this as a starting point, I'll show you how I tackle these problems:

(Kf - Ki) + (Vf - Vi) + (Uf - Ui) = 0 <--------------This is the same as ΔE = ΔK + ΔV + ΔU. f meaning final and i meaning initial.

Your change in Kinetic energy was 720N, so:

(720N) + (Vf - Vi) + (Uf - Ui) = 0

Your change in Potential energy was -1,080N(negative because the object goes down), so:

720N - 1,080N + (Uf - Ui) = 0 ( Notice the minus sign before 1080. Remember the change in Potential energy was negative. Pay attention to the signs)

You don't know the change in thermal energy because that is what you are trying to find. So you clean the formula up a bit and you end up with:

ΔU = 1,080N - 720N

ΔU = 360N

And there you go. Hope it's clear enough. Any other questions, just shoot!

Vic

[niccolo98]

I understand that 360N may be lost. What I do not understand, is how we know that this is due to thermal energy. We do not have the surface are, the drag coefficient, nor the atmospheric conditions. To me, the 360N are Total energy loss, not energy loss due to kinetic energy turning into thermal energy. 1080N is the total potential energy, the energy loss of 360N, would be ONE THIRD of the total energy lost to heat. I am pretty sure that at those speeds, the thermal energy loss is not that great. thanks.
nick