Understanding Work and Energy Transfer: The Relationship and Implications

[beatrix kiddo]

chroot, doesn't the Earth have kinectic and potential energy as it revolves around the sun? would this mean that there is work being done? or, because there is conservation, no work gets done?

 

[urtalkinstupid]

Wouldn't that be:

$$E=\Delta q +W$$

The change in heat would be more reasonable. You take the final temperature and subtract from the initial, right? This gives the total amount of heat put into the system.

 

[chroot]

That's the point I'm trying to argue. Energy is created in order to keep that force constant. In order to make a force applicable, there is a needed input of energy.

Quite simply, no.

As Chrono's said, there is the same problem with permanent magnets, except much greater, because they hang on a refrigerator door like mountain climbers hang on the side of a cliff.

Both magnetism and gravity are conservative forces, which means they conserve energy. If they were not conservative, there would be serious consequences in the behavior of such systems in translations in time, which would pretty much rewrite physics from the ground up -- and it'd be entirely wrong. The conservation of energy and the concommitant invariance of physical laws to translations in time are perhaps the most fundamental properties of physics in this universe.

This all ties into the Work equation.

Which equation? $W = \vec F \cdot \vec d$? The one that you don't even understand?

- Warren

 

[chroot]

The change in heat would be more reasonable. You take the final temperature and subtract from the initial, right? This gives the total amount of heat put into the system.

Temperature and heat are not the same thing at all.

- Warren

 

[Chronos]

If it's any consolation, the sun will go red giant and vaporize the Earth long before have a chance to crash into it's cold, dead corpse. But what the heck, a creationist could have told you that.

 

[nickdanger]

Wouldn't that be:

$$E=\Delta q +W$$

The change in heat would be more reasonable. You take the final temperature and subtract from the initial, right? This gives the total amount of heat put into the system.

You make a good point. But in fact neither q nor w need to be thought of as a change. The heat in particular is just a quantity passing the boundary between the system and the surroundings. We sum the the heat and the work to get the quantity, E. This is certainly a delta if you want to think about absolute energy of the system, but there are problems with this and it is not necessary to reference the change from absolute energy content for most thermodynamic problems.

 

[nickdanger]

Warren is correct. It's energy by heat and not temperature at all

 

[urtalkinstupid]

Yes, temperature is a measure of heat. Sorry for that. There is a required source of energy to exert work on an object. Energy is related to force. We've established that through a poorly derived equation. I'm sure a little more work we can get a nice relationship.

If a magnet is said to do no work, how is that possible? We know it requires a force to act against gravity to stay on the refrigerator, but no work is done, because it doesn't move anything. In examples ithat nvolved pushing stuff, the energy is transferred into heat, if nothing is moved. What is the case with the magnet?

 

[JoeWade]

you seem to have a problem with work not being done unless movement is involved.

since that's how we DEFINE "work" maybe you'd better use a different term.

 

[chroot]

since that's how we DEFINE "work" maybe you'd better use a different term.

It is not how we define work.

- Warren

 

[chroot]

There is a required source of energy to exert work on an object.

You've yet to substantiate this claim.

If a magnet is said to do no work, how is that possible? We know it requires a force to act against gravity to stay on the refrigerator, but no work is done, because it doesn't move anything. In examples ithat nvolved pushing stuff, the energy is transferred into heat, if nothing is moved. What is the case with the magnet?

Magnetism is a conservative force. A human's muscles pushing something do not constitute a conservative system.

- Warren

 

[JoeWade]

It is not how we define work.

- Warren

come again? Force Distance, no?

 

[chroot]

Work is defined differently in different contexts. In ideal gas mechanics, for example, it's $W = p \Delta V$.

- Warren

 

[JoeWade]

well let's not get sidetracked. for the system at hand distance is the factor.

in an orbital system, the distance between the object and its satellite is what we're calculating with. in a perfect circular orbit, no work would ever be done. in an elliptical orbit work is done one way by the primary drawing the satellite closer and increasing its velocity, then the other way back to apex by the sattelite, using up the velocity to get back to where it started.

hence conservation of energy in the system.

no NET work

 

[chroot]

Thanks JoeWade, that's already been said.

- Warren

 

[JoeWade]

i said it using smaller words

 

[rayjohn01]

Beatrix
If the distance between them increases ( which I am told it does ) and their masses do not change , then their center of mass does not change but the force of gravity decreases ( between them ) , to balance this the common( and opposite) centifugal force must decrease which means both orbital velocities decrease wrt their center of mass. Since the mass of the Earth is much greater than the moon it's orbital velocity is much smaller but the percentage changes to each are the same and equal to half the percentage change of the distance.

 

[Alkatran]

Here's what you're having trouble understanding:

The velocity never changes. The distance between the two objects never changes.

Now, notice that energy is NOT a vector. This means that a change in direction WILL NOT CHANGE ENERGY AT ALL. You can rotate the speed all you want, you have the same kinetic energy.

Now, you also need to understand that there are different 'types' of energy and they are calculated differently. For example, kinetic energy is calculated via the equation you've already given: d*F = Ek. However, potential gravity is another type of energy and has a different equation associated to it: d*F=Eg.

Oh wait! They're the same thing! or are they?
The distance in the kinetic energy is distance TRAVELLED. The distance in the potential gravity equation is the distance BETWEEN two objects.

Since (assuming circular orbit) the moon is not changing speed, its kinetic energy is unchanging, so we can calculate the potential gravity change:
Since the moon is the same distance from the Earth on both sides, and the same force is (generally) being exerted on it, potential gravity is conserved as well.

Now, since the orbit is actually elliptical, and potential gravity energy drops as the moon approaches earth, this energy needs to go somewhere. It ends up as kinetic energy, making the moon move faster as it approaches earth, and, therefore, moving slower as it recedes.


Understand?

 

[beatrix kiddo]

it's ok, tran.. my post was old...

Now, since the orbit is actually elliptical, and potential gravity energy drops as the moon approaches earth, this energy needs to go somewhere. It ends up as kinetic energy, making the moon move faster as it approaches earth, and, therefore, moving slower as it recedes.


Understand?

yeah.. that's why i didn't do a follow up post...

Now, notice that energy is NOT a vector. This means that a change in direction WILL NOT CHANGE ENERGY AT ALL. You can rotate the speed all you want, you have the same kinetic energy.

u really, REALLY don't have to tell me energy isn't a vector..

 

[Alkatran]

it's ok, tran.. my post was old...
yeah.. that's why i didn't do a follow up post...
u really, REALLY don't have to tell me energy isn't a vector..

I wasn't arguing specifically towards you, more along the lines of this:

Yes, temperature is a measure of heat. Sorry for that. There is a required source of energy to exert work on an object. Energy is related to force. We've established that through a poorly derived equation. I'm sure a little more work we can get a nice relationship.

If a magnet is said to do no work, how is that possible? We know it requires a force to act against gravity to stay on the refrigerator, but no work is done, because it doesn't move anything. In examples ithat nvolved pushing stuff, the energy is transferred into heat, if nothing is moved. What is the case with the magnet?

Because electromagnetic forces are calculated in the same way gravity is, the same "different energy" argument applies. However, as was already stated, the force/work/energy/I'm really not sure, not my area from gravity is spread through the molecular bonds in the atoms that the magnet is 'latching' onto. (right?)

 

[urtalkinstupid]

Force is related to Energy, Force is related with work, so Energy and Work are related.

I was not saying energy was a vector that can be applied in a direction. The energy is spread out (scalar), while force is applicable in a direction vector). Are you trying to tell me that work is not scalar, so I can't relate Energy to it, because it contains a vector? I'm not getting it.

 

[Chronos]

Alkatran has it right, as does chroot. You don't need an energy supply if no work is being done. Force and energy are not the same thing. Energy is conserved, force is not.

 

[Alkatran]

Force is related to Energy, Force is related with work, so Energy and Work are related.

I was not saying energy was a vector that can be applied in a direction. The energy is spread out (scalar), while force is applicable in a direction vector). Are you trying to tell me that work is not scalar, so I can't relate Energy to it, because it contains a vector? I'm not getting it.

I'm not exactly sure how to explain this to you, because you have a history of misinterpreting explanations. But let's just say that whatever it is that gets rid of the vector part of a force is the multiplication by distance.

You can't add vectors to scalar quantities, so if work isn't scalar you can't add it to energy... and we do. So.. work is scalar.

 

[urtalkinstupid]

You can convert scalar to vector and vice versa.

 

[Chronos]

You can convert scalar to vector and vice versa.

I will try to refrain from getting sucked into arguing against your flawed theories and the precious little math you use, but clearly do not understand, to support them.

 

[russ_watters]

This is as far as I got in the thread and it bears repeating:

This is not difficult to understand.

- Warren

This is not difficult to understand. I'm no longer amused and I no longer believe there is any chance you kiddies are making an honest effort here.

 

[beatrix kiddo]

chronos u can convert from scalar to vector.. and russ we aren't trying to amuse u..

 

[Math Is Hard]

chronos u can convert from scalar to vector.. and russ we aren't trying to amuse u..

why would you do that? I am not sure I follow...? For instance, if you take away the directional component of a force and only have a magnitude of acceleration (times mass) remaining, doesn't that leave you with a push or pull in no direction?

But maybe I don't understand what you mean by "convert".

 

[Alkatran]

why would you do that? I am not sure I follow...? For instance, if you take away the directional component of a force and only have a magnitude of acceleration (times mass) remaining, doesn't that leave you with a push or pull in no direction?

But maybe I don't understand what you mean by "convert".

When you use the work equation, you multiply by sin(A), which breaks it down to one of the components of the vector. This component is scalar. (You do it all the time when summing vectors)

In essence, multiplying by a trig function tends to remove the vectoriel properties of a vector.

To Urtalkinstupid:
My mistake, it's the multiplication by the angle, not the distance, which makes Work scalar.

 

[russ_watters]

...and russ we aren't trying to amuse u..

No, you're probably just trying to amuse yourselves. Regardelss, the only reason you two are still members here is you are amusing to us (except a small possibility others are learning from your mistakes). Our patience, however, has limits.

The concept being explained here is (again) junior high science class simple (and I'm pretty sure I could make an average elementary school student understand it*). Its simply not possible for you two to not be smart enough to understand it. That means you guys are choosing to not understand it. Why, we're not sure, but regardless, you are not making an effort to help yourselves and you are not making a positive contribtion to this board.

*Elementary school analogy: A car stationary at idle does no useful work. All of the energy input by the engine is wasted as heat, either dissipated by the radiator/engine block or blown out the tailpipe.

 

[jcsd]

You can convert scalar to vector and vice versa.

You can get a scalar from a vector (by taking the modulus for example), but you can't get a vector from a scalr without introducing another vector.

 

[beatrix kiddo]

No, you're probably just trying to amuse yourselves. Regardelss, the only reason you two are still members here is you are amusing to us (except a small possibility others are learning from your mistakes). Our patience, however, has limits.

basically u see stupid and me as a couple of clowns here for ur enjoyment... and ur patience is limited... what's going to happen when it runs out? are u going to close this thread down, too?

Its simply not possible for you two to not be smart enough to understand it. That means you guys are choosing to not understand it. Why, we're not sure, but regardless, you are not making an effort to help yourselves and you are not making a positive contribtion to this board.

we choose not to understand?! oh we understand.. we ALL understand...

*Elementary school analogy: A car stationary at idle does no useful work. All of the energy input by the engine is wasted as heat, either dissipated by the radiator/engine block or blown out the tailpipe.

...

You can get a scalar from a vector (by taking the modulus for example), but you can't get a vector from a scalr without introducing another vector.

agreed

 

[Entropy]

so Energy and Work are related.

Work and energy are the same thing.

$$E = {{kg} \cdot {m^2}} / {s^2}$$

$$W = F \cdot m$$

and...

$$F = {{kg} \cdot {m}} / {s^2}$$

so substitute F and you get...

$$W = {{kg} \cdot {m}} / {s^2} \cdot m =$$

$${{kg} \cdot {m^2}} / {s^2} = E$$

Therefore...

$$E = W$$

 

[Alkatran]

You can get a scalar from a vector (by taking the modulus for example), but you can't get a vector from a scalr without introducing another vector.

You use two scalar values and an angle to get a vector. (Unless the components of a vector aren't considered scalar?)

Anyways, it stands to reason that if you can go one way, you can go the other by doing the opposite thing. If multiplying by sin(30) eliminated the vector.. well divide by it... or use arcsin.

 

[Entropy]

basically u see stupid and me as a couple of clowns here for ur enjoyment... and ur patience is limited... what's going to happen when it runs out? are u going to close this thread down, too?

Errr... More like ban you guys from the board.