# Energy Conservation in Quasistatic process

1. Jul 30, 2011

### mishrashubham

Alright, this might sound stupid, but let us imagine a scenario. I bring an object from height 10 m to 4 m such that the velocity of the object remains constant. Due to this the potential energy of the object decreases. Now my question is, where does this energy go?

2. Jul 30, 2011

### Dickfore

In the work done by the external forces that oppose the force of gravity.

3. Jul 30, 2011

### mishrashubham

I do not not understand how energy can go into work. Isn't work the name given to the quantity that tells us the amount of energy transferred between bodies? So in this case, negative work is being done by the external force on the body. Therefore the body is losing energy. But where is this energy transferred to?

4. Jul 30, 2011

### Staff: Mentor

If an object (A) free falls from 10 m to 4 m it will gain KE. In order to keep A from gaining KE there must be a second object (B) exerting a force on A. By Newton's 3rd law A also exerts a force on B. This force does work on B. The extra energy is transfered to B.

5. Jul 30, 2011

### Philip Wood

For example, if you lower the object gently by hand, you are exerting an upward force on the object (so the net force on the object is zero, and it doesn't gain speed), so the object exerts an equal and opposite force, i.e. a downward force, on your hand. This force does work on you, specifically on your arm muscles, as you lower your hand. Your muscles acquire internal energy and become a small amount hotter.

6. Jul 30, 2011

### Dickfore

Have you heard about the work-energy theorem and conservative forces?

7. Jul 30, 2011

### mishrashubham

So that means if A is applying a force equal in magnitude and opposite in direction to that of the gravitational force on a body B at every instant, such that B descends towards the surface of the earth with constant velocity, the potential energy lost by B is transferred into A. Am I right?

Yes, I am aware of the theorem, but that is only concerned with kinetic energy isn't it?

8. Jul 30, 2011

### Staff: Mentor

The direction of the energy transfer depends on the direction of the force relative to the displacement (w=f.d).

Last edited: Jul 30, 2011
9. Jul 31, 2011

### Philip Wood

So (illustrating Dalespam's remark) the object that you are lowering exerts a downward force on your hand and the point of application of the force moves through a downward displacement so does a positive amount of work on your hand, so you gain energy (though as I said earlier, you can't do anything useful with this energy; it's 'dissipated' in your arm muscles, making them slightly hotter).

But you exert an upward force on the object, and the point of application of the force moves downwards, so the work done by you on the object is negative, and the object loses energy (to you).

The net result is that energy is transferred from the Earth-object system to you, as the object approaches the Earth, and gravitational potential energy decreases.

Last edited: Jul 31, 2011
10. Jul 31, 2011

### mishrashubham

Alright thanks. I got it.

11. Jul 31, 2011

### Dickfore

True, but, there are two (external) forces acting on the body:
1. The gravitational force, which is conservative and, therefore, the work done by it is simply the difference in the potential energy between the initial and final position
2. The external force that opposes gravity.

When you apply the work-energy theorem in this case, what do you get? The point is that when we say an object has potential energy, this energy is not property of the body, but a property of the body being in an external force field that exerts a force on it and does work on any trajectory that is simply equal to the difference of the potential energy between the intial and final positions.

Furthermore, for heat processes, there is another form of energy that the system (body) can exchange with the environment, namely, heat. In our case, there is the possibility that due to friction with the air, the body may accept heat. However, the force of friction is increasing with increasing speed. In the quasistatic limit, the velocity tends to zero, therefore, the heat transfer should also tend to zero.

And, finally, the kinetic energy of a body is only due to the motion of its center of mass as a whole. Nevertheless, there might be a motion of its internal constituents relative to the center of mass as well as potential energy due to their interactions. All of these contribute to the internal energy of the body. The internal energy is a function of the thermodynamic state of the body. If this state had not changed during the process, one can conclude that the internal energy had not changed as well.

12. Jul 31, 2011

### mishrashubham

The work-energy theorem in this case tells me that the net work done on the body is zero since there is no change in kinetic energy. Does this mean that that total energy of the body remains constant. But surely it has lost some energy since it is closer to the surface than before.

When a body is lets say, kept on a table, it does not move from its position relative to the surface due to an opposite force exerted by the table on the body equal to its weight. We say that it possesses potential energy with reference to the surface. And this remains constant because its height remains constant. Another way to tell this is that the net work done on it is zero. (Am I right?)

But in my original situation it is the same case isn't it? Except that now the body now has some initial velocity. But potential energy changes even though net work is zero.

13. Jul 31, 2011

### Dickfore

It seems you are confused about the concept of potential energy. I wrote this:

14. Jul 31, 2011

### mishrashubham

I know; something is seriously wrong with my understanding of potential energy. I will have to clear my basics before proceeding. So please help.

Could you tell me what exactly potential energy is? And what do we mean when say "potential energy of a system is X"? Does is mean that every component of the system has X amount of energy, not counting its kinetic energy? Or is it something else? Can potential energy be possessed by a body just like kinetic energy?

15. Jul 31, 2011

### Dickfore

For me, it is just a bookkeeping convention. Let me clarify. The work-energy theorem says that:
$$W_{\mathrm{tot}} = K_{f} - K_{i}$$
All the forces that act on a system can be split into two broad categories.
1) Forces whose work is independent on the path between two points, or, equivalently, is zero on any closed trajectory. These forces are called conservative. For them, one can assign a function of coordinates (up to an arbitrary additive constant), $V = V(\vec{r})$ such that the work done by the conservative force on the system is:
$$W_{\mathrm{cons}} = V_{i} - V_{f}$$
The force is derivable from this function through the negative gradient:
$$\vec{F}_{\mathrm{cons}} = -\vec{\nabla} V(\vec{r})$$

2) Forces that do not satisfy the above criterion. They are called non-conservative forces. There are several subcategories, among which I can think of the following:
a) Dissipative - the work done on any trajectory is negative;
b) Gyroscopic - the work is zero on any trajectory, because the force is always perpendicular to the velocity vector

Now, let us rewrite the work-energy theorem by splitting the work done by conservative and non-conservative forces:
$$W_{\mathrm{cons}} + W_{\mathrm{noncons}} = K_{f} - K_{i}$$
$$(V_{i} - V_{f}) + W_{\mathrm{noncons}} = K_{f} - K_{i}$$
$$W_{\mathrm{noncons}} = (K_{f} - K_{i}) + (V_{f} - V_{i})$$
$$W_{\mathrm{noncons}} = (K_{f} + V_{f}) - (K_{i} + V_{i})$$
Now, we can define a quantity:
$$E = K + V$$
called total mechanical energy. The function $V(\vec{r})$ is called potential energy of the system in a force field. Then, the work-energy theorem simply states that the change of the total mechanical energy is equal to the work done by non-conservative forces:
$$W_{\mathrm{noncons}} = E_{f} - E_{i}$$
If the work of the non-conservative forces is zero, we arrive at the conclusion that the total mechanical energy is conserved, hence the name.

So, it is a matter of convention. The only important thing is to not double count. If you attribute the potential energy to the system, then do not count the work done by the conservative forces in the work-energy theorem. If you use the work-energy theorem in its original form (with only the kinetic energy), then by all means count the work of the conservative forces. It is a simple difference of the initial and final potential energy.

According to the first option, the answer to your question is that the total mechanical energy is decreased because of the negative work done by the external force that opposes the gravitational force.

According to the second option, the answer is that the kinetic energy remains unchanged (and zero) because the external force is equal by magnitude and opposite in direction to the gravitational force. Thus the works done by these forces are exactly opposite. But, the work done by the gravitational force is equal to the difference of the initial and final potential energy.

16. Jul 31, 2011

### Staff: Mentor

I wouldn't say that the potential energy belongs to any individual component of the system, but rather to the system as a whole.

17. Aug 2, 2011

### mishrashubham

Alright thanks for the explanation. So rephrasing my question; Worknoncons= change in mechanical energy. In my original scenario, the mechanical energy has changed due to change in potential energy even though kinetic energy remains constant. That is explained by the fact that A nonconservative force i.e. my hand has done some work on the body. Since I am doing negative work on the body, does this mean that the potential energy lost by the body is transferred into my body i.e do I gain the energy that the body has lost?

So does it mean that sum of the energies of all the components of a system is equal to the potential energy of the system?

18. Aug 7, 2011

### mishrashubham

*bump*

19. Aug 7, 2011

### Staff: Mentor

I don't know. You would have to define carefully what you meant by "the energy of a component of a system". Here is a Wikipedia link that I like on the topic:
http://en.wikipedia.org/wiki/Binding_energy

20. Aug 13, 2011

### mishrashubham

Alright...

Does it make any sense if I say, "A particle X possesses Y joules of energy."? I mean do I have to provide any additional information just like when I need to specify the frame when I say "Particle A moves with velocity B m/s."?