Conversion of EPE to Kinetic Energy in Magnetism

In summary, the conversation is discussing the concept of electric potential energy between two objects with positive charges and different masses. The participants are trying to figure out the correct formula for calculating the final speeds of the objects after being released from rest. They realize that they need to use both mechanical energy and momentum conservation laws to solve the problem and that the final speeds depend on the ratio of the masses of the objects.
  • #1
lluke9
27
0
So suppose we have two objects, both with positive charge Q Coulombs and X meters apart.
And now suppose Object 1 has a mass of m1 and Object 2 has a mass of m2.
In other words, the charges are identical, but masses are not.

So each of these will have an Electric Potential Energy of -QEΔX, because EPE = -QEΔd.
Easy peasy, right? EPE = KE, just like PE = KE for a falling object! Just like mgh = (1/2)mv^2, -QEX = (1/2)mv^2. So:

EPE = KE, so for Object 1 it's
-QEΔX = (1/2)m1v2.

and for Object 2 the formula is
-QEΔX = (1/2)m2v2.

Apparently, that's wrong and I was actually supposed to work it out this way:
-QEΔX = (1/2)m1v2 + (1/2)m2v2

It does make sense... but it doesn't. I mean, each object has each of their own potential energy, right? And each object loses that EPE and changes it 100% into KE, right? Isn't that what the potential energy is?

I mean, I don't remember doing PE = KE1 + KE2 for a falling object!
As far as I know, I've always done:
mgh = (1/2)mv2.

I don't remember ever making the mass of the Earth share the potential energy.
If I apply the same concept to gravity, it would be like:
mgh = (1/2)mEarthv2 + (1/2)mobjectv2.
What!? I don't remember ever doing that!

Help me out here, guys! I am greatly befuddled! :cry:
 
Physics news on Phys.org
  • #2
lluke9 said:
I mean, each object has each of their own potential energy, right?
Wrong. See my response (post #2) here.
It's probably not your fault that you are befuddled. To solve problems of this kind, where you start with an initial configuration and speeds of two objects and you want to know (say) the speeds of the two objects in their final configuration, you need to use two conservation laws, mechanical energy and momentum. That's because you have two unknowns, namely the speeds, and energy conservation is not enough. In most textbooks energy conservation comes before momentum conservation. Why? Because in the case of a small object falling near the surface of the Earth, for every 1 Joule of energy gained by the Earth, the object gains energy of order ##\mathrm{10^{24}}##.
Here is the correct calculation using the conservation laws for a rock of mass ##m## released from rest by ##h##. Initially, the rock and stone are at rest relative to each other. we are looking for the final speeds of the two objects. We get them by using the two conservation laws. Subscript ##e## stands for "Earth".

1. Momentum conservation, ##P_{before}=P_{after}##.
$$0+0=M_eV_e+mv=0~\rightarrow~V_e=-\frac{m}{M_e}v$$
2. Mechanical energy conservation, ##ME_{before}=ME_{after}##.
$$mgh=\frac{1}{2}M_eV_e^2+\frac{1}{2}mv^2=\frac{1}{2}M_e \left( -\frac{m}{M_e}v \right)^2+\frac{1}{2}mv^2=\frac{1}{2}\left(1+\frac{m}{M_e}\right) mv^2$$Now you see that for a mass of order 1 kg, the ratio of the masses is ##m/M_e=10^{-24}## which means that to an extremely super good approximation ##mgh\approx \frac{1}{2}mv^2##, i.e. all of the potential energy of the two-body system goes into kinetic energy of the rock. Clearly that's not the case when the masses are of the same magnitude; you have a 50-50 split when the masses are equal. But let's find the final speeds anyway. We have $$v=\sqrt{\frac{2gh}{1+m/M_e}}\approx \sqrt{2gh}$$and $$V_e=\frac{m}{M_e}\sqrt{\frac{2gh}{1+m/M_e}} \approx 0.$$ No wonder you are befuddled. Once you learned about momentum conservation, it would have been a good idea for your instructor to have shown you what's under the hood.
 
Last edited:

What is the definition of EPE in magnetism and how is it related to kinetic energy?

EPE stands for elastic potential energy, which is the energy stored in a system due to the deformation or stretching of an object. In magnetism, EPE is the energy stored in a magnetic material due to its orientation in a magnetic field. This energy can be converted into kinetic energy when the magnetic material moves or changes position in the field.

How is the conversion of EPE to kinetic energy in magnetism related to the concept of work?

The conversion of EPE to kinetic energy in magnetism follows the principle of work-energy theorem, which states that the work done on an object is equal to the change in its kinetic energy. In this case, the work done in moving or changing the position of a magnetic material against the force of the magnetic field is converted into kinetic energy.

What factors affect the conversion of EPE to kinetic energy in magnetism?

The conversion of EPE to kinetic energy in magnetism is affected by the strength of the magnetic field, the orientation and properties of the magnetic material, and the distance and speed at which the material moves or changes position in the field. These factors determine the amount of work done and the resulting kinetic energy.

Can EPE be completely converted into kinetic energy in magnetism?

No, the conversion of EPE to kinetic energy in magnetism is not 100% efficient. Some of the energy is lost as heat due to friction or resistance in the movement of the magnetic material, and some may be stored as potential energy again if the material returns to its original position in the magnetic field.

What are the real-life applications of the conversion of EPE to kinetic energy in magnetism?

The conversion of EPE to kinetic energy in magnetism is used in various technologies such as electric motors, generators, and magnetic levitation systems. It is also essential in the functioning of devices like speakers, hard drives, and MRI machines. Additionally, the principles of EPE and kinetic energy in magnetism are used in the design and development of renewable energy sources like wind turbines and hydroelectric generators.

Similar threads

  • Electromagnetism
Replies
11
Views
1K
  • Introductory Physics Homework Help
Replies
10
Views
2K
  • Introductory Physics Homework Help
Replies
21
Views
1K
  • Introductory Physics Homework Help
Replies
3
Views
236
  • Other Physics Topics
Replies
13
Views
3K
Replies
8
Views
5K
  • Introductory Physics Homework Help
Replies
15
Views
301
  • Introductory Physics Homework Help
Replies
6
Views
937
  • Special and General Relativity
2
Replies
55
Views
3K
  • Classical Physics
Replies
29
Views
2K
Back
Top