Work-Energy Theorem: Bar, Resistor, & Magnetic Field

In summary: The power dissipated in the resistor is P=Wv. You are asked to find the work done by gravity alone and this is certianly not zero!
  • #1
Dahaka14
73
0
Okay, I just took a test where there was a loop with a bar and a resistor in a magnetic field going into the screen as follows
____/\/\/\_____
lxxxxxxxxxxxxxxxlxxx
lxxxxxxxxxxxxxxxlxxx
lxxxxxxxxxxxxxxxlxxx
lxxxxxxxxxxxxxxxlxxx
l____(bar)_____lxxx
lxxxxxxxxxxxxxxxlxxx
lxxxxxxxxxxxxxxxlxxx
lxxxxxxxxxxxxxxxlxxx
lxxxxxxxxxxxxxxxlxxx

where the bar starts at rest, and begins to accelerate vertically down due to gravity. We were asked to find the final velocity (which I understand), and then AFTER IT HAS REACHED TERMINAL VELOCITY find the power dissipated through the resistor as a function of time and the work done by gravity as a function of time.

At first I set P=IR^2, and solved using the terminal velocity I found. But then I thought that the current isn't changing, because the velocity is constant, so I assumed it would be zero. I am very shaky on this one, and it's probably wrong, right?
Then, for the work done by gravity, I set W(t)=mgh=mgvt. But then I remembered the work-energy theorem, and there is no change in kinetic energy due to its constant velocity, and there is no net force on it anyhow. What are the answers?
 
Physics news on Phys.org
  • #2
Dahaka14 said:
Okay, I just took a test where there was a loop with a bar and a resistor in a magnetic field going into the screen as follows
____/\/\/\_____
lxxxxxxxxxxxxxxxlxxx
lxxxxxxxxxxxxxxxlxxx
lxxxxxxxxxxxxxxxlxxx
lxxxxxxxxxxxxxxxlxxx
l____(bar)_____lxxx
lxxxxxxxxxxxxxxxlxxx
lxxxxxxxxxxxxxxxlxxx
lxxxxxxxxxxxxxxxlxxx
lxxxxxxxxxxxxxxxlxxx

where the bar starts at rest, and begins to accelerate vertically down due to gravity. We were asked to find the final velocity (which I understand), and then AFTER IT HAS REACHED TERMINAL VELOCITY find the power dissipated through the resistor as a function of time and the work done by gravity as a function of time.

At first I set P=IR^2, and solved using the terminal velocity I found. But then I thought that the current isn't changing, because the velocity is constant, so I assumed it would be zero. I am very shaky on this one, and it's probably wrong, right?
I don't understand your sentence.What is "it" when you say "it should be zero"??
If the current is constant it does not mean it must be zero!
Then, for the work done by gravity, I set W(t)=mgh=mgvt. But then I remembered the work-energy theorem, and there is no change in kinetic energy due to its constant velocity, and there is no net force on it anyhow. What are the answers?
There is no net force and the net work is zero. But you are asked to find the work done by gravity alone and this is certianly not zero!

The work done by gravity is mgh indeed. The power is mg dh/dt = mgv.
 
  • #3
I typed too short of a post. I meant 0 for both the power dissipated and work done by gravity. I see your point about work done BY GRAVITY, I was just confused by thinking about the work-energy theorem with respect to the system. This is one concept that I've never really gotten a solid ground on; if there is no change in kinetic energy, then how is there work done by gravity? I ask this because last semester there was a question posed to the class during lecture saying that if one moves a box from the ground to a table, there is no work done by gravity because the change in kinetic energy is zero. I first responded that there was work done by gravity because there was a change in potential energy due to its position in the field, but he said I was wrong...

Now your answer to the power dissipated in the resistor...are we only considering gravity when we say that P=Fv? What about the magnetic field's force? How do we determine to use gravity as the source of the power dissipation? This was on our exam and we never went over this concept.
 

1. What is the work-energy theorem?

The work-energy theorem states that the work done by all the forces acting on an object is equal to the change in its kinetic energy. This means that the net work done on an object will result in a change in its motion, either by speeding up or slowing down.

2. How does the work-energy theorem apply to a bar?

In the context of a bar, the work-energy theorem can be used to calculate the change in kinetic energy of the bar as it moves. This can be useful in analyzing the motion of a bar in various scenarios, such as when it is being lifted or swung.

3. What role does a resistor play in the work-energy theorem?

A resistor is a device that resists the flow of electric current, and therefore, it can convert electrical energy into heat. In the context of the work-energy theorem, a resistor can be included as a force that does work on an object, resulting in a change in its kinetic energy.

4. How does a magnetic field factor into the work-energy theorem?

A magnetic field can also be included as a force acting on an object in the work-energy theorem. This is because a moving charged particle, such as an electron, experiences a force when it enters a magnetic field. The work done by this force can then be calculated to determine the change in kinetic energy of the particle.

5. What are some real-world applications of the work-energy theorem with a bar, resistor, and magnetic field?

The work-energy theorem with a bar, resistor, and magnetic field can be applied in various real-world scenarios, such as in the design and analysis of electrical circuits, particle accelerators, and electromagnetic motors. It can also be used to study the motion of objects in magnetic fields, such as a charged particle moving through a magnetic field in a cyclotron.

Similar threads

Replies
4
Views
935
Replies
4
Views
1K
Replies
65
Views
3K
Replies
34
Views
3K
  • Mechanics
Replies
2
Views
963
Replies
16
Views
1K
  • Mechanics
Replies
1
Views
1K
Replies
3
Views
2K
Replies
5
Views
813
  • Mechanics
Replies
33
Views
2K
Back
Top