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

[Dahaka14]

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?

 

[nrqed]

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.

 

[Dahaka14]

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.