Work done by external agent in attractive potential

[Waxterzz]

Here's the deal:

You have an attractive potential like the gravitational one.

-∫F.dr = U2 - U1 = ΔU

If you go from U1 to U2, an external agent has to provide positive work against the direction of the Force Field. But your gravitational potential has to deliver negative work in order to raise its potential to U2. The F in the equation is gravitational force. So an attractive potential does negative work to lower its potential.
U2-U1 is positive, in combination with the minus sign, your work by the gravitational force is negative.
How to rewrite your equation if you want F to be the Force exerted by the external agent?

∫Fext.dr = U2 - U1 = ΔU

I would say drop the minus sign, but the general form for a conservative force is F = -∇U
Is my error of thought because of the fact that the F in F = -∇U is your gravitational force due to your gravitational potential and so F = -∇U is not valid for you're external agent? Because the your external agent is in the opposite direction of your force.

Now

If you go back from U2 to U1

-∫F.dr = U1-U2

Right side is negative, so your work done by the gravitational field is positive. So an attractive potential does positive work in order to lower its potential.

But what does your external agent do in this case? Now the external force is aligned with the force field, so I guess that the external agent also delivers positive work?

So equation for the external force:

-∫Fext.dr = U1-U2 = ΔU

So the equations for your field -∫F.dr = ΔU, always stays the same.
But the equations for your external agent differ by a sign.

Is this normal?

But I thought work done by external agent = -1 * work by the field you're acting in?

Am I thinking too much. Semantics, I hate it.

Excuse me for my horrible paint drawing. :)

Also, If you choose youre Potential to be zero at infinity. An attractive potential is always negative and a repulsive potential is always positive right?

 

[xAxis]

Your post is a bit confusing. Usual convention is that work done by the field is positive, and work done by external agent against the field is negative, so that the two equations integral and deferential are always consistent for attractive and repulsive potential.

Also, If you choose youre Potential to be zero at infinity. An attractive potential is always negative and a repulsive potential is always positive right?

right.

 

[Waxterzz]

Your post is a bit confusing. Usual convention is that work done by the field is positive, and work done by external agent against the field is negative, so that the two equations integral and deferential are always consistent for attractive and repulsive potential.


right.

If you go from potential U1 to U2 in my silly paint drawing, the field does negative work? The external agent that causes this shift in potential is doing positive work.

F.dr is negative, since your Force is directed to the right and your displacement is to the left.

And what if the force exerted by an external agent is aligned with your force field?

If you lower a pen from your desk to the ground, you're doing positive work as does the gravitational field?

 

[sophiecentaur]

If you go from potential U1 to U2 in my silly paint drawing, the field does negative work? The external agent that causes this shift in potential is doing positive work.

F.dr is negative, since your Force is directed to the right and your displacement is to the left.

And what if the force exerted by an external agent is aligned with your force field?

If you lower a pen from your desk to the ground, you're doing positive work as does the gravitational field?

Not really. You have to 'do work' (i.e. positive work) when you lift the pen. Work is 'done on' your arm by the pen as you lower it. The pen is 'helping you' on the way down.

I think you have got yourself in a bit of a twist here. Just keep at it and you will finally be able to accept what is only a convention when you get down to it - and that convention is consistent.

 

[Waxterzz]

Not really. You have to 'do work' (i.e. positive work) when you lift the pen. Work is 'done on' your arm by the pen as you lower it. The pen is 'helping you' on the way down.

I think you have got yourself in a bit of a twist here. Just keep at it and you will finally be able to accept what is only a convention when you get down to it - and that convention is consistent.

I think my error is in my first post:

∫Fext.dr = U2 - U1 = ΔU

If I lift an object, I do positive work indeed, and my potential energy(stored in my muscles) decreases. So the ΔU is not about the gravitational/any field anymore, but about the potential energy of the external agent, in this case.

So it has to be

∫Fext.dr = - ΔUext

And ΔU = -ΔUext (since the gravitational field gains energy, and the external agent loses it)

I think that's my error of thinking? Identifying what is what.

 

[xAxis]

If you go from potential U1 to U2 in my silly paint drawing, the field does negative work? The external agent that causes this shift in potential is doing positive work.

If the external agent is causing this shift then the force is not conservative and the equations are not valid.

F.dr is negative, since your Force is directed to the right and your displacement is to the left.

No, F.dr is always opposite of the gradient of potential, which is a rate of change of potential, not potential itself which in this case decreases as x increases, but for repulsive potential wouldn't be the case.

And what if the force exerted by an external agent is aligned with your force field?

If you lower a pen from your desk to the ground, you're doing positive work as does the gravitational field?

The external agent, especially human makes this really a nonconservative field, so it depends on the way that you apply the force. But then
-∫F.dr = ΔU really doesn't apply.