# Work Done confusion

1. Oct 28, 2009

### Chewy0087

Hm just having a little problem with this which should only take a second to check, trust me I have used the search function/wikipedia but it's difficult to find a concise answer.

If i were to raise a 1kg weight 1 meter, according to $$W = F.d$$ I will have done 1J of work, fine however I think this is 1J against gravity. Because according to the work-energy theorum no work is done, as the kinetic energy remains the same, fine. from what I understand this stems from the fact that as I raise it i (providing an upwards force + 1m upwards) am doing positive work, however gravity (acting downwards but moving it upwards) is doing equal negative work.

So total work done $$mgh - mgh = 0$$ However, now it has gained potential energy, and if I were to drop it gravity does positive work on it equal to the potential energy it had. Obviously energy is still conserved so I have no issue there but the real question is simply this;

does work done not have to be conserved? or is there a better way of thinking about it? because potential energy is not taken account of as far as i can see with Work...

thanks, I know i've gone on abit here and wasn't very clear, and it is a very simple question but would be nice to have it cleared up!

Last edited: Oct 28, 2009
2. Oct 28, 2009

### mikelepore

You mean a 1 N object, not a 1 kg object.

You have to apply 1 N of force upward to cancel out the 1 N of downword force exerted by gravity, so that there will be a net force of zero on the object. It is with a net force of zero that you raise the object, assuming that you raise it with negligible acceleration, that is, constant velocity. There is no NET work done on the object when you raise it. That agrees with the fact that the final KE is zero.

Last edited: Oct 28, 2009
3. Oct 29, 2009

### jmb

For clarity I'm going to refer to your 1kg object as a box in what follows, but obviously it can be anything you want.

The problem is that the gravitational potential energy belongs to the earth-box system not to the box itself. It is a failure to take this into account that leads you to what you rightly identify as a worrying conclusion: that no work has been done on an object but its energy has increased.

Try instead thinking of it like this:

You apply a force of 1N to the box over a distance of 1m. This transfers energy to the box. At the same time the box does work against (it's moving in the opposite direction to the force) the gravitational force between the earth-box system. This transfers energy into the gravitational potential of the earth-box system. Since the force you apply to the box and the gravitational force it works against are equal, the same amount of energy is trasferred out of the box (and into the gravitational potential) as is transferred in by you. Thus, looking just at the kinetic energy of the box, there has been no net work done on it. However, looking at the earth-box system as a whole, the mechanical work you have done (which you are correct in thinking of as a form of energy) has been transferred into the potential energy of the combined system and so energy remains conserved.

I hope that makes it clear, please feel free to ask further questions if it doesn't!

4. Oct 29, 2009

### Staff: Mentor

The concept of potential energy is just a way of keeping track of the work done by a conservative force. You either include potential energy or you discuss the work done by gravity, but not both. (If you use both concepts together, you'll be double counting!)

Considering gravity as a force: The net work is zero, thus there's no change in KE.

Considering gravitational PE: You do some work in lifting the mass, thus increasing the PE. The work you do just accounts for the increase in PE, so there's no change in KE.

5. Oct 31, 2009

### Chewy0087

hmmm when you say double counting what do you mean? that they're the same thing?

this is confusing!

6. Oct 31, 2009

### Chewy0087

also just to clarify, if I applied a force upward and when it had been raised 1m and was going 5m/s would the work done been

$$W = m(g + \frac{5^2}{2})$$ ?

because if so, that would imply that

[tex] Work Done (by someone) = \delta U_{e} + \delta K_{e} [/itex]

however the work-energy theorum clearly states that the work done is only the change in Ke, the only reason I can think of is that it dismisses Pe due to the fact that Work done for Pe by you = - work done by gravity...

this is getting me down... =P

edit: OH i see! so basically the work-energy theorum doesn't account for conservative/non conservative forces? only motion? therefore we need the concept of potential energy etc?

Last edited: Oct 31, 2009
7. Oct 31, 2009

### Staff: Mentor

Right. The work done by you increases the total mechanical energy. Note that the work done by you does not represent the total work done by all forces. Edit: As Bob S points out, that equation written out fully would be W = m(gh +v2/2). But I know that you plugged in h = 1 m and v = 5 m/s.

Right. The "work-energy" theorem says that the net work done by all forces will equal the change in KE. That includes the work done by gravity. If you prefer to represent the force of gravity in terms of a potential energy, then you can say that the net work done by all forces (except gravity) will equal the change in KE plus gravitational PE. Your choice, but you can only pick one at a time (that's what I meant by not "double counting"):

Work done by you + work done by gravity = change in KE [the Work-Energy Theorem]
Work done by you - mgh = change in KE
Work done by you = change in KE + mgh = change in KE + GPE [conservation of mechanical energy]

Make sense?

Last edited: Oct 31, 2009
8. Oct 31, 2009

### Bob S

You are mixing units inside the brackets. g has units meters/sec2, while the second term has units meters2/sec2
Bob S

9. Oct 31, 2009

### Staff: Mentor

He's plugging in numbers, so the units are not obvious. (Not a wise policy.) He's using a distance of 1 m, so it doesn't appear.

10. Oct 31, 2009

### Chewy0087

that's fantastic, much obliged

so out of interest, wouldn't the gain/loss in PE be equal to the work done by gravity?

Last edited: Oct 31, 2009
11. Oct 31, 2009

### Staff: Mentor

Yes, that was the point of my last two equations. The work done by gravity would be -mgh, which is equivalent to an increase in GPE of mgh.