Why is work done expressed as F.dx instead of x.dF?

[Mr Real]

My question is:why is small work done,dW taken as F.dx instead of x.dF because integrating the latter should also give the same result(taking x out of the integral and integrating dF gives x (F) ).Visualising this integration graphically seems to suggest the same.
Similar is the case with quantities like electric flux (=E.dA),why isn't it dE.A ?
p.s.There are similar questions here but answers to them didn't help.

 

[Dale]

I am not even sure what dF would mean in this context. In most cases I can think of, F is a function of x, but x isn't a function of F.

 

[Mr Real]

I am not even sure what dF would mean in this context. In most cases I can think of, F is a function of x, but x isn't a function of F.

Okay I got what you mean but we can make the force dependent on distance (i.e. a function of distance) by varying it with increasing distance (e.g. force is given by the relation:F=3x,where x is distance )??
And can't all of it be seen in the graph? dF can be seen as an infinitesimal portion of the graph, so we get a very small area under the curve and summing up all these areas still gives the total area(which I think,will be the total work done..)?

 

[Dale]

we can make the force dependent on distance (i.e. a function of distance)

Sure. Since F can be written F(x) then F(x).dx makes sense. In contrast x(F).dF doesn't make sense precisely because x often cannot be written as x(F), such as when F is constant.

dF can be seen as an infinitesimal portion of the graph, so we get a very small area under the curve and summing up all these areas still gives the total area

No, that doesn't work. Suppose for the sake of argument we have some invertable force profile such that we can write both F(x) and x(F). Then F(x)dx is an infinitesimal element of the area between the curve and the x-axis while x(F)dF is an infinitesimal element of the area between the curve and the F axis. They are opposite parts of a rectangle, not the same area.

 

[Mr Real]

About getting different areas under the curve,that is a part of my doubt too.Why do we only consider the area that the curve makes with the displacement axis (x-axis here ) and not the area it makes with the force axis (y-axis here)?

Sure. Since F can be written F(x) then F(x).dx makes sense. In contrast x(F).dF doesn't make sense precisely because x often cannot be written as x(F), such as when F is constant.

But isn't displacement always a function of force(because it changes with the force applied)? And if it isn't always true, then what happens when we consider the case when it is?Then, wouldn't x.dF would be meaningful?

 

[pixel]

Work is defined in a way so that the work done is equal to the change in kinetic energy. That's true for dW = F dx. It's not true for dW = x dF.

 

[Dale]

But isn't displacement always a function of force(because it changes with the force applied)?

Not in general, no. If you have F=3x then you can indeed invert that and obtain x=F/3, both of which are valid functions. But if you have a constant force, F=3, then you cannot invert it to solve for x. Remember the mathematical definition of a function. A function has one output value for each of its input values, but in the case of a constant force, if you input the force then you have multiple output positions. So position is not a function of force.

About getting different areas under the curve,that is a part of my doubt too.Why do we only consider the area that the curve makes with the displacement axis (x-axis here ) and not the area it makes with the force axis (y-axis here)?

That other area isn't useful for analyzing physics as far as I know.

 

[Mr Real]

Not in general, no. If you have F=3x then you can indeed invert that and obtain x=F/3, both of which are valid functions. But if you have a constant force, F=3, then you cannot invert it to solve for x. Remember the mathematical definition of a function. A function has one output value for each of its input values, but in the case of a constant force, if you input the force then you have multiple output positions. So position is not a function of force.

Yeah, but if we consider the case when displacement Is a function of force; then wouldn't the product dF.x be meaningful and then it could be integrated?

That other area isn't useful for analyzing physics as far as I know.

That's what I don't understand. Why do we only take the area made with displacement axis and not the one with force axis?

 

[Dale]

Yeah, but if we consider the case when displacement Is a function of force; then wouldn't the product dF.x be meaningful and then it could be integrated?

Yes. Your example of F=3x is one such case.

That's what I don't understand. Why do we only take the area made with displacement axis and not the one with force axis?

Because that is the only one that has proven to be useful. As @pixel mentioned, it is useful because it is equal to the change in KE.

The other one is certainly not equal to the change in KE, and as far as I know it isn't equal to any other interesting quantity either.

 

[Chestermiller]

The integral of Fdx is not the same as the integral of xdF. What made you think that they are the same. I think a little review of calculus would be helpful here.

 

[Mr Real]

Yes. Your example of F=3x is one such case.

So even in this case can't the work done be given by the integration of dF.x?

Because that is the only one that has proven to be useful. As @pixel mentioned, it is useful because it is equal to the change in KE.

The other one is certainly not equal to the change in KE, and as far as I know it isn't equal to any other interesting quantity either.

Is there some convention regarding this or can you provide a link where it's clearly mentioned (or even the history regarding this derivation). I want this as I have seen a similar question on Quora by a user named suraj tiwari ( I don't know how to provide the link) and there different reasons were given, therefore I want the right reason.

 

[Mr Real]

The integral of Fdx is not the same as the integral of xdF. What made you think that they are the same. I think a little review of calculus would be helpful here.

But on integrating x.dF, x being constant comes out of the integral and integrating dF gives F; now dot product is commutative so F.x=x.F . I don't know where I am doing the mistake because thinking about it and visualising it graphically seems to suggest the same.

 

[Chestermiller]

The only way that Fdx=xdF is if F = kx, where k is a constant.

 

[Mr Real]

The integral of Fdx is not the same as the integral of xdF. What made you think that they are the same. I think a little review of calculus would be helpful here.

But on integrating x.dF , x being constant comes out of the integral and integrating dF gives force so that we get x.F; Now dot product being commutative F.x=x.F so according to me it should also give work done. I don't know what I'm doing wrongly because thinking about it or visualising it graphically seem to suggest the same.

 

[Dale]

So even in this case can't the work done be given by the integration of dF.x?

I am sure there will be some cases where, by coincidence, it happens to work. But the relationship does not hold in general. I don't know which cases would coincidentally work and which would not. I would guess that F=3x would work but F=3x+1 would not.

Is there some convention regarding this or can you provide a link where it's clearly mentioned (or even the history regarding this derivation). I want this as I have seen a similar question on Quora by a user named suraj tiwari ( I don't know how to provide the link) and there different reasons were given, therefore I want the right reason.

Often there are multiple valid reasons for the same thing, so I wouldn't get worked up by this. Every valid reason is "right".

I definitely do not have a reference that clearly states that x.dF is not useful (I cannot imagine why an author would explicitly make that statement, it would be a distraction that would confuse the majority of readers and any good editor would remove it). I can provide many references that show that F.dx is useful.

But on integrating x.dF , x being constant comes out of the integral and integrating dF gives force so that we get x.F; Now dot product being commutative F.x=x.F so according to me it should also give work done. I don't know what I'm doing wrongly because thinking about it or visualising it graphically seem to suggest the same.

I would recommend working it out. Try F=3, F=3x, F=3x+1, and F=3x^2. Integrate all of those from x=0 to x=2. Invert all of them, adjust your limits of integration, and integrate. Report the results, I am particularly curious about the F=3x+1 case.

 

[Chestermiller]

If x is constant (at say x1), then the integral of Fdx is not Fx. It is equal to x1(F2-F1), where F1 and F2 are the initial and final values of F.

 

[Dale]

The only way that Fdx=xdF is if F = kx, where k is a constant.

I also think that this is correct.

 

[Mr Real]

I think I got some of it. Please correct me if I'm wrong. In post #12 I had said that we can take x out of the integral as it is constant; this, in fact is not true if we assume displacement constant we would get a straight line parallel to the force axis and then work done(area under the curve will be zero). Therefore the 2 integrals will be different.

 

[Mr Real]

I definitely do not have a reference that clearly states that x.dF is not useful (I cannot imagine why an author would explicitly make that statement, it would be a distraction that would confuse the majority of readers and any good editor would remove it).

But at the beginner level I think it's imperative to state the reason that why is F.dx considered and not x.dF as most(if not all) beginners will probably get confused about it. I cannot imagine that there did not arise this issue when the formula was first derived or even that is an explanation about this confusion in authors' books.

 

[Mr Real]

Now, as a result of this discussion i know that integral of x.dF has been proven not to give the net work done. But some parts of my original question still stands, why doesn't it give the work done instead of F.dx or is not even a recognised quantity even though it seems to be closely connected with work? And also similar quantities like electric flux ( why is it equal to integral of E.dA and not that of dE.A ?). Is the experimental result the only evidence? Or was it decided as the convention for some unknown reason?
(I know, now it seems as though I'm asking things that are not known but I can't think of a possible reason of it not being discussed anywhere!)

 

[Dale]

most(if not all) beginners will probably get confused about it

Actually, I have never had anyone else besides you ask this question. That is not to say that it is a bad question, but it is very uncommon.

I don't think that textbook authors can be expected to anticipate such rare questions. In any case, it would more likely be a topic for a calculus text than for physics text. It is a mathematical property that the integrand x.dy is not generally the same as y.dx irrespective of its use in physics.

 

[Mr Real]

It is a mathematical property that the integrand x.dy is not generally the same as y.dx irrespective of its use in physics.

Yeah, now I am clear with that.
I know it is not a common question but doesn't the quantity x.dF signify anything? Even though it is also a product of force and displacement like F.dx and its integral too can be seen on the same graph (it is the area the curve makes with the force axis) but still it's not a meaningful quantity? And isn't there any other proof of why the integral of F.dx(instead of the integral of x.dF) gives the work done, apart from the experimental evidence( like using the work-energy theorem to compare it with the change in kinetic energy)?

 

[pixel]

...And isn't there any other proof of why the integral of F.dx(instead of the integral of x.dF) gives the work done, apart from the experimental evidence( like using the work-energy theorem to compare it with the change in kinetic energy)?

Here is a proof: http://www.sparknotes.com/physics/workenergypower/workpower/section4.rhtml
Refer to the section: Full Proof of the Work-Energy Theorem

As I mentioned above, work in physics has a specific definition. That definition is chosen so that the work done is the change in kinetic energy. It's okay to ask whether your alternate expression might have some meaning, but if it does, it is not work.

 

[Dale]

doesn't the quantity x.dF signify anything

I don't even know what the quantity dF would signify physically. It is a weird quantity. Yes, you can write it down mathematically, but it's meaning escapes me. I cannot think of any situation where it arises naturally.

Similarly with your other example of dE.

 

[sophiecentaur]

why is F.dx considered and not x.dF

Could you suggest a physical interpretation of xdF? It's an incremental change of force over a distance. If the force is unchanging (say friction) then dF would be zero and what would your limits be for F? They would be ' from F to F' and so the Definite Integral would be zero. =x(F-F)
Doesn't this demonstrate how you need to keep one foot on the ground when you start playing Maths out of context of its Physics application?

 

[nasu]

Physically, xdF models a small change in force at a fixed position.
And Fdx will be a small change in position at a constant force.
So they don\t have the same meaning at all.

If you start with the definition of work for constant force, it is quite easy to see which one of the two will be useful to define work for a force that depends on position.

 

[Mr Real]

Could you suggest a physical interpretation of xdF? It's an incremental change of force over a distance. If the force is unchanging (say friction) then dF would be zero and what would your limits be for F? They would be ' from F to F' and so the Definite Integral would be zero. =x(F-F)

But who said that the force is unchanging? Sure the work done would be zero if one considers the special case when the force is constant, but x.dF doesn't mean that the force F is constant.

 

[Dale]

Sure the work done would be zero if one considers the special case when the force is constant

And that is clearly wrong.

but x.dF doesn't mean that the force F is constant.

There are many physically interesting examples where F is constant. So a restriction to only non constant forces severely limits the usefulness of a quantity

 

[Mr Real]

And that is clearly wrong.

I was talking about the formula x.dF that if we integrate this for a constant force then the integral would be zero. I was not talking about F.dx.

 

[sophiecentaur]

I was talking about the formula x.dF that if we integrate this for a constant force then the integral would be zero. I was not talking about F.dx.

I really have no idea what point you are trying to make. Random maths doesn't necessarily have any relation to Science. There are a million maths arguments that will tell you that 0=0 or 1=2 but they have no meaning.