Is the Force-Displacement Relationship Conserved in Energy Conservation?

[Algren]

How would one prove W = F.s and why would only F.s be conserved (in energy conservation) and why not F.s^2, or F.s^3? One possibility is by experimental observation and verification. But how would one go about it without experimentation?

s = Displacement vector
F = Force vector
and by dot product, we are basically trying to find: (Component of Force along the displacement)*(displacement)

 

[Ryoko]

Not sure what 's' is. I'm assuming distance since Work can be defined as Force * Distance.

To answer why it won't be s^2 or some other factor, you need only look at the units and do a little dimensional analysis. The units for work are N.m and not N.m^2.

 

[Algren]

The units for work are N.m and not N.m^2.

I guess that HAS been derived from F.distance itself. Cyclic.

I'm assuming distance since Work can be defined as Force * Distance.

Thats what i am trying to prove. Instead, one could use Conservation of Energy to prove it, and i am asking whether or not it is necessary, and how to do so.

 

[Dale]

How would one prove W = F.s

You don't prove that. That is the definition of work. You don't prove definitions, you just define them.

and why would only F.s be conserved (in energy conservation)

This is called the work-energy theorem. Here is a proof of that:
http://faculty.wwu.edu/vawter/PhysicsNet/Topics/Work/WorkEngergyTheorem.html

There is also, of course, a proof in Wikipedia:
http://en.wikipedia.org/wiki/Work_(...ion_of_the_work-energy_theorem_for_a_particle

 

[Algren]

There is also, of course, a proof in Wikipedia:
http://en.wikipedia.org/wiki/Work_(...ion_of_the_work-energy_theorem_for_a_particle

Thank you for the links on Work energy theorem and it's proofs. It answers the question, but it somewhat fails to satisfy my purpose.

Although, this might seem a bit odd, but i also want to prove that KE of a particle of mass m is in fact 1/2 mv^2 without using F=ma or F.s = W.

But, for a pirticular system of forces, why would the quantity ƩFS be used in transfer of energy and why not ƩFs^2 , why not ƩF^3 . S^7. Now, using 1/2 mv2 or any potential energy equations to state it would lead to cyclic proofs. And to use the definition of work done, well, arent we trying to prove that in the first place?

Now, what work done signifies is the energy transferred into the body by the system of forces (by work energy theorem) in the form of KE and Potential. Using Potential energy equations (all are derived from the definition of work) would lead to cyclic.

Otherwise stating it(definition of work) as an Axiom of sorts would be more suitable.

 

[Dale]

The work energy theorem contains all of the physics of this discussion. It proves that $f \cdot s = \frac{1}{2}m v^2$.

Then we DEFINE the word "work" to mean the quantity on the left and we DEFINE the word "kinetic energy" to mean the quantity on the right. You don't prove definitions. The reason that we are interested in the quantities that we have defined as "work" and "kinetic energy" is because they are related through the work energy theorem.

You certainly could take the quantity $f^3 \cdot s^7$ and DEFINE the word "flubnubitz" to refer to that quantity. Then you could talk about the flubnubitz done by a force, just like you could talk about the work done. The only thing is that we don't care about flubnubitz since there is no flubnubitz-energy theorem nor any other similar theorem. It is a definition, but it is a physically useless definition since it doesn't tell us anything about anything other than flubnubitz.

 

[256bits]

why not ƩFs^2

Proof Why Not

Let F be a force pushing on an object.
Let x be the distance traveled by the object.
Let W = F s^2

1. Move the object a distance 2s, in two intervals of s
First interval
W1a = F s^2

second Interval
W2b = F s^2

Total work = W1a+W1b= 2 F s^2

2. Now move the object a complete distance 2s
For a distance 2s
W = 4 F s^2

Obviously it takes twice as much work to move an object the same distance by doing a half interval start and stop, rather than moving the same distance without stopping.

conclusion
W ≠ F s^2

 

[Algren]

conclusion
W ≠ F s^2

Well, i guess that does it. So, we could say we've proved that the definition of work is correct.

Now, as i said, i wanted to prove F=ma. Which i guess is impossible, what do you say?

 

[Jorriss]

Well, i guess that does it. So, we could say we've proved that the definition of work is correct.

Now, as i said, i wanted to prove F=ma. Which i guess is impossible, what do you say?

Just to be clear, he didn't prove W =/= F*s^2. He showed it is a bad definition of work. Work is defined.

 

[WannabeNewton]

Now, as i said, i wanted to prove F=ma. Which i guess is impossible, what do you say?

If your basis is Newtonian mechanics then you cannot prove this. It is a postulate that is empirically verified.

 

[Fredrik]

Well, i guess that does it. So, we could say we've proved that the definition of work is correct.

That's not what 256bits did. He just showed you that the alternative definition associates the term "work" with something that doesn't have the properties we want it to have.

You have already been told this a couple of times in this thread, but you seem to have ignored it: Definitions can't be proved. All a definition does is to specify what word we are going to use for a mathematical concept. A definition is a choice of what word to use. You can't prove these choices any more than you can prove your choice of what color socks to wear tomorrow.

Now, as i said, i wanted to prove F=ma. Which i guess is impossible, what do you say?

It's part of the definition of Newtonian mechanics that the motion of every particle satisfies a differential equation of the form $mx''(t)=F(x(t),x'(t),t)$. Now we can simply choose to call the thing on the right "force".

The "F=ma" assumption is part of the definition of the theory because we want a theory in which a particle's motion is completely determined by a specification of the position and the velocity at some specific time t₀. Such a specification is called an initial condition. So it would make sense to ask how to prove that a differential equation of this form has a unique solution for each initial condition. Unfortunately this is very hard. It's proved in books on differential equations.

 

[Dale]

Now, as i said, i wanted to prove F=ma. Which i guess is impossible, what do you say?

As WannabeNewton said, if you are using Newtonian mechanics then this is not something you prove, it is a postulate. And as Fredrik said, that postulate is generally taken as the definition of the term "force".

However, what you can do is to take a more fundamental theory, such as quantum mechanics, as your starting point and then derive F=ma as a classical limit of quantum mechanics. That is a seriously deep rabbit hole, so I won't join you on that journey, but you are certainly free to go there if you choose.