Deep meaning of the work integral formula

[Aleoa]

I want to understand very deeply the meaning of the work integral formula:

\int m\frac{d\bar{v}}{dt}d\bar{l}

It is not enough for me to know that it was defined in this way, I want to know why it was defined in this way.
To start, what is the physical meaning of m\frac{d\bar{v}}{dt}d\bar{l}, what we are looking for when we calculate this term ?

 

[Doc Al]

Try to give us some context for your question. What aspect are you asking about? For example, are you familiar with the "work-energy" theorem and how it is derived? (See: Work–energy principle)

 

[Dale]

work integral formula:

\int m\frac{d\bar{v}}{dt}d\bar{l}

This is a strange form of the formula. The usual formula for work is: $W=\int \vec F \cdot d \vec l$. They appear to only be equivalent if there is only a single force so $F=F_{net}$

I also would appreciate some context. Where did you get this form and what are you looking to understand about it? “Deeply” is more of a description of the amount of understanding but not a description about what it is that you want to understand. What frustrates you about your current understanding?

 

[Aleoa]

My intuition is that in the integral formula of the work \int m\frac{d\bar{v}}{dt}d\bar{l} the vectors dt and dl are linked, and that the integral simply represents the sum of all the mdv / dt along a path. Is this true ?
How dl and dt vectors are linked ?

 

[Dale]

the integral simply represents the sum of all the mdv / dt along a path. Is this true ?

Definitely not. $\int y \; dx$ is the sum of all of the $y\; dx$, not the sum of all the $y$.

It is customary to actually respond to the posts of the other participants, and not simply ignore them. It strongly discourages people from participating

 

[Aleoa]

So what is the fundamental quantity we are summing using the integral ? Doing some calculation I get the basic quantity we sum is mvdv, but physically what this quantity represents? It seems so an arbitrary quantity to me...

 

[A.T.]

I want to know why it was defined in this way.

To be useful in making quantitative predictions. Like everything in physics.

 

[Aleoa]

When we write \int m\frac{d\bar{v}}{dt}d\bar{l} , what is the physical meaning of weighting the m\frac{d\bar{v}}{dt} with the dl ?

 

[Ibix]

It isn't a "weighting", it's a dot product.

The work done by a force is the component of the force in the direction of motion times the distance travelled. For an elementary displacement $d\vec{l}$, that quantity is $\vec F.d\vec l$. Then you integrate along the path to get the total work done.

 

[Aleoa]

It isn't a "weighting", it's a dot product.

The work done by a force is the component of the force in the direction of motion times the distance travelled. For an elementary displacement $d\vec{l}$, that quantity is $\vec F.d\vec l$. Then you integrate along the path to get the total work done.

This means that Fdl is a very important quantity, i suspect it's a conservative quantity. How can i analyze Fdl matematically (or physically) just to understand what means that it's a conservative quantity ?

The fact that the integral of Fdl has the name "work" say nothing to me, they could also have called it "X". What i want to understand is what is special about Fdl

 

[Orodruin]

What i want to understand is what is special about Fdl

It describes things we can observe to a good approximation and let's us make accurate. That is the basic requirement of any theory. Whether it has a ”deep” meaning or not is irrelevant.

 

[gmax137]

What i want to understand is what is special about Fdl

Which is a tougher assignment:
Carry the box of copier paper from the ground floor up to the 2nd floor, or
Carry the box of copier paper from the ground floor up to the 3rd floor?

That is where my intuition on integrating F dot dS begins.

Leaving off the numerous caveats on human physiology vs simple physics.

 

[Ibix]

This means that Fdl is a very important quantity, i suspect it's a conservative quantity.

No. $\vec F$ may or may not be a conservative force. And energy is conserved, but that is not shown by this statement alone. So $\vec F.d\vec l$ is neither conserved (but the work must come from somewhere and the total energy is conserved) nor conservative (though $\vec F$ may be).

What i want to understand is what is special about Fdl

As Orodruin says, it's useful. That kind of post hoc justification is all there is in science, really.

 

[jtbell]

The simplistic verbal definition of "work" for a constant force is "force times distance." This definition goes back to about 1600 and apparently arose in the analysis of simple machines such as levers. I descibed this a bit and gave a reference in the following post several years ago:

https://www.physicsforums.com/threads/what-is-up-with-work.584812/#post-3804085
It's post #9 in that thread, in case the link doesn't take you directly to it.