Force and work

  • #26
dextercioby
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DocAl said:
I'm missing your point, Daniel. If I define work in the usual way,[itex] dW=\vec{F}\cdot d\vec{r}[/itex],why can't I consider work to be a function of time? Certainly its rate of change with time is not (necessarily) zero!
First of all,the notation you've given is wrong.
[tex] \delta W=:\vec{F}\cdot d\vec{r} [/tex]
,simply because it's not a differential,but a one-form.
You cannot define "work" by the (wrongly written) formula,you define "differential work" which is not the same thing,obviously.

Even if i repeat myself,this is the definition of work:
[tex] W_{1\rightarrow 2} [\vec{r}] =:\int_{1}^{2} \vec{F}(\vec{r}(t),\dot{\vec{r}}(t),t)\cdot \vec{n} dl [/tex]

It is a functional of [itex] \vec{r}(t) [/itex].

My question is,to you and to Marlon:
What is the mathematical significance of these "animals"?
[tex] \frac{dW_{1\rightarrow 2}[\vec{r}]}{dt} [/tex]
and the most interesting by far:
[tex] \frac{\delta W}{dt}[/tex] ??

Daniel.
 
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  • #27
quasar987
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quasar987 said:
What is work in respect to time?

Anyway, here's what dF/dt is in term of work:

[tex]\frac{dF}{dt} = \frac{d^2p}{dt^2}[/tex] (time derivative of the second law)

And since

[tex]T = \frac{p^2}{2m}[/tex]

and

[tex]\frac{dW}{dt} = \frac{dT}{dt}[/tex]

then

[tex]\frac{dW}{dt} = \frac{1}{m}\frac{dp}{dt} \Leftrightarrow m\frac{dW}{dt} = \frac{dp}{dt} \Leftrightarrow m\frac{d^2W}{dt^2} = \frac{d^2p}{dt^2} = \frac{dF}{dt}[/tex]
I guess what marlon was saying is that the logic is correct ?

Because there's an error in the differentiation (I forgot the chain rule). It should have been

[tex]\frac{dW}{dt} = \frac{p}{m}\frac{dp}{dt} = m\frac{dv}{dt}v=Fv[/tex]
 
  • #28
dextercioby
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Marlon was saying everything was correct.He never mentioned logics/computations.
Don't missinterpret the words written.

Daniel.

PS.Anyways,i'm still waiting for their answers...
 
  • #29
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quasar987 said:
I guess what marlon was saying is that the logic is correct ?

Because there's an error in the differentiation (I forgot the chain rule). It should have been

[tex]\frac{dW}{dt} = \frac{p}{m}\frac{dp}{dt} = m\frac{dv}{dt}v=Fv[/tex]
Indeed that is what i meant. Don't mind dextercioby with his useless remarks. He is just angry and not willing to see he's just regurgitating self-adapted physics...

let's not get into that...

YOU ARE RIGHT...

marlon
 
  • #30
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dextercioby said:
First of all,the notation you've given is wrong.
[tex] \delta W=:\vec{F}\cdot d\vec{r} [/tex]
,simply because it's not a differential,but a one-form.
You cannot define "work" by the (wrongly written) formula,you define "differential work" which is not the same thing,obviously.

Even if i repeat myself,this is the definition of work:
[tex] W_{1\rightarrow 2} [\vec{r}] =:\int_{1}^{2} \vec{F}(\vec{r}(t),\dot{\vec{r}}(t),t)\cdot \vec{n} dl [/tex]

It is a functional of [itex] \vec{r}(t) [/itex].

My question is,to you and to Marlon:
What is the mathematical significance of these "animals"?
[tex] \frac{dW_{1\rightarrow 2}[\vec{r}]}{dt} [/tex]
and the most interesting by far:
[tex] \frac{\delta W}{dt}[/tex] ??

Daniel.
Daniel, stop abusing this classical trick of blowing everything out of proportion (you like that don't you?)... You know very well what Doc Al and I are trying to say.

All this mumbo jumbo about functionals is just not relevant here...Drop it, since it is very sad and nobody will learn anything from this...

Besides the total differential expresses that a process is reversible so when describing such processess, it will happen in terms of total differential and Pfaffian differential equations. If energy-dissipations occur during some proces (irreversible), you cannot use total differentials and therefore the [tex]\delta[/tex] will be used.


Besides, your motivation for the formula for W that you use is not necessary here. In this context (as well as in most classical mechanics-contexts) the formula of Doc Al will do... Again you are misusing your "knowledge" of theory in order to make a redundant point. Pardon the French but it really is...

regards
marlon
 
  • #31
dextercioby
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marlon said:
Indeed that is what i meant. Don't mind dextercioby with his useless remarks. He is just angry and not willing to see he's just regurgitating self-adapted physics...

let's not get into that...

YOU ARE RIGHT...

marlon
:rofl: I'm not angry... :tongue2: For the last couple of days,i've been laughing my a$$ out... :rofl: I ain't got nobody to be angry at...If somebody proves me wrong,i'll accept it,even if it's Marlon... :tongue2: So far,i haven't had sufficient evidence to admit i was/am/will be wrong...

Self-adapted physics u say... :tongue2: I wish i could believe you... :tongue2:

Daniel.
 
  • #32
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dextercioby said:
:rofl: I'm not angry... :tongue2: For the last couple of days,i've been laughing my a$$ out... :rofl: I ain't got nobody to be angry at...If somebody proves me wrong,i'll accept it,even if it's Marlon... :tongue2: So far,i haven't had sufficient evidence to admit i was/am/will be wrong...

Self-adapted physics u say... :tongue2: I wish i could believe you... :tongue2:

Daniel.
good for you...

marlon
 
  • #33
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dextercioby said:
This is wrong,if 'W' stands for potential energy.If it stands for work,then the force should not depend on time:
[tex] \frac{dW}{dt}=\frac{d}{dt}(\vec{F}\cdot \vec{r})=\vec{F}\cdot \vec{v}=P [/tex](1)
,where P is the mechanical power.
this is just plain wrong again...

[tex]\Delta W = \int mvdv = \Delta E _{kinetic}[/tex]

from this formula you need to start...

marlon
 

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