Force and work

  • #1
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I know that the integral of force (in respects to distance) is work.

What about the derivativeof force (in respects to time)? is that equal to work in respects to time?
 
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  • #2
quasar987
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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]
 
  • #3
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I thought that momentum = m*v. You have m*W
 
  • #4
dextercioby
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quasar987 said:
[tex] \frac{dW}{dt}=\frac{dT}{dt} [/tex]
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.
[tex] \frac{dT}{dt}=\frac{m}{2}\frac{d}{dt}\vec{v}^{2}=m\frac{d\vec{v}}{dt}\cdot \vec{v}=\vec{F}\cdot \vec{v}=P [/tex](2)


quasar987 said:
[tex] \frac{dW}{dt}=\frac{1}{m}\frac{dp}{dt} [/tex]
Compute the RHS:it gives the acceleration,right (for constant mass)????Is the acceleration equal to power (v.(1))???Never.
It's wrong.And the implications that follow are wrong as well.

Daniel.
 
  • #5
quasar987
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I tought that looks weird too.. :P
 
  • #6
quasar987
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I don't understand how you get your equation (1) dexter. It's as if you're saying that

[tex]W=\vec{F}\cdot \vec{r}[/tex] ?!

The way I would do it is this:

[tex]W=\int_{t_1}^{t_2}(\vec{F}\cdot \vec{v})dt[/tex]

[tex]\Rightarrow \int_{t_1}^{t_2}dW = \int_{t_1}^{t_2}(\vec{F}\cdot \vec{v})dt[/tex]

[tex]\Rightarrow dW = (\vec{F}\cdot \vec{v})dt[/tex]

[tex]\Rightarrow \frac{dW}{dt} = \vec{F}\cdot \vec{v}[/tex]

And so it would appear that dW/dt = P = dT/dt no matter if F is time dependant or not.

Do you see what is wrong in those steps?
 
  • #7
dextercioby
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Yes,at the second equation.You didn't take into account that the force could depend upon time.
[tex] W=\int_{t_{1}}^{t_{2}} \frac{d}{dt}(\vec{F}\cdot\vec{r}) dt [/tex]

Do you see the difference?I think it's obvious that your calculations will hold,iff the force is time independent...

Daniel.
 
  • #8
quasar987
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I do understand that for

[tex]\frac{d}{dt}(\vec{F}\cdot\vec{r})dt [/tex]

to be equal to

[tex](\vec{F}\cdot\vec{v})dt[/tex]

we need that F be time-independant. Otherwise, it would go like this...

[tex]\frac{d}{dt}(\vec{F}\cdot\vec{r})dt = (\frac{d\vec{F}}{dt}\cdot\vec{r})dt+(\vec{F}\cdot\vec{v}})dt[/tex]

... But at what point do we need to take this derivative in the definition of work?

Symon gives the integral

[tex]\int_{t_1}^{t_2}(\vec{F}\cdot \vec{v})dt[/tex]

as the most general definition of work, and it is obtained by noting that

[tex]\frac{dT}{dt} = \vec{F}\cdot\vec{v}[/tex] (for any force F)

, by multiplying each side by dt, and by integrating.
 
  • #9
dextercioby
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It doesn't have to do with the definition of work,but with equating its derivative with the time derivative of the KE
[tex] \frac{dT}{dt}=\vec{F}\cdot\vec{v}\neq \frac{dW}{dt} [/tex]
,for time dependent forces.Okay?

Daniel.
 
  • #10
quasar987
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Things are starting to fall into place now I think...

Just to make sure: does this also means that for a force depending on the position of the (moving) particle,

[tex]\vec{F}\cdot\vec{v}\neq \frac{dW}{dt}[/tex]

because the time derivative of force wont be 0 ??


But that seems absurd because for F function of position and conservative, we know that

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

a contradiction. :yuck: Arrrgh, curse you work!
 
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  • #11
dextercioby
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quasar987 said:
[tex]\frac{dW}{dt}=-\frac{dV}{dt}=\frac{dT}{dt}[/tex]

a contradiction. :yuck: Arrrgh, curse you work!
:rofl: What you stated is correct,unless the force depends on time.If it doesn't,then the energy of the system is a prime integral,therefore is constant on the solutions of the Newton equations.
Call the total energy E.
Then:
[tex] E=T+V [/tex](1)
Differentiate this relation wrt to time:
[tex] 0=\frac{dT}{dt}+\frac{dV}{dt}\Rightarrow -\frac{dV}{dt}=\frac{dT}{dt} [/tex] (2).
But for a conservative force:
[tex] W=-V [/tex](3)
Then from (2) and (3),it follows:
[tex] \frac{dW}{dt}=\frac{dT}{dt} [/tex](4)
But if the force depends on time,then the total energy of the system is not constant in time,therefore relations (1) pp. (4) don't hold.

Daniel.
 
  • #12
quasar987
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Do you have the book "Mechanics" by Symon? At page 52, talking about the forced hamonic oscillator under a time-dependant force F(t), he writes, "In the steady state, the rate at which work is done on the oscillator by the applied force is

[tex]F(t)\frac{dx}{dt}=...[/tex]"

Doesn't this contradict the statement

It doesn't have to do with the definition of work,but with equating its derivative with the time derivative of the KE
[tex] \frac{dT}{dt}=\vec{F}\cdot\vec{v}\neq \frac{dW}{dt} [/tex]
,for time dependent forces.Okay?
?!?!?!
:confused:
 
  • #13
quasar987
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Also, when we say that

[tex]\frac{dW}{dt}=\vec{F}\cdot\vec{v}[/tex]

for a time-independant force, do we mean it in the following sense...

[tex]W_{1 \rightarrow 2} = \int_{t_1}^{t_2}(\vec{F}\cdot\vec{v})dt[/tex]

Where [itex]W_{1 \rightarrow 2}[/itex] stands for the work done by the force F in going from point 1 to point 2, along a certain definite path; a process which begins at time [itex]t_1[/itex] and ends at time [itex]t_2[/itex].


[tex]\Rightarrow \frac{dW_{1 \rightarrow 2}}{dt}=\frac{d}{dt}\int_{t_1}^{t_2}(\vec{F}\cdot{\vec{v})dt =\int_{t_1}^{t_2}\frac{d}{dt}(\vec{F}\cdot{\vec{v}dt)=\int_{t_1}^{t_2}(\vec{F}\cdot d\vec{v})=\vec{F}\cdot \int_{t_1}^{t_2}d\vec{v}=\left[\vec{F}\cdot \vec{v} \right]_{t_1}^{t_2}[/tex]

???
 
  • #14
dextercioby
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Nope,in the last line u have the derivative of a number.Which is zero.Capisci?The way u defined it (which is correct,BTW),
[tex] W_{1\rightarrow 2} [/tex]
is a number and taking time derivative from it should yield zero.

So the whole line is wrong...

Daniel.
 
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  • #15
quasar987
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Well if I just substitute [itex]t_2[/itex] for the more general variable t, is it o.k. then?

(And position 2 for the more general position x)
 
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  • #16
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Yes, it is okay.
 
  • #17
quasar987
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You know vector calculus Stargate but can't say what happens to the acceleration if the force is doubled?! :bugeye:

Dexter, can you comment of post #12? Don't you find that troubling? :grumpy:
 
  • #18
dextercioby
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quasar987 said:
You know vector calculus Stargate but can't say what happens to the acceleration if the force is doubled?! :bugeye:

Dexter, can you comment of post #12? Don't you find that troubling? :grumpy:

You're being bad to the boy. :mad:

The short unofficial version is:i'm right,and he's wrong.I was wrong with the notation however,i got carried away. :frown: There's no such thing as time derivative of the work,or,if there is,is identically zero.
Work is a functional.It is defined as application from the space of solutions of the motion equations into R.
There's no such thing as 'W'.It should always be:[itex] W_{1\rightarrow 2} [/tex].In thermodynamics we refer to work as a 'process quantity'.It doesn't have derivatives in the ordinary sense.Maybe Ga^teaux or Fréchet derivatives,but not ordinary ones.The differential work is not a differential.It is a one-form.In thermodynamics,where it usually appears under 'differential' notation,it appears with a 'd-bar' (similar to 'hbar'),just as 'differential' heat does,explicitely to show it is not a differential.

For Newtonian forces (that depend only on coordinates,time and momentum,and not on acceleration and time derivatives of acceleration),it is defined,for a particle which is acted on by a force [itex] \vec{F}(\vec{r}(t),\vec{v}(t),t) [/itex],as a line-integral (curvilinear integral of the second kind),viz.
[tex] W_{1\rightarrow 2} =:\int_{1}^{2} \vec{F}(\vec{r}(t),\vec{v}(t),t)\cdot\vec{n} dl [/tex]
,where (1),(2) are points in the ordinary [itex] E_{3}[/itex] between which the body's movement takes place:
[tex] (1):\{x(t_{1}),y(t_{1}),z(t_{1})\} [/tex]
[tex] (2):\{x(t_{2}),y(t_{2}),z(t_{2})\} [/tex]
,"t" is a parameter along the path/classical trajectory the body follows under the influence of the force "F";customarily,in phyisics,it's the old-rusty 'time'.
[itex]\vec{r}(t);\vec{v}(t) [/itex] are the position vector and the velocity vector of the body of mass "m".[itex]\vec{n}[/itex] is the unit-vector tangent to the path/classical trajectory the body follows in its movement and 'dl' is the line element of the same path/classical trajectory.

That's that.'Work' is a functional,it is a number,it's time derivative is identically zero.
I'm sorry again,i didn't realize what i was doing... :frown:


Daniel.
 
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  • #19
quasar987
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But in every classical mechanics textbook, it is written

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

What about that ?! What does it mean then?
 
  • #20
dextercioby
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quasar987 said:
But in every classical mechanics textbook, it is written

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

What about that ?! What does it mean then?
Three possible things.
1.These guys are idiots (are aware of what i've written,yet they write something else).
2.They're not aware of what i've written.They haven't read books like Arnold or Marsden.
3.They're idiots.Even for constant forces (that would 'come out' from under the line-integral),the result is still zero,because it would be the derivative of a number (the magnitude of force times the length of the trajectory).

Daniel.
 
  • #21
quasar987
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How about this? It is wrong formally to write dW/dt = dT/dt because W is just a number and is only defined over a path, etc. (all you said).

But we define the rate at which work is being "compiled" (as the particle progresses along the path) as

[tex]\frac{dW}{dt} \equiv \frac{dT}{dt} = \vec{F}\cdot \vec{v}[/tex]

and it is a definition that makes sense since multipying both sides by dt and integrating from t1 to t2, we get the "Work-scalar" (real work) [itex]W_{1 \rightarrow 2}[/itex].

By the way, I am disapointed that nobody else gives his opinion on this topic. (Thanks for the effort Stargate :tongue2:)
 
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  • #22
Doc Al
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dextercioby said:
3.They're idiots.Even for constant forces (that would 'come out' from under the line-integral),the result is still zero,because it would be the derivative of a number (the magnitude of force times the length of the trajectory).
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!

I don't know what you mean by saying "it would be the derivative of a number". It's a number all right (a scalar) but that number is not a constant, it's a function. (Sure it depends on the path and how the force depends on time, but so what?)
 
  • #23
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Doc Al 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!

I don't know what you mean by saying "it would be the derivative of a number". It's a number all right (a scalar) but that number is not a constant, it's a function. (Sure it depends on the path and how the force depends on time, but so what?)
indeed, i think dexter is once again submerging us with his personalized vision on physics. The fact that F is time dependent has nothing to do with this all...

marlon
 
  • #24
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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]
According to me this is correct...

marlon
 
  • #25
dextercioby
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marlon said:
quasar987 said:
[tex]\frac{dW}{dt}=\frac{1}{m}\frac{dp}{dt}\Rightarrow ... [/tex]

According to me this is correct

marlon

Compute the RHS:it gives the acceleration,right (for constant mass)????Is the acceleration equal to power (v.(1))???Never.
It's wrong.And the implications that follow are wrong as well.

Marlon is wrong...He didn't even read the formulas.If he did,then it's :surprised :surprised :surprised

Daniel.
 

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