\int_{t_1}^{t_2} \rightarrow [F.d]_{t<t_0}^{t>t_1}

where the F.d is the dot product.

........................wtf am i doing wrong?
since

F=\frac{\partial p}{\partial t}=\partial (mv)

= M(\frac{\partial v}{\partial t}+\frac{F.v}{c^2_o}(v)

which gives the expressiom of A rest energy.

if a=\frac{v}{t} allows v=c, if acceleration permits t \rightarrow \infty , with asymptotically evaluated under a approaching c=\sqrt{E}{M}.

I will have more to say, that is, if you will let me.

The third equation isn't working for me. Can someone help me there?

The first equation doesn't make sense. Are you integrating F.d?

What do you want the third equation to say?

Thanks ben for being intuitive towards this. I will try and explain... latex is hardly my solution of my more compatible ability. I will try to rephrase this all.

\int (F.d)^{\frac{1}{2}

Right, this is how the first part is supposed to work out.

Then i have, (F.d)^{\frac{3}{2} over \frac{3}{2}(2)

Once this fraction has been done, c is added to the fraction. How does one use this in latex?

Equivalently, then follows,

i have \frac{1}{3}(F.d^{\frac{3}{2})+c.

So how do i put this all together? Latex totally fucks me. lol

Your integral does not make sense, Reiku. You need a dummy variable. What are you performing the integration with respect to? Time? Distance? Area? Something else?

Your first integral looks like the formula for work along some curved path C,

W=\oint_C {\mathbf F}\cdot d{\mathbf l}

It's a bit hard too tell, though, since you haven't expressed things correctly. Please explain what you are trying to do.

The first equation doesn't make sense. Are you integrating F.d?

What do you want the third equation to say?

Well, as you know, power equals energy over time. An integral of this expression (which is obviously expressed as a change in the energy over time), equals the dot-product of Force and Distance, which is algebraicly expressed as an equivalance:

[F.d]^{t_1}_{t_0}

Is this not right?

Your integral does not make sense, Reiku. You need a dummy variable. What are you performing the integration with respect to? Time? Distance? Area? Something else?

Your first integral looks like the formula for work along some curved path C,

W=\oint_C {\mathbf F}\cdot d{\mathbf l}

It's a bit hard too tell, though, since you haven't expressed things correctly. Please explain what you are trying to do.

Well, if i have this right, and i am sure most of you have caught on that these are highly related to momentum, i would have thought it was more accurate to express:

\Delta K_e=W=\int^{t_1}_{t_0} F.dr ?

What ineterests me most however is that F.v using the dot product, we find:

\frac{dW}{dt}=F.v

which leads to

F.v=0

How do I see \frac{dW}{dt} = 0? This says that power (the rate of change of work) = 0, identically. Does this make sense to you?

Well, if i have this right, and i am sure most of you have caught on that these are highly related to momentum, i would have thought it was more accurate to express:

\Delta K_e=W=\int^{t_1}_{t_0} F.dr ?
No, for a couple of reasons.

\Delta K_e=W is wrong. Work is change in energy, not just change in kinetic energy.

W=\int^{t_1}_{t_0} F.dr is also wrong, assuming those integration limits are times t0 and t1. You are integrating with respect to distance. The integration limits must have units of length, not time.

Work is change in energy, not just change in kinetic energy.

What about the work-kinetic energy theorem?

Raise a 1 kg block 1 meter vertically in a uniform 10 m/s2 gravity field, with the block starting and ending at rest. There is no change in kinetic energy (the block starts and ends at rest), but there is a 10 joule change in potential energy. Work was required to accomplish that 10 joules change in potential energy.

D H:

Your example considers a process with at least two forces - gravity and the force causing the initial acceleration of the 1 kg block upwards. Each of these forces does work individually during the motion. The net work done over the entire motion by the combination of forces is zero, and so is the change in kinetic energy.

The work-kinetic energy theorem tells us what the work done by a force does - it changes the kinetic energy of an object.

Right?

The work-kinetic energy theorem is only valid in a regime where total mechanical energy is conserved (i.e., conservative forces). It ignores non-mechanical energy such as chemical potential energy and non-conservative forces such as friction.

If you want to include all forms of energy: Mass, thermodynamic energy, chemical potential energy, ..., then yes the work-kinetic energy theorem is tautologically true in a closed system -- and also not particularly useful.

How is kinetic energy defined/derived, D H?

That is not the right question, James. The question is how is force defined. With conservative forces, force is the gradient of some potential. That total energy is conserved in a conservative force situation is a mathematical tautology.

Thermal energy is not conservative (what is the potential function?), yet we often talk about work done against friction. Pick up any engineering text. We often talk about work done against gravity. Pick up any introductory physics text. Work as a useful concept requires a useful system boundary.

D H:

Ok. I'll stop now. I actually think you're right that work can change types of energy other than kinetic. Just apply the first law of thermodynamics to a system, with no heat transfer to or from the system. For example, in your example of lifting a block, the first law becomes:

\Delta E_{int} = \Delta U_g + \Delta K + \Delta X = W + Q

where X is other kinds of energy the system has (chemical, nuclear, other types of potential energy, whatever), K is kinetic energy, and \Delta K = \Delta X = Q = 0 in this example.

The work-kinetic energy theorem, as I understand it, defines what we mean by "kinetic energy". The conservation of the mechanical energy of an isolated system acted on only by conservative forces then follows.

The concept of "work done against friction" is problematic in a number of ways. For example, for a simple situation of a block sliding along a rough plane, if the block is considered as the system of interest, then the kinetic energy transferred as a result of friction does not appear entirely in the block, but is also transferred partly to the environment.

The work-kinetic energy theorem, as I understand it, defines what we mean by "kinetic energy".
I disagree. Kinetic energy is a well-defined concept and the definition has nothing to do with the work-kinetic energy theorem. Kinetic energy is a function of velocity. I see it the other way around: the work-kinetic energy theorem defines work as total energy less kinetic energy. For this for the alternative definition of work, W=\oint {\mathbf F}\cdot d{\mathbf l}, to both be true one must have constant total energy and {\mathbf F} = -{\mathbf \nabla}W. In other words, a conservative force in a closed system.

Welll defined in what sense... you do know that when v=c, even momentum is an approximation?

How do I see \frac{dW}{dt} = 0? This says that power (the rate of change of work) = 0, identically. Does this make sense to you?


Whether it makes sense to me is irrelevelent, however, is is a consequence of math, which i am happy to explain.

Welll defined in what sense... you do know that when v=c, even momentum is an approximation?'Well defined' is typical math-speak for it being clearly defined, without ambiguity or dodgy method (and I'm aware my definition of 'well defined' isn't exactly well defined). And momentum most certainly is defined for massless objects.
Whether it makes sense to me is irrelevelent, however, is is a consequence of math, which i am happy to explain. I don't think Ben's issue is with algebra but the entire premise itself.

For instance, a swinging pendulum has a varying amount of potential energy and so its work done against gravity is constantly varying. It's an explicit counter example to dW/dt = 0.

'Well defined' is typical math-speak for it being clearly defined, without ambiguity or dodgy method (and I'm aware my definition of 'well defined' isn't exactly well defined). And momentum most certainly is defined for massless objects.
I don't think Ben's issue is with algebra but the entire premise itself.

For instance, a swinging pendulum has a varying amount of potential energy and so its work done against gravity is constantly varying. It's an explicit counter example to dW/dt = 0.

Idiot, and i mean that res[ectively.

Ben in his original question posed an argumenent of relevence. Now, does this forum hold relevence, albietly-said, whether it is a sequitor or not?