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Surely there must be a reason we decided to use this as a metric for mechanical energy.How was it developed and what made it more acceptable than other work formula candidates (Like force over time, for example).

By work formula, I'm referring to $W=F\;d\cos\theta$

Thank you.

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    $\begingroup$ Are you aware of the work-energy theorem? $\endgroup$ – Omar Nagib Aug 28 '15 at 5:53
  • $\begingroup$ No! What would that be? $\endgroup$ – Striker Aug 28 '15 at 6:01
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    $\begingroup$ You could try Googling it to find out ... $\endgroup$ – John Rennie Aug 28 '15 at 6:29
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    $\begingroup$ I did - and what I found strange was, if I'm interpreting it correctly, its use in strengthening the argument for the work formula, when the kinetic energy formula is in itself often derived from the work formula. $\endgroup$ – Striker Aug 28 '15 at 6:33
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The notion of work in physics was first formulated by the French mathematician Gustave Coriolis in Calculation of the Effect of Machines, or Considerations on the Use of Engines and their Evaluation published in 1829. Coriolis defined work as "weight lifted through a height". He was concerned with developing a term that could measure the units of work accomplished by men, horses, or steam-powered machines without discriminating among them. Prior to his definition of work, the term "force-displacement" had been in use, but it was ambiguous because it included mechanical power, quantity of action, and dynamic effect, depending on the context, without unifying them.

Coriolis proposed the term "dynamode" (1,000 kilograms through 1 meter) as a unit of work, but although it didn't catch on, the technical term "work" was accepted as a precise way of quantifying the combination of force and displacement.

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Our equation for work follows from the conservation of energy. If we consider some object then we expect that if we do work $W$ on it then its kinetic energy must increase by $W$.

So the requirement for the equation for work is that it must be equal to the change in kinetic energy. Proving this is usually done using integral calculus, but since you give the equation for a constant force let me just consider this since it's a lot simpler. I'll also assume the force is directed along the direction of motion so $\theta = 0$ and $\cos\theta = 1$. The expression for work then simplifies to $W = Fd$.

We know that $F = ma$ (Newton's second law) so we get:

$$ W = Fd = mad \tag{1} $$

And we all learned at school the simple SUVAT equation:

$$ v^2 = u^2 + 2ad $$

Let the object start at rest, so $u = 0$, and use equation (1) to substitute for $ad$, and we get:

$$ v^2 = 2\frac{W}{m} $$

which rearranges to:

$$ W = \tfrac{1}{2}mv^2 $$

And the right hand side is just the kinetic energy, so this tells us that the work $W$ is equal to the change in kinetic energy, just as we expected.

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There are some physical quantities that are usefull (and this is under statement), like energy. It is conserved, it is a function of some other very important quantities that can help you describe the motion of the body etc. If you can justify energy, there should be no problem in justifying work, which is energy transfered to a body by some force. Quantity known as energy has its own right to exist, it is preserved, it can change its form, it can be transfered from one system to other. Neutrino was proposed on the basis of a reaction that seemed not to obbey conservation of energy. So, if this quantity has such great properties what more motivation you need? Quantities that are conserved in closed systems are very usefull. It can be proen that amount of work done is equal to a change of the energy of the system. So work is used to describe these interactions and energy exchange. Other formulae, like the one you proposed, force over time, is just rate of change of force over time, it can not be connected directly to the energy of a system. Force times time is called impulse or momentum (linear). So force acting over some time t on a body gives that body some ammount of momentum and energy. If you double the time you will double the momentum but wont double the energy. You will increase it but wont double it. You will actually increase it 4 times. But, energy given to a body is directly proportional to a force acting on it and a distance over which it is acting. So: work done=energy change and energy change=forcexdistance. Why you need cosine? Because of course, only the force component that acts along the direction of displacement can change the energy of the body. Cosine projects the force along that direction. If force vector is along the displacement vector, angle is zero and cosine is 1, so you just get F*d=W.

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  • $\begingroup$ This does not provide an answer to the question. To critique or request clarification from an author, leave a comment below their post. $\endgroup$ – Jon Custer Aug 28 '15 at 14:17
  • $\begingroup$ :-( :-/ :-P heh... $\endgroup$ – Žarko Tomičić Aug 28 '15 at 15:55

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