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Why is work defined as force dot displacement? I know that it is defined like that based on the observational fact - we do more work when we apply greater force or move to a greater distance. But I wanted to know if there was a more physical reason to back this fundamental relation.

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up vote 2 down vote accepted

Actually work as you mentioned has meaning only in classical mechanics, in this physics there are classical (vector) forces and meaning for displacement, but in modern Quantum Physics there are no more such things, or at least their meaning is totally different, there are no more force as this simple vectors or meaning to say that our particle has been displaced 5 meters.

So what I want to say, is that actually energy is much more fundamental that "work" , while other answers explained the relation between work and energy in classical sense, they didn't mention why energy is more fundamental, and that because energy is a concept that deeply related with the structure of our space & time, more precisely because our space & time are homogeneous and isotropic (this means that rotating or displacing a box and applying on it the same experiment will not change the results, rather you will do experiment now or after 100 years), this nature according to "First Noeather theorem" makes energy & momentum conserved.

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$dW = F \cdot ds$

Where $dW$ is an infinitesimal amount of Work, and $ds$ is an infinitesimal displacement. With this,

$dW = m\left(\frac{dv}{dt}\right) \cdot ds $

since $F = ma = m\left(\frac{dv}{dt}\right)$. Therefore,

$dW = m\left(\frac{ds}{dt}\right)dv = m \cdot vdv$

Now from the chain rule it follows that

$vdv = \frac12 d(v^2)$

where $d(v^2)$ means a small change in $v^2$. Therefore

$dW = m\left(\frac{ds}{dt}\right)dv = m \cdot \frac12 d(v^2)$

Integrating over $W$ for constant acceleration yields

$W = \frac12 m(v^2 - u^2)$

where $v$ is the final velocity and $u$ is the initial velocity. Therefore,

$W = \frac12 mv^2 - \frac12 mu^2 = $ final KE - initial KE.

So work being defined as Force $\cdot$ Displacement makes work equivalent to the change in kinetic energy.

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+1 Nice, but not yet a complete answer. For example, pushing a large mass over a rough surface will require a lot of work, but will not necessarily change the kinetic energy of the mass. – Rody Oldenhuis Sep 20 '12 at 10:53
But kinetic energy itself is defined on the concept of work. It is like proving something with a thing, which itself has been proved by the former thing. – pagla Sep 20 '12 at 11:02
Actually KE came first. The origin of KE = 0.5mv^2 came from experiments where different weights were dropped into clay. It was found that penetration depth was proportional the square of the impact velocity, and mass. – Mew Sep 20 '12 at 11:17
What about the 1/2? – Andres Salas Jul 29 '14 at 15:15
@AndresSalas 1/2 came from the fact that in the 18's when the concept of work was introduced , the only mathemetically described (in relative sense) was gravity and from the tempetation of d = at^2/2 , mv^2 was made mv^2/2 – Faiz Iqbal Feb 23 at 7:15

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