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I am trying to make sense of the meaning of the definition of work. The original definition of work was also known as "the weight lifted through a height." I was hoping that our mathematical definition of work would correspond to this notion in a well-defined way but it seems like it doesn't.

According to our mathematical definition of work, if I "lift the weight through a height" at two different constant accelerations, say, I get two different values for the amount of work I have done.

Why do we make the definition like this? Why don't we make the definition such that the work done by "lifting a weight through a height" doesn't depend on the velocity/acceleration with which it is lifted. It seems the definition was originally intended for this and it got changed along the way somewhere.

Consider a path $x(t): x(0) = 0, x'(0) = 0$ taking values on the real line and $x''(t) = a = F$. Then if $T_a = \frac{\sqrt{2}}{\sqrt{a}}$ then $x(T_a) = 1$. Also $W = \int_0^{T_a} F \cdot x'(t) dt = \int_0^{T_a} a \cdot at dt = a$.

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  • $\begingroup$ What you're saying isn't true. Lifting a fixed mass over a fixed distance always needs the same work, regardless of how fast you do it. But for differing accelerations, you have to consider that the final velocity, and hence kinetic energy, of the mass may be different. $\endgroup$
    – ACuriousMind
    Dec 1, 2015 at 0:34
  • $\begingroup$ en.wikipedia.org/wiki/Work_(physics) $\endgroup$
    – user83548
    Dec 1, 2015 at 0:38
  • $\begingroup$ What mathematical definition of work were you given that doesn't give the same result as "weight lifted through height"? $\endgroup$
    – The Photon
    Dec 1, 2015 at 0:47
  • $\begingroup$ Consider the example I just gave in the question. $\endgroup$
    – Joe
    Dec 1, 2015 at 1:04
  • $\begingroup$ @The Photon I am using the line integral definition. See the example above. Does it not make sense? $\endgroup$
    – Joe
    Dec 1, 2015 at 1:10

1 Answer 1

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The boundary conditions and units (setting $m=1$ and choosing the length $l$ of the path also $1$) obscure the connection to the "lift" definition, but in fact they coincide for the given situation.

The formula you derived for the work done against a constant force is $ W = a $, which should really read

$$ W = m\cdot a\cdot l$$

If the force on the body is due to gravity and height above ground doesn't change too much, gravitational acceleration is basically constant $a=g$ giving back the familiar formula $mgh$ with $h$ identified with $l$.

Note that "lift the weight through a height" is only a sensible definition for work done against gravity if the force can be assumed to be constant. It is a limiting case of the general expression of work as the line integral of a force.

Also be aware that work cannot only be done against gravity, but also e.g an electric field. Defining work through lifting is somewhat nonsensical, less one defines "lifting" as "moving against a force", which brings us back to the integral formulation.

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