Re: Muscle work

[Tim Folkerts <tfolkert@FHSU.EDU>]

No! This isn't right.  You are being Aristotelian.


> Consider a heavy box on frictionless wheels on a level surface.
> Someone is pushing it to the left with a constant force of F.
> From rest, you push it a distance X to the right and bring it to
> rest again.  You must exert a force greater than F to move it.  Let's
> call the force you exert G.  The work done by you is GX.  You hold it
there.
> While it is there, you exert a force equal to F but do no work (no
> displacement).  Then you return the box to its original position.  You
> must exert a force less than F to move it.  Let's call the
> force you exert E.  The work done by you is EX.  Since E<F<G, the work
> done by you on the way back (EX) is less than the work done by you on the
> way there (GX).

*Momentarily* you must apply a force greater than F to get it moving.  Once
it is moving, a force of exactly F will keep it moving at constant speed.
At the end, a force of just under F will allow it to come to rest.  The
work you do is FX.

On the way back, the situation is reverse, at you do exactly -FX.  The
magnitude is the same, but the sign is different.

If you include friction, then the idea works.  Call f the frictional force.
On the way over, your force is F+f, since you need to overcome friction.
On the way back, your force is F-f, since your rival needs to overcome
friction.  Your total work is then +(F+f)X - (F-f)X = 2fX


Tim