Re: Work/Energy Theorem

[SCIAMANDA@edinboro.edu]

> . . . Imagine standing still on a frictionless surface and then
> pushing off a wall to get yourself moving.  Since the points of
> application of either the force on you or the force on the wall do not
> move, no work is done either on or by you.  However, you do gain kinetic
> energy as you slide away from the wall. 
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Eugene (Gene) P. Mosca           Phone 410-293-6659 (Fax 3729)
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I think here is where the confusion and the "need" for the idea of
"pseudo" work come from.  The work-energy theorem for a particle comes
from taking the dot product of both sides of the vector equation F=ma
with the vector displacement (dr) of m.  The result is then integrated
over the trajectory of m.  The "point of application" of F never comes
into consideration (in the general case it is not always easy to even define
this notion).  The same can be said when the theorem is derived for a system 
of particles; here dr is the vector displacement of the system CM and again 
the point(s) of application of the force(s) never come up.

Our problem is that we want to read more into the W.E. theorem than is there. 
When you push yourself away from a wall, the W.E. theorem simply states 
that there is a numerical equality between the wall force integrated over YOUR 
C.M. displacement and the increase in your kinetic energy (in the simple, 
frictionless case).  We then call that force integral the work done by the 
wall and (perhaps subconsciously) expect that this should mean a transfer 
of some kind of energy from the wall to your body.  The W.E. theorem says 
nothing of the kind.  Energy conservation (which in the general case cannot 
come from only mechanics) is satisfied by your body's metabolism;  the wall 
is used as an enabling mechanism, as Dario explained so well.  
( If the wall had a much smaller mass so that it would move, 
would we still invoke "pseudo work"?)

In short, I think we want to use the adjective pseudo whenever the agent 
doing the work is not the physical source of the resulting kinetic energy 
change . . . nothing in the work energy theorem says that it need be!  
Like Bernoulli's theorem, it states a numerical equality which we are tempted 
to interpret as always implying a causal relation of some kind. 

Bob Sciamanda           sciamanda@edinboro.edu
Dept of Physics
Edinboro Univ of PA     http://www.edinboro.edu/~sciamanda/home.html