Re: work done by pushing

[Bob Sciamanda <trebor@VELOCITY.NET>]

This is precisely my point.  Simply integrate F=mA (where F=the net
external force on a system of total mass m and A is its CM acceleration)
over the center of mass trajectory. The CM work-energy theorem results.
It is simply an implication of F=mA and says that the line integral of F
over the CM path is numerically equal to the system's change in KE.  It
says NOTHING about the source or sink or flow of that energy;
specifically, it does not imply that there is some flow of energy from the
agent of F to the system.  Call that line integral what you will - the
statement is as true as F=mA,  no more, no less.

Bob

Bob Sciamanda (W3NLV)
Physics, Edinboro Univ of PA (em)
trebor@velocity.net
http://www.velocity.net/~trebor

----- Original Message -----
From: John Denker <jsd@MONMOUTH.COM>
To: <PHYS-L@lists.nau.edu>
Sent: Tuesday, October 26, 1999 12:22 PM
Subject: Re: work done by pushing


> At 01:16 AM 10/26/99 -0400, Bob Sciamanda wrote:
> > When I push myself away from a wall, the force from the wall does
> > work
>
> That's not right.
>
> > (defined as its line integral over the displacement of my CM),
>
> That's not the right definition of work; see below.
>
> > but the wall is not an energy source.
>
> That's true.
>
> ============================
>
> When we say "A does work on B" it means that energy is flowing across
the
> boundary between A and B.  This is not happening when you push away from
a
> wall.  At the boundary, there is no significant "F dot dx".  The main
place
> where there is some "F dot dx" is at your shoulder, and we all agree
that
> your arm is doing work on your torso.
>
> ______________________________________________________________
> copyright (C) 1999 John S. Denker             jsd@monmouth.com