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Work-energy theorem



A theme within the recent energy thread related to the applicability or
otherwise of the work-energy theorem to the example of somebody pushing
away from a wall.

The work-energy theorem, as usually stated, applies only to the motion of a
particle. It says that the works of all the forces on the particle are
equal to the change in the particle's KE. Work is defined as the line
integral, with respect to displacement, of a force. Since the path is the
same for all the forces acting on a particle, it is fairly easy to derive
the result, by collecting all the integrands (one for each force) and
replacing the sum of the forces by ma. The line integral of the
acceleration gives the change in one-half the velocity squared.

The theorem breaks down for extended bodies because the paths for the
points of application of the forces are no longer the same - it may not
even be possible to recognise the paths.

If you want to rescue the theorem by considering your extended body as a
collection of particles, you have to consider all the works for all the
forces on all the particles. That includes all the internal force-pairs
representing all the interactions among all the particles. Although
Newton's third law may tell you that all the internal forces cancel, the
internal works need not cancel. Indeed if there is any change in the
separation of two interacting particles, there will be non-zero work for
that force pair. The result is then that the sum of all the works on all
the particles is equal to the sum of the changes in KE for all the
individual particles. It is entirely possible therefore that the total
change in KE could be attributable to internal work only. The extended
theorem remains "true" but it is useless for practical calculations.

A trivial result that looks like work-energy can be obtained by following
the centre of mass (CM) of a system of particles (or a body). Take the line
integral of the acceleration of the CM and you will get the change in half
the velocity squared for the CM. Multiply both sides by the mass and you
get something that looks like KE on one side of the equation but the other
side is essentially meaningless. (I think that some people have called it
pseudo-work.)

Returning to the person pushing off the wall, if you could measure the
magnitude of the contact force pair between hand and wall and integrate
that over time you ought to find that it will be equal to the change in the
body's (horizontal) momentum. The experiment could be done with a Vernier
force probe and a Microcomputer Laboratory. As Brian McInnes pointed out,
analysis in terms of forces is more penetrataing than an energy argument.
Ian Sefton
I.Sefton@physics.usyd.edu.au