Re: Energy, etc

[Brian McInnes <bmcinnes@PNC.COM.AU>]

Robert, comments on your reply to my posting on the non-applicability
of the "work-energy theorem" to extended, articulated bodies.


> As far as I can tell, an object won't speed
> up unless there is a net force acting in the direction of the object's
> displacement.

I agree with you.

> In other words, an object's kinetic energy can only
> increase when there is work done on it,

No, I don;t agree.

> where I define work as
>
>     dW = F dot dx = m d(v^2) / 2

Yes, as far4 as it goes
> In this expression, F is the net force on the object, the same F as in
> F=ma. "dx" is the displacement of the object that is being accelerated
> (not the force, as if anyone can define the displacement of a force).

No.  F dot dx is the line integral of the force.  One can define and
observe the displacement of a force quite easily.  For the interaction
between the skater and the wall that displacement in zero: the force
begins located at the wall-hand contact point and stays there during
the interaction.  The displacement of the centre of mass is very
different, and different from the different displacements of most of
the points on and in the skater's body.
There is a similar looking equation coming for Newton's second law
being applied to the changing motion of the skater's centre of mass.
It is:  F(net) delta s(centre of mass) = delta (1/2 mv(center of
mass)*2.  That relates the change in kinetic energy of the body to
something that, on quick inspection, could be mistaken for the work
done on the skater but is not that.  Ian Sefton, in his useful
contribution to this thread noted that this term is sometimes called
psuedo-work.  Whatever we choose to call it, it certainly is not the
work done by the net force (which happens to be equal to the force
applied by the contact force from the wall) because that work is zero.

> If the object speeds up, then there must be a net force in the direction of
> the object

Yes

> (and thus F dot dx must be positive).

No, It's zero; see above.
>

> Just as F=ma can be applied to extended objects, so can dW=dK.

Yes, but to quote Albert Arons: "This is a subtle matter...Do not be
diminished if you find initial difficulty with the idea; you will be
in good company."  As I remarked earlier in this thread, it took me
quite some time to come to terms with this idea.

> If the
> extended body's KE changed, there must've been work on on it (i.e., there
> must be a net force in the direction of the motion, as in the case of the
> skater pushing off the wall).

No, the reckoning for the extra kinetic energy is from a reduction in
chemical energy, rather than any work being done on the skater.  It is
a conservation of energy situation.
Two other nice examples of zero-work, non-zero external forces and
conservation of energy are acceleration of an automobile and a block
sliding to rest on a rough level surface.



> Again, I think this is misleading, as it ignores the force of the floor
> (friction), which is crucial for the walker to speed up.

No, I'm not at all ignoring that force.  It is responsible for the
acceleration of the centre of mass and is used in the expression I
referred to earlier to find what the change in kinetic energy is.  But
that frictional force does no work and ,
>
A decrease in
> chemical energy will occur even if the walker attempts to walk on a
> frictionless surface, yet there will be no increase in the translational
> kinetic energy.

I guess it will.  The question is where will it show up in our
consideration of the first law of thermodynamics.  Initially as an
increase in thermal energy, which will also occur if the surface is
rough and the walker can make progress on it.
> So...the question is "what did the work on the skater pushing against the
> wall?"  After all, the normal force cannot do work, can it?
>
> I answer it thusly...
>
> As the skater pushes against the wall, it makes more sense to consider the
> arm and wall as one system and the rest of the skater as another.  The
> arm/wall exerts a force on the skater's body, which experiences a
> displacement.  The arm/wall does work on the skater's body, much as a
> spring does work.  As the arm "detaches" from the wall, the body does work
> on it as the arm experiences a displacement in the direction of the body.
>
> The work/energy theorem doesn't break down as far as I can tell. [thank
> goodness!]

Robert, you are trying to preserve the work-energy theorem at all
cost.  There are still major difficulties with the difference between
displacement of the centres of mass (now twofold: the arm and of the
rest of the body) and the displacement of the point of application of
the force (which does happen to be non-zero at the attachment point of
the arm to the rest of the body).

Regards,
Brian McInnes