On 05 Oct 2003 14:36:23 GMT, a (Dave Baker) wrote:
From: Gary Coffman
On 05 Oct 2003 02:31:09 GMT, a (Dave Baker) wrote:
From: Gary Coffman
Hmmmm. Force is a vector. In the definition of mechanical work given in
Machinery's Handbook, the distance used to calculate work is the total
displacement *in the direction the force is applied*. I think that makes it
a vector too. If I recall my elementary physics correctly, it doesn't matter
what path is taken to the final position when calculating energy states
(and work is alternatively defined as the difference between the beginning
and ending energy states of a closed system). That's consonant with the
summation of an arbitrary number of vectors, so the distance should
indeed be considered a vector quantity.
snip
The laws of conservation of energy determine that work is always a scalar
quantity. If we simultaneously apply two forces of magnitude 10 units each to
an object we clearly need to know the directions to obtain the resultant net
force. This can be as high as 20 units if the forces are in the same
direction,
as low as zero if they oppose or anything in between. However, If we do 10
units of work to the object by moving it in a particular direction and then
another 10 moving it back again the net work is always 20 units and never
zero. The equation has to be balanced by the total expenditure of energy
consumed in doing the work.

That *sounds* good, but I'm not sure it holds up to analysis. For example,
if I raise a 1 kilogram mass 1 meter, I do work, requiring energy to accomplish.
But if I then let the mass return to its original position, I regain the energy.
So we have a situation where there has been a zero net change in energy.
Doesn't that also require that negative work was done so that net work is
also zero? Otherwise, we seem to have a violation of the conservation laws
because we'd have net work with no net energy expenditure.

I was anticipating that reply and if it hadn't been so late last night I was
going to mention it in advance. The reason being is you had alluded to it in
the bit about energy states at the top of your earlier post. Potential energy
states and work musn't be confused. You don't do work to an object when it
falls against gravity, the earth does. Whether you arrange to have that energy
recovered, hydroelectricity etc, doesn't affect the amount of work that got
done during the process. Work isn't something that gets conserved, only energy
is.

Well, Feynman disagrees (at least partly). He gives examples in his
"Lectures on Physics" of cases where work must be negative. One
of those cases is an orbiting body, another is a frictionless pendulum.
The equation w = F dot D turns out not to be general enough to cover
all cases. The more general equation is:

w = [definite integral from a to b] F cos(theta) dS

Where w is work, a and b are the end points of the displacement,
F is a force vector, theta is the angle between the force and the
direction of displacement, and S is the unit displacement vector.

Now for straight line displacements, and where theta = 0,
cos(theta) = 1, and we have the simple degenerate case of
w = F dot D. But consider the case where theta = 180.
Cos(theta) is then -1, and the resulting work is -w, negative
work.

Another way of looking at this is that work = delta Ke or
Ke(b) - Ke(a). In cases where Ke(a) is larger in magnitude
than Ke(b), you get a result which is negative work.

This is *why* bodies in orbit, or frictionless pendulums,
show zero net work. The positive work done during one
part of the motion is canceled by the negative work done
through another part of the motion, resulting in zero
net work being done.

Feynman shows that for any closed conservative system
which is in principle reversible, net work will always be zero,
no matter how torturous the path taken to close the loop.

Gary