On Sun, 05 Oct 2003 00:56:09 -0400, Tom Quackenbush wrote:
On Sat, 04 Oct 2003 20:22:54 -0400, Gary Coffman
wrote:
SNIP
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.

Now if distance were considered scalar, there'd be no question, a vector
times a scalar is a vector, and so work would be a vector quantity. But if
the distance is considered a vector too, then there are two possibilities.
The cross product of a vector and another vector is a vector. But the dot
product of two vectors is a scalar.

In this particular case of aligned vectors, the *magnitude* of the answer is
the same, but one resultant is vector and the other scalar. So we are left to
wonder why we would apply the dot product instead of the cross product in
calculating work. I'd be interested in hearing others' thoughts on that.
SNIP
I don't know.

The way I learned, with w=mv^2, work must be scalar because we're
squaring the velocity.

That's not obvious. Ke = (mv^2)/2, and it isn't obvious that work should
always be twice kinetic energy. Kinetic energy and work aren't the same
thing, though they share the same units. A meteor in orbit around the Sun
has kinetic energy, but is doing no work by simply traversing its orbit.

Work is defined as the product of a force and the displacement it causes.
So

w = F * d

And

F =m * a

So we can substitute and get

w = m * a * d

in units, that equates to

kg * m/s^2 * m

which can be simplified to

kg * (m/s)^2

which has the same units as your formulation.

It arises from a different cause, and may differ numerically,
ie the 'm' in 'a' and the 'm' in 'd' may have different values
(they almost certainly will in most calculations), so you can't
just square a single numerical value for 'm' in reality.

This is like the case of torque, where the units are the same,
but we already know torque isn't work, and that torque *is*
a vector. Kinetic energy also has the units of work, but a
body with kinetic energy most certainly has a direction, so
is it vector or scalar? I'm afraid we can't tell by the way the
units work out.

Gary