In article , Gary Coffman says...

On 5 Oct 2003 07:45:54 -0700, jim rozen wrote:
Kinetic energy of a moving object and the work
done to produce that motion are defined as the
same thing - and yes the units are the same
as they should be. This comes under the general
heading of 'conservation of energy.'

But we have just been told (in the parts you snipped) that
work is w = mv^2 and kinetic energy is Ke = (mv^2)/2. So
they aren't the same, by a factor of 2. If both equations
are correct, we need an explanation for the factor of 2
difference.

I missed that. Work done on an object is 1/2 mv(sq).
But that's a conservation of energy thing, from a
force vector standpoint, its F(vector) x D(vector) x cos (theta)
where the cosine of the angle is there because only
the component of displacement in the direction of the
force counts.

But back to the original question, why is mechanical work
calculated with the dot product w = F dot D instead of the
cross product w = F X D?

Because it's only the component of the work in the
direction of the displacement that adds to the energy
of the particle.

The dot product is just the magnitute of each vector
times the cosine of the angle between them. For
example, if you push down really hard on a roller
skate, but it only moves slowly along the floor, it's
only the component of your hand's force *parallel*
to the floor that is actually pushing the skate to
speed it up. There the value of the dot product is
small as it should be.

I've heard some handwaving
about the conservation laws (and I'm a firm believer in
them), but no specific derivation which requires dot over
cross for this calculation.

OK, I managed to get a copy of halliday and resnick and
looking it over, it looks like they use the term "speed"
when calculating 1/2 mv(squared), not the vector velocity.

They proceed to derive an example of the "work - energy
theorem" for the case of a constant force acting on an
object. But they don't use vector velocity, but rather
the scalar speed instead. So where I put Vs below, they're
scalars, not vectors.

begin quote:
======================================

The simplest situation to consider is that of a constant
force F(vector). Such a force F(vector) will produce a
constant acceleration a(vector). Choose the x-axis to be
the common direction of F(vector) and a(vector). What is the
work done by this force on the particle in causing a
displacement x? For constant acceleration, the relations are

a= (V - Vo)/t

and

x= (t) X (V + Vo)/2

where Vo is the particle's *speed* [emphasis mine] at
t = 0 and V is its speed at time t.

Then the work done is:

W = (F)(x) = (ma) (x) [newton's law]

W = (m) [(V - Vo)/t] [(V + Vo)/2] (t)

W = 1/2 mV(sq) - 1/2 mVo(sq)

We call one half the product of the mass of a body and the
square of its *speed* [again, emphasis mine] the kinetic
energy of the body. If we represent kinetic energy by the
symbol K then

K = 1/2 mV(sq)

The work done by the result force acting on a particle is
equal to the change in kinetic energy of the particle.

Although this is proved for a constant force only, it holds
whether the resultant force [again, the dot product between
F(vector) and D(vector)] is constant or variable.

================================================== ===========
(end quote)

They go on to give some examples, where the force is not
in the direction of accelleration (circular motion, for
example, where cos(theta) is equal to zero) and where the
force varies in magnitude as well.

So basically according to H&R I was wrong in saying that
work is proportional to the dot product of two vectors,
it's not. It's proportional to the square of the *speed*
which is not a vector.

Sorry for the confusion.

Jim

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