Chronology Current Month Current Thread Current Date
[Year List] [Month List (current year)] [Date Index] [Thread Index] [Thread Prev] [Thread Next] [Date Prev] [Date Next]

Re: positive and negative work



On 11/12/01 6:31 PM, "Brian Whatcott" <inet@INTELLISYS.NET> wrote:

At 10:18 AM 11/12/01 -0500, Gene Mosca wrote:

I first need to thank Gene for not visibly noticing that I was
addressing the wrong person.

Brian Whatcott wrote:

We use Professor Leigh's favorite model: the car rolling on an incline.
Let us suppose that Michael is sitting in this car, which has started
to roll backwards, downhill. He wants to proceed slowly astern,
so he applies a little throttle while he is in a forward gear.
This prevents the car from careering away.

The work done by the car, we say, is negative. There is no
discussion, no debate. [!!!] This is the convention.

I disagree. The car exerts a force on the road surface, which is
stationary, and the tire is rolling so the frictional force is static, not
kinetic. It seem to me that the car is not doing any work on the road
surface. Is there anything else the car is doing work on?

Gene

--
Eugene P. Mosca
301 Constitution Blvd.
Kutztown, PA 19530
(610) 683-3597
emosca@ptd.net


Hmmmm...we agree that a car would accelerate down an incline,
if a force acting 'uphill' were not holding its speed steady.

I agree. That would be the force exerted on the tire tread by the road
surface.

We see that this retarding force varies with the throttle so we are
confident that the car is supplying this retarding force.

I disagree. The acceleration of the car is proportional to the sum of the
external forces acting on it. The car can push the road harder down the
slope. According to Newton's third law the road will simultaneously push
the car up the slope harder.


I can agree with you, but STILL I can say that the car is controling the
retarding force - if "supplying the force" makes you uncomfortable.
That's what drivers do, after all!


We see that this force acts opposite the motion, so we say by convention
the work is negative.

The definition of work I am using is: Work is the integral of F dot v dt,
where v is the velocity of the point of application of the force. You
stated:
The work done by the car, we say, is negative.
Since we are talking about the work done "by" the car, I presume we are
talking about the work done by the forces exerted by the car on its
surroundings. The force the car exerts on the stationary road surface does
no work on it.


Yes, I understand now that saying that the car is retarded does not mean
that the CAR is doing negative work - in this point of view. I think you
too will agree that we are applying a force to the road surface which
contributes to accelerating the Earth's movement in the direction of the
car's movement.
So here, I should say that the car's tires are doing work on the Earth.

Can you agree with this?


I agree, but with caution as I am unsure what you mean by work. The Earth's
displacement is quite miniscule. Thus, the work done on it is also
miniscule--and usually negligible. (I am assuming a reference frame in
which initially the Earth is at rest.) The work done by the tire on the
Earth is equal to the integral of F dot v dt, where v is the velocity of the
road surface and F is the force exerted on the road by the tire. The
pseudowork done by the car on the Earth is the integral of F dot v_cme dt,
where v_cme is the velocity of the center of mass of the earth. It is
unclear, but doubtful, that the work you are talking about is either of
these.


But this is another way of saying that if the car's
work on the road is negative, the road's work on the car is positive.


Now, to reverse my assertion: if the braked tires are doing positive work
on the Earth, then the Earth is doing negative work on the car. The Earth
is reacting a force uphill, while its point of application is moving down
hill.
Does this now seem like a sensible physics model for negative work?

I would not agree that the point of application of the force is moving
downhill. The part of the tire in contact with the road is at rest relative
to the road, so it has the same velocity as the road. The road is not
moving downhill and neither is the part of the tire in contact with it.

Since the tires do not slip on the road, the velocity of the tire tread in
contact with the road is v, the velocity of the road. The work done is the
integral of F' dot v dt, where F' is the force by the road on the tire.
Since F' = -F, the work done by the road on the tire is the negative of the
work done by the tire on the road. The pseudowork done by F' is the
integral of F' dot v_cmc dt, where v_cmc is the velocity of the center of
mass of the car. This velocity is downhill in your example.

Perhaps you are referring to center-of-mass work (pseudowork). Pseudowork is
the integral of F dot v_cm dt, where v_cm is the velocity of the center of
mass of the system the force acts on.

I can assure you that no concept of pseudowork or quasi work entered my
thoughts.

Carl's model does not seem to offer a mechanism. Would it be clearer if
the tires were locked, and the car were slowly skidding down hill?

Perhaps not. What if the car were stationary, and the road were a
treadmill running uphill? The treadmill is then applying a force in the
direction
of its motion at a given velocity. This seems to describe a rate of
[positive] working.

What did I miss? :-)

It seems to me you neglected to define work. It appears there are several
working definitions of work, and I have defined two here.

Gene



**************************
* Eugene P. Mosca x


<smile> Gene, I don't think I'm smart enough to be choosing between
viable alternative models for a physics concept. I usually satisfy myself with
supposing that I can find a useful product of force and displacement acting
in the same direction as a measure of positive work.

My problem is that I am not smart enough to discern what you mean unless I
can parse it into something that is well defined. When you refer to the
work done by the car, I interpret that literally. However, I now presume
what you mean is the work done by one part of the car, say the transmission,
on another part of the car, say the drive shaft.

But this leads to some curious speculations wherein I see that the sign of
the work changes from place to place inside the object of intrerest - the car.

I expect that this is why there is a natural urge expressed on the list to
stick with particles - they don't reverse in this disconcerting way.

Thank you for patiently rehearsing the usual arguments.
Would you say you are orthodox, in this respect?


I believe I am orthodox. Do you find that a good thing or a bad thing. I
do see the urge to stick with particles in phys-l discussions. I also see
the need for lucid discourse, if misunderstandings are to be avoided. If
the term work is used, a definition of its meaning is needed. To me, the
default definition is the one associated with the first law of
thermodynamics. This is the work that is associated with energy transfer.
To say that work is the integral of F dot dr, where dr is the displacement,
is inadequate unless we are also told what force and what displacement are
being referred to.

Gene

--
Eugene P. Mosca
Physics Department
MI350-B
United States Naval Academy
(410) 293-6668