Re: Car acceleration

["John S. Denker" <jsd@MONMOUTH.COM>]

"Carl E. Mungan" wrote:
> He clearly had a circle
> around the car indicating the choice of system and an arrow on his
> diagram labeled "work of the road on the car." I put up my hand at
> the end and asked him about it and he quietly said something to the
> effect of "maybe that wasn't a very good choice of system."

IMHO he's trying to cover one error with another.
IMHO drawing a box around the car (not including
the road) is a fine system, well worth analyzing.
He just analyzed it wrong.  The work of the road
on the car is zero.  F dot dx.  What could be simpler?

> 1. I can choose the system any way I like. Different choices lead to
> different insights about the forces, work, momentum & energy
> transfer, etc. I have discovered that certain textbooks insist that
> systems must be chosen in only certain ways. In particular, some
> books baldly state that one must never choose the system boundary to
> be the "open" interface between two objects across which friction
> acts. My response to this is: You are missing an opportunity to gain
> additional, valuable insights. Who are you to tell me I cannot choose
> my system other ways?

Well said.

> 2. According to the pseudowork-kinetic-energy theorem, things are
> clear. The net external force F on the car is static friction forward
> minus air drag backward. (If you want to include rolling friction, as
> Herb seems to want, toss that in as another backward term. Of course,
> there is only one overall friction, but there is no harm in
> artificially splitting it into pieces for discussion purposes.) The
> net center-of-mass displacement dx is forward in some time interval
> dt. Therefore, we have Fdx = d(mv^2/2). Simple and clear to students.

I find pseudowork arguments unnecessary at best,
often unclear and confusing.

The key concept hear is _momentum_.  Momentum is
transferred across the aforementioned system
boundary.  Students can easily enough calculate
the effect this momentum-change has on the
energy budget, without introducing the additional
and not-very-useful notion of pseudowork.

If you analyze the pseudowork carefully, you
will discover that it
is connected only to pseudo-KE, not the real
genuine total KE.  In the case of the car, there
will be real genuine KE associated with the rotation
of the tires, engine machinery et cetera, which
cannot be even guessed at based on pseudowork
arguments.

In any case, this is primarily a force-balance
problem.  Bringing in any form of work (pseudo
or otherwise) or any form of energy is a red
herring IMHO.  It is at most an afterthought,
if some student asks how there can be a transfer
of momentum with no transfer of energy.

Hint: Reduced mass.

Hint #2:  Consider the case of pushing against
something that is not hugely massive compared
to the car.  Scott's file-folder is an excellent
extreme case, but intermediate cases (comparable
mass) are interesting, too.

> If you additionally assume F is constant, this agrees with an
> analysis using Newton's second law plus the equations of kinematics
> for constant acceleration.

But what if it isn't?  What about my punter who
only intermittently pushes on the pole?
(This is a minor gripe, not much more than a
nitpick, so don't feel obliged to respond.  The
rotational KE issue is incomparably more profound.)

> 3. Introducing thermodynamics does not help.

That's for sure.

> (just tooting the "pseudowork" horn again)

Not music to my ears :-)