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Re: [Phys-l] Energy Question Negative Reps.



If no net change in KE occurs from beginning to end, and no net change in PE
occurs from beginning to end, then the lifter did no work on the barbell -
pure and simple. Why is there a discussion on this? Are we making up an
alternate definition of work?

The fact that this doesn't jibe with our intuition probably just means that
the work-KE theorem is probably not the clearest way to present physics to
students.

Bob at PC

-----Original Message-----
From: phys-l-bounces@carnot.physics.buffalo.edu [mailto:phys-l-
bounces@carnot.physics.buffalo.edu] On Behalf Of rtarara@saintmarys.edu
Sent: Friday, January 12, 2007 2:06 PM
To: Forum for Physics Educators
Subject: Re: [Phys-l] Energy Question Negative Reps.

All kinds of complications here--but one key is to concentrate on the work
done on the bar. Using the traditional intro-level work/energy theorem,
the NET work on the bar is zero both going up and coming down because
there is no change in the KE. However, since there are two forces acting
(weight and the lifter) there is 500J of work done by the lifter going up
and again coming down. Now the work is negative coming down--making the
total work BY THE LIFTER zero, but the energy involved still has to be
transferred from the lifter. So, 1000 Joules of energy to lift and
replace the bar. Of course this is not the energy expendended
(transferred) from the lifter. There is a rather large factor to be
considered in terms of efficiency. If the human body is modeled as a
simple heat engine, the thermodynamic efficiency would be on the order of
only 5%. We do a little better than that, but not a whole lot. The bulk
of the energy 'burned' by the lifter becomes thermal energy.

We always calculate the amount of exercise needed to burn off 1 kg of fat.
I use running the stadium steps over at Notre Dame as the exercise. When
you get something like 500 repetitions from a simple mechanical equivalent
of heat calculation, students are quick to look for some mitigating
factor. Efficiency (metabolism) is the answer.

Work really is fairly complex for intro students because we need to keep
track of work done ON an object and work done BY and agent. That the work
can be either positive or negative complicates things and that each force
acting may be doing work even more. The work/energy theorem helps a lot
if one can concentrate on the NET WORK DONE ON AN OBJECT. Positive net
work if the object speeds up, negative if the slows, and zero if there is
no change in speed.

So, having said all this (which you probably already know), I think the
key here not confusing the work done by the lifter with the energy
expended (transferred) by the lifter.

Rick






----- S. Goelzer <sgoelzer@coebrownacademy.com> wrote:
Continuing on John Denker's theme of energy that cannot do work when

is energy is the ability to do work:

This problem has bothered me for some time -

How much work is done during one repetition of a weightlifting
exercise?
Keeping it simple, let's say a 1000 N barbell is bench pressed 0.5 m

up and then down at roughly constant speed. The bar was replaced on
the rack with a polite minimum of clanking.

The change in GPE for the bar is simple, it increases 500 J and then

decreases 500 J.

What energy expenditure does the person make? During the positive
part of the rep. upwards the person's chemical potential energy
decreased by 500J or they transfered 500 J to the bar (everyone happy

with the terminology of energy changes? no! - didn't think so).

On the negative stroke downward ( considered by many weightlifters to

be the more important in developing muscle mass and strength) the bar

is losing GPE, but the lifter is still pushing and decreasing their

supply of CPE to lower the bar in a controlled manner. This negative

controlled stroke is called the burn when the muscles are fatigued
and almost at the point of failure; it is believed by most to be the

process by which growth is stimulated. The decrease in GPE of barbell

is not available to help lower the weight as it might be it the bar
was raised and then lower by a counter weight system.

It would seem that the lifter must do work to gain energy.

How do I total out the work done by the lifter? If I am traditional
with my signs and I use W = Fd with d defined as displacement (as it

is in many but not all books), I get the lifter did no work - this
makes no sense to me and seems physically unreal.

As I see it conceptually, the decrease in CPE of the lifter due to
the exercise (barring other metabolic functions and efficiency
losses) was 1000 J from one meter of pushing 1000N. In addition, the

lifter increased in unusable thermal energy 500 J from the decrease
in GPE of the bar. It will now be necessary for the lifter to
transfer 1500J of thermal energy to the surroundings to maintain body

temp.

Has this ever been tested on a man sized calorimeter?


Comments?


Scott


**********************************
Scott Goelzer
Physics Teacher
Coe-Brown Northwood Academy
Northwood NH 03261
603-942-5531e218
sgoelzer@coebrownacademy.com
**********************************


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_______________________________________________
Forum for Physics Educators
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