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

[rtarara@saintmarys.edu]

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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