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From: Jeff Liebermann <jeffl@cruzio.com>
Newsgroups: rec.bicycles.tech
Subject: Re: Petential Energy doing Work
Date: Mon, 15 Jul 2024 18:43:55 -0700
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On Mon, 15 Jul 2024 21:13:00 -0400, Frank Krygowski
<frkrygow@sbcglobal.net> wrote:

>On 7/15/2024 2:42 PM, Jeff Liebermann wrote:
>>  
>> By slowly lowering your center of mass, you are lowering your
>> potential energy.  Potential energy is where nothing is moving.  You
>> have the potential of moving but there's little or no kinetic (moving)
>> energy produced.  In other words, you cannot dissipate potential
>> energy without first converting it to kinetic energy.
>
>Actually, by using your body strength to lower a mass, you do dissipate 
>the potential energy without (necessarily) converting it to kinetic 
>energy. You do it by applying opposing work.

Yeah, that's more accurate.  I intentionally didn't include that
because it would have required that I include the velocity at which
the center of mass is raised and lowered.  Since you mentioned it and
since Tom wanted to know where the energy went, it's easy enough to
explain.  If you raise yourself up with your feet on the pedals,
you're also compressing the tires and increasing the size of the
contact patch.  That's where the energy went.  If you start with your
legs straight and are out of the saddle, lowering yourself into the
saddle will momentarily reduce the load on the tires, thus allowing
the tires to expand (slightly) and momentarily decreasing the size of
the contact patch.  When you stop moving up or down, the tire pressure
and contact patch sizes are the same for both cases (up and down).

We haven't even gotten Tom to realize that jumping up and down on the
saddle does not change the direction of the forces involved and
therefore does not produce any forward motion.

>Mechanical work is defined as essentially force times distance. (I'm 
>omitting details Tom wouldn't understand.) The work done in lowering a 
>mass to a position of rest is equal and opposite to the mass's initial 
>potential energy. At the end of the process, the energy would be zero.
>
>Again, there are some complications (variable forces, accelerations and 
>decelerations, various possibilities for the PE datum, etc.) which 
>people other than Tom might want to discuss. But the simple case should 
>make the physics clear.

The static model is VERY simple because it ignores any states where
something is moving, accelerating, bouncing, storing or releasing
energy, or involving relativistic complications.


-- 
Jeff Liebermann                 jeffl@cruzio.com
PO Box 272      http://www.LearnByDestroying.com
Ben Lomond CA 95005-0272
Skype: JeffLiebermann      AE6KS    831-336-2558