I apologize to the list for my previous post. I hadn't originally intended
to do the "Aristolelian" nor any other interpretation. I just thought it
would simplify matters to look at the example of pushing an object and
remove the object being "worked on" from the object "doing work". I was
rushed and at the last minute added something I ought not have.
In any event, Tim Folkerts pointed out the point I was trying to make: that
one does positive work on the object one way and negative work on the object
the other way. How much work I do on the object is not directly related to
how much effort I need to apply. Just holding the object steady requires
effort but I do no work on it.
I'll shut up now.
Robert Cohen
-----Original Message-----
From: Tucker Hiatt
To: PHYS-L@lists.nau.edu
Sent: 7/23/01 12:24 PM
Subject: Re: Muscle work
I've explained why this question bugs me. (Although Tim Folkerts
recent posting has helped considerably. Thanks!) Please let me also
explain why it bugs my students.
We do a classroom experiment that's designed to give students a
decent intuitive understanding of work and power. In the course of
performing various calisthenics/exercises, students measure muscle
force, displacement, and time, and then they roughly calculate the
muscle work done and the power output. Students are often amused to
find that their power in ascending a small flight of stairs, for
example, exceeds 1 horsepower (746 watts). They also come to
appreciate why turning off an unused 100-watt bedroom lamp is worth
doing. (Someone somewhere can stop doing pull-ups!)
Often, the muscle motion under examination is repetitive. Push-ups
and pull-ups are particular favorites. I instruct the students to
complete each half-cycle of their exercise at constant speed, if
possible. (And I acknowledge that we are ignoring the work done
during turn-around times.) This means that a student does work
during the downstroke as well as during the upstroke. On the
upstroke of a constant-speed pull-up, the average muscle force
applied is roughly the student's weight. On the constant speed
downstroke, however, the average muscle force is ...??? Mustn't it
also be the student's weight? And therefore the (magnitude of the)
work done is the same? Doesn't that vivify the question: Why do the
upstroke and downstroke FEEL different?
Tim's refusal to ignore periods of acceleration (in his second
starred "*" comment) may be the key. I'll also try his nice
suggestion that we compare "pull-up time-to-exhaustion" to
"mid-pull-up HANG-time-to-exhaustion".