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Re: spring compression





From W. Barlow Newbolt Oct 30, 97 11:45


...I think that the calculation with the work done due to gravity and the
potential energy of the spring is correct. If you apply your weight to a
spring loaded platform the platform will accelerate downward through the
postion and then vibrate around. I would expect it to come to rest at the
equilibrium position, however.

Who is dissipating energy for this to happen? Only the quilibrium
position is here well definited, where one gets the first result.
If the damping is so strong that the platform reaches the
equilibrium position without any oscillation, half of the work
done by gravity should be dissipated, and both ways would agree.
isn't it?

M.A.Santos
msantos@etse.urv.es

On Wed, 29 Oct 1997, Neil D. Adams wrote:

Hi
I came across a problem in my physics class that I haven't been able to
resolve. Perhaps it is the result of mental overload, or failing memory
circuits. The problem is from University Physics by Young and Freedman,
Chapt 7, no. 60, the b part of the problem. The student is to calculate
how far a sping supported platform will compress if a 90 kg man steps
gently on it. I worked this by setting his weight (w=mg) equal to
Hooke's law force (F=kx) leading to x=mg/k. The instructors manual
worked the problem by setting the gravitational potential energy (U=mgx)
equal to the compressional energy (one-half k (x squared)) (sorry but I
don't know how to do superscripts with this program). This leads to the
expression x=2mg/k. What gives? Thanks for your help.


W. Barlow Newbolt 540-463-8881 (telephone)
218 Howe Hall 540-463-8884 (fax)
Washington and Lee University newbolt.w@fs.science.wlu.edu
Lexington, Virginia 24450 wnewbolt@liberty.uc.wlu.edu

"Prediction is very difficult, especially about the future."

Neils Bohr