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Re: spring compression



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.

You are right; the book is wrong. If the man steps *quickly* on the
platform it will deflect as they say, but it will then oscillate about
the point you calculated and settle there. The law of conservation of
(mechanical) energy is inapplicable to this problem. The reason is
simple. Energy is conserved in the process (or rather the work energy
theorem is applicable), but it must be recognized that as the man
gently transfers his weight to the platform the average force he
exerts downward on it is less than his weight since his other foot is
on the ground. Thus he only pushes it down part way. The details will
require calculus, but it is very simple.

Leigh