Re: spring compression

[Joe Darling <jdarling@emh1.otc.cc.mo.us>]

I'll hazard an answer.  If you let yourself down carefully, holding on to
the wall or whatever, you would reach the equilibrium position where the
two forces balanced.  However, if you just stepped onto the platform,
initially the force of gravity would be greater than the force of the
spring and you would accelerate downward to the distance x = 2mg/k at
which time the work done by gravity would be equal and opposite to the
work done by the spring and your kinetic energy would have returned to its
original value, zero.  However, you would not be in equilibrium, as now 
the force of the spring would be greater than that of gravity, and you
would begin to accelerate upward and neglecting friction, would oscillate
back and forth between your original level and x = 2mg/k.

Joe D. Darling   jdarling@emh1.otc.cc.mo.us
Instructor of Physics and Physical Science  
Ozarks Technical Community College
1020 East Brower   Springfield, MO 65802
(417) 895-7907   (417) 895-7085 FAX

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.