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real leaf spring.) Now, F=k'*phi=k*x. Using the same kind of reasoning asHere is where I think you have run amuck. F=k*x=k'phi only for d constant.
above, (double d, F const => double phi) I get F=K*c*phi/d=K*x/d^2. But
then the work integral gives E = (K*c^2/2)*phi^2 = (K/2)*(x/d)^2.
--
--James McLean
jmclean@chem.ucsd.edu
post doc
UC San Diego, Chemistry