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If the 200 grams is then stretched a distance x from the new equilibr=
ium point, the potential energy relative to the new equilibrium point=
is equal to 1/2 kx^2. What do you call the quantity 1/2 kx^2? The =
difference in the elastic-gravitational potential energy?
The change in gravitational energy over a vertical distance of a meter or
two is very close to zero.