OK, so I've been wracking my brain for the past hour trying to figure out how to calculate k in a problem like this:
A mass of 10 kg is attached to a spring hanging from the ceiling. It is released, allowed to oscillate, and comes to rest at a new equilibrium point 10 meters below the spring's natural length. What is the value of the spring constant k for this spring?
There's two approaches that are giving me different answers:
1: We can use forces: at equilibrium, the force of gravity will equal the spring force, so mg = kx. This gives a value of (10)(9.8) = k (10) or k = 9.8.
2: We can use potential energy. Before the mass is released, it has gravitational potential energy; at the new equilibrium, it has LESS gravitational PE, but more elastic PE, since it is now stretched. The elastic PE must have been converted from gravitational PE, so dPE (elastic) = dPE (gravitational). Since the change in height is the same as the change in stretch, the h in mgh = the x for spring stretch. So:
dPE (elastic) = dPE (gravitational), h = x
.5kx^2 = mgx
.5kx = mg
k = 2mg/x
Plugging in, we get k = 2(10)(9.8)/10 or k = 19.6, which is twice as much as the k found through the other method.
I must be missing something here, why am I getting two different values for k depending on which approach I use?