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?
Best Answer
Energy isn't conserved in this experiment -- you release the spring, and the spring oscillates, loses energy while settling down, and ends in a new equilibrium after having lost energy. you calculated the turnaround point of the block if the spring was frictionless, and correctly determined that this is further down than the new equilibrium.