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## Homework Statement

A mass of 0.5 kg hangs motionless from a vertical spring whose length is 0.90 m and whose unstretched length is 0.45 m. Next the mass is pulled down to where the spring has a length of 1.15 m and given an initial speed upwards of 1.4 m/s. What is the maximum length of the spring during the motion that follows?

## Homework Equations

L=length of unstretched spring; (0.45m)

d=length of spring in equilibrium; (0.9m)

x=stretch=(d-L); (0.45m)

mg=kx ---> k=mg/x; (10.89N/m)

Equilibrium position = y = (L+x); (0.9m)

Usi+Ki=Usf

1/2k(1.15-y)^2+1/2mv^2=1/2k(max stretch)^2

## The Attempt at a Solution

What I did at first was measure the stretch from the equilibrium position (0.25m) to get Usi and I solved for (max stretch) and added the (max stretch) to the length of the spring in equilibrium. I found some sources online, however, that say I should measure the stretch from the unstretched length? How does that even make sense? I would look up the answer in an answer key but I don't have one!