- #1

clope023

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**[SOLVED] Potential Energy on an ideal spring**

## Homework Statement

An ideal spring of negligible mass is 12.00cm long when nothing is attached to it. When you hang a 3.15kg weight from it, you measure its length to be 13.30cm.

If you wanted to store 10.0J of potential energy in this spring, what would be its total length? Assume that it continues to obey Hooke's law. (also says there might be 2 answers)

## Homework Equations

K1 + U1 + Wother

= K2 + U2

or

Wel = 1/2ky1^2 - 1/2ky2^2

= Uel1 - Uel2

U = mgy

## The Attempt at a Solution

12cm = .12m = y1

13.30cm = .133m = y2

10J = Wother

1/2ky1^2 = 10J + mgy1 + 1/2ky2^2

1/2k(.0144m^2) = 10J + (3.15kg)(9.8m/s^2)(.133m) + 1/2k(.0177m^2)

-14.11J = 1/2k(.0177m^2-.0144m^2)

-14.11J = 1/2k(.0033m^2)

-28.22J = k(.0033m^2)

-28.22m = (.003m^2)

is this right? would I just divide -28.22m by .003m^2 and get a #? I'm pretty sure it's not right but I'm not sure of other equations to use, should I use kinematics as well?

edit: or perhaps I should have another potential energy in there for y1? or use the potential energy for mass in y1 and solve for y2?

or perhaps use this equation?

K1 + Ugrav1 + Uel1 + Wother = K2 + Ugrav2 + Uel2

seems like this would work better than what I was using?

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