Posted by **zianna** on Wednesday, August 22, 2012 at 11:50pm.

an ideal spring of negligible mass is 12.00 cm long when noting is attached to it. When you hang a 3.15kg weight from it, you measures its length to be 13.40 cm.

if you wanted to store 10.0 J of potential energy in this spring, what would be its total length?

Assume that it continues to obey hooke's Low.

- physics -
**Ajayb**, Thursday, August 23, 2012 at 9:02am
First find K of the spring:

Mg = Kx

here, x = elongation of the spring

= 13.40 - 12.00

= 1.40 cm

= 0.014 m

K = Mg/x

= 3.15*9.8/0.014 = ??

Having got K, find X from:

KX^2/2 = 10.0J

where X is the elongation when stored PE is 10.0 Joules.

The total length of spring then would be:

X + 0.12 m

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