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**Elastic Potential Energy Question! tough**

An 86.0 kg climber is scaling the vertical wall of a mountain. His safety rope is made of nylon that, when stretched, behaves like a spring with a spring constant of 1.20 x 10^3 N/m. He accidentally slips and falls freely for 0.750 m before the rope runs out of slack. How much is the rope streched when it breaks his fall and momentarily brings him to rest?

This is my approach:

First i found the energies...

KE = 1/2(86)(0) = 0J

PE = (86)(9.8)(0) = 0 J

PE(elastic) = = 1/2ky^2 = 1/2(1.2 x 10^3) (.75)^2 = 338J

THEN, i put The total energy in this equation to solve for elastic distance..

E = KE + PE + PE(elastic)

338J = 1/2(86kg)(0m/s) + (86)(9.8 m/s^2)(.750m) + 1/2(1.2 x 10^3 N/m) (x)

x = 1.62 m (i know I am wrong b/c the book answer is different)

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