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?(adsbygoogle = window.adsbygoogle || []).push({});

My Approach (I know it's gonna be wrong...i just don't know any other way to solve it!!):

First: I used PE(initial) + KE(initial) = PE(final) + KE(final) --- to find the initial speed!

(9.8 m/s^2)(86 kg)(0m) + 1/2(86 kg)V(initial) = (86 kg)(9.8 m/s^2)(.750 m)

V(initial) = 14.7 m/s

Second: I used the initial velocity, 14.7 m/s, and found the acceleration using Vf^2 = Vi^2 = 2ay

I got a = 144.1 m/s^2

Third: I used the acceleration, 144.1 m/s^2, to find the Force

F=(86 kg)(144.1 m/s^2) = 12392N

Lastly: i found the x (distance it traveled after 0.750m) -- This is where i think i have my plugging messed up (you can even assume it the same on the work above)

F=kx

12392N = (1.20 x 10^3 N/m)x

x=10.4 m

^^answer doesn't correspond with book answer!

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