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? 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!