Calculating the Stretch of a Climber's Safety Rope: A Simple Harmonic Question

[cheechnchong]

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 going to 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!

 

[Hootenanny]

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 going to be wrong...i just don't know any other way to solve it!):

Your approach is indeed wrong. You need to consider the conservation of all forms of energy, including gravitational potential, kinetic and elastic potential. Can you rewrite your initial equality including the above quantities?

Just as a side note, why the SHM title?

 

[Aidan130791]

possible answer?

Okay,

After he falls, he travels with F=ma N. Acceleration is g=10N/kg=10m/s^2. Therefore, the 86kg climber falls with a force of 860N.

Now, to make the rope stretch 1m, you need 1,200N. So, the distance the rope stretches can be expressed as a percentage. i.e.

860N/1200N = 0.716666666m = 0.72m

Right?

 

[Mentz114]

Try equating the kinetic energy gained during the free-fall, to the potential energy in the rope when he comes to rest dangling on the end.