Question

A 86.8-kg climber is scaling the vertical wall of a mountain. His safety rope is made of a material that, when stretched, behaves like a spring with a spring constant of 1.76 x 103 N/m. He accidentally slips and falls freely for 0.580 m before the rope runs out of slack. How much is the rope stretched when it breaks his fall and momentarily brings him to rest?

Answer 1

The spring constant of the rope, k = 1.73 x 10³ N/m
The height with which the climber falls is 0.580 m
The potential energy of the climber is the maximum spring potential energy.
m g h = (1/2) k x²
Where x is the distance up to the rope stretches.
The height h = 0.580 + x  
m g (0.580 + x) = (1/2) k x²
86.8 kg x 9.80 (0.580 + x) = (1/2) 1.76 x 10³ x x²
493.37 + 850.64 x = 880 x²
880 x² - 850.64 x - 493.37 = 0
Solving this quadratic equation for x, we get
x =  1.374
&
x = -0.4079
We take the positive root, x = 1.374 m