Question

A 85.3-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 1.16 x 103 N/m. He accidentally slips and falls freely for 0.757 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

Work done by rope = ∆KE of climber + ∆PE of climber

As the climber falls 0.815 meters, the climber’s velocity increases at the rate of 9.8 m/s each second. Since we don’t know the time, use the following equation to determine the velocity at that point.
vf^2 = vi^2 + 2 * a * d, vi = 0, a = 9.8, d = 0.757
vf^2 = 2 * 9.8 * 0.757
vf = √14.8347

Now we can use conservation of energy to determine the distance the rope stretches. As the rope stretches, the climber’s velocity decreases to 0 m/s.
∆KE of climber = ½ * 85.3 * 14.372 = 612.96

As the rope stretches, the climber moves s specific distance downward. During this time, the climber’s potential energy is decreasing.
∆PE = 85.3 * 9.8 * ∆h = 835.94 * h, h is the distance the climber falls.

As the rope stretches, its potential energy increases.
∆PE = ½ * 1.29 * 10^3 * d^2 = 645 * d^2
The distance the rope stretches is the same as the distance the climber falls.
∆PE = ½ * 1.16 * 10^3 * d^2 = 580 * h^2

Work done by rope = ∆KE of climber + ∆PE of climber
580 * h^2 = 612.96+ 835.94 * h
5 80* h^2 – 835.94 * h – 612.96 = 0

=0.7331