Question

A 94.9-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 1.98 x 103 N/m. He accidentally slips and falls freely for 0.692 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

mass of the climber (m) = 94.9 kg
Spring constant of rope (K) = 1.98*10³ N/m
Initial potential energy = mg(H+X)
Where H is the distance of fall = 0.692 m
and X is the stretch of the spring
= 94.9*9.81*(0.692+X) = 930.969(0.692+X)-----------------(1)
Now final energy
=Potential energy of the rope= (1/2)KX² -------------(2)
Applying energy conservation
(1/2)KX² = 930.969(0.692+X)
(1/2)*(1.98*10³)X² = 930.969(0.692+X)
990X² - 930.969X - 644.23 = 0
On solving this we get
x = 1.404 m
hence the rope will stretch by 1.404 m .