Question

A 79.7-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 1.72 x 103 N/m. He accidentally slips and falls freely for 0.968 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

along verrical

for free fall

initial velocity voy = 0

acceleration ay = -g

displacement y = -0.968 m

final speed at y = vy = ?

vy^2 - voy^2 = 2*ay*y

vy^2 - 0 = 2*9.8*0.968

vy^2 = 18.97

the rope stretches by distance d

the total energy before stretching

Ei = (1/2)*m*vy^2 + m*g*d

the total energy after the climber momentarily comes to rest

Ef = (1/2)*k*d^2

from energy conservation

Ef = Ei

(1/2)*k*d^2 = (1/2)*m*vy^2 + m*g*d

(1/2)*1.72*10^3*d^2 = (1/2)*79.7*2*9.8*0.968 + 79.7*9.8*d

d = 1.496 m <<<----------ANSWER