Question

A 67.3-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 1.23 x 103 N/m. He accidentally slips and falls freely for 0.921 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

Answer:

$d=0.59m$

Explanation:

From the question we are told that:

Mass $m=67.3 kg$

Spring constant $\mu=1.23 * 10^3 N/m$

Fall Height $h=0.921m$

Generally the Energy theorem equation for momentum is mathematically given by

Change in KE=Work done by gravity + work done by spring

$0=mg*(h + d) - \frac{\mu d^2}{2}$

$0=(67.3 * 9.81 (0.921 + d)) -\frac{(1.23 * 10^3 * d^2}{ 2}$

$0=608.1-660.213d-615d^2$

Solving Quadratic equation

$d=0.59m$