Question

A 50 kg rock climber accidentally falls from the side of a rock and freefalls until she is stopped by her 9 m saftey rope. Assuming that the rope stops her completely with no rebound, what is the impulse impared on her body by the rope? If the rock climber's rope took 0.224s to stop her completely after she fell, what was the average force felt by the rock climber as she stopped?

Answer 1

Answer:

a) 665 kg.m/s

b) 2.97 kN

Explanation:

The impulse is given by:

$J=\Delta p\\J=m(v_a-v_b)$

where

va=velocity after

vb=velocity before

The velocity just before is given by:

$v_b=\sqrt{2*9.81m/s^2*9m}=13.3m/s(-\hat{j})$

The velocity just after is zero because there wasn't a rebound.

So the impulse is:

$J=50kg*(0-13.3m/s(-\hat{j}))\\J=50kg*(0+13.3m/s(\hat{j}))=665kg.m/s$

The impulse is also given by:

$J=\Delta p=F* \Delta t\\so:\\F=\frac{J}{\Delta t}\\\\F=\frac{665kg.m/s}{0.224s}=2.97kN$