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Problem 7: The length of nylon rope from which a mountain climber is suspended has a force constant of 1.55 x 10^4 N/m. Mass of his equipment is 92kg f = 2.065 Part (b) How much would this rope stretch, in centimeters, to break the climber’s fall if he free-falls 1.6 m before the rope starts to stretch? f2 = 1.460 Part (d) How much would this rope stretch, in centimeters, to break the climber’s fall if he free-falls 1.6 m before the rope runs out of slack if the rope was twice the length?

Problem 7: The length of nylon rope from which a mountain climber is suspended has a force constant of 1.55 x 10^4 N/m. Mass of his equipment is 92kg f = 2.065 Part (b) How much would this rope stretch, in centimeters, to break the climber’s fall if he free-falls 1.6 m before the rope starts to stretch? f2 = 1.460 Part (d) How much would this rope stretch, in centimeters, to break the climber’s fall if he free-falls 1.6 m before the rope runs out of slack if the rope was twice the length?

College Physics
College Physics
11th Edition
ISBN:9781305952300
Author:Raymond A. Serway, Chris Vuille
Publisher:Raymond A. Serway, Chris Vuille
Chapter1: Units, Trigonometry. And Vectors
Section: Chapter Questions
Problem 1CQ: Estimate the order of magnitude of the length, in meters, of each of the following; (a) a mouse, (b)...
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Problem 7: The length of nylon rope from which a mountain climber is suspended has a force constant of 1.55 x 10^4 N/m. Mass of his equipment is 92kg f = 2.065 Part (b) How much would this rope stretch, in centimeters, to break the climber’s fall if he free-falls 1.6 m before the rope starts to stretch? f2 = 1.460 Part (d) How much would this rope stretch, in centimeters, to break the climber’s fall if he free-falls 1.6 m before the rope runs out of slack if the rope was twice the length?
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Step 1: GIVEN DATA

k equals space 1.55 cross times 10 to the power of 4 space N divided by m m equals space 75 space k g f r e q u e n c y equals 2.065 space h z

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