8:48 PM iPad 43% O edugenwileyplus com Chapter 10, Problem 34 A 76.2 kg climber

Question

8:48 PM iPad 43% O edugenwileyplus com Chapter 10, Problem 34 A 76.2 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 1.75 x 103 N/m. He accidentally slips and falls freely for 0.745 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? Number Units the tolerance is +/-2% LINK TO TEXT LINK TO TEXT Question Attempts: Unlimited SAVE FOR LATER SUBMIT ANSWER Copyright 2000-2017 by John Wiley & Sons, Inc. or related companies. All rights reserved.

Explanation / Answer

Change in gravitational potential energy = change in elastic potential energy.

mg(L + x) = (1/2)kx2

=> 76.2(9.8)(0.745 + x) = (1/2)(1750)x2

=> 875x2 - 746.76 - 556.3362 = 0

solving this quadratic gives, x = 1.33 m and x = - 0.477 m

discard the negative value since extension should be positive.

=> x = 1.33 m

So, the rope stretches by 1.33 m before momentarily bringing the jumper to rest.

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