The length of nylon rope from which a mountain climber is suspended has a force

Question

The length of nylon rope from which a mountain climber is suspended has a force constant of 1.7 x 10^4 N/m. What is the frequency, in hertz, at which he bounces given his mass and the mass of his equipment is 84 kg? How far, in centimeters, would this rope stretch to break the climber's fall if he free-falls 12 m before the rope runs out of Assume that tying two ropes of the same length together would have the same effect on the spring constant as two springs connected in series. What would the frequency of oscillation be, in hertz, if the rope was twice as long? If the climber falls the same distance before running out of slack, how far would the rope stretch, in centimeters, if it was twice as long?

Explanation / Answer

(b) Applying conservation of energy
Potential energy of the person = potential energy of the spring
mg(1.2+x) = (1/2)kx2
where m is the mass of person , x is the stretch of the rope and k is its force constant.
84*9.81(1.2+x) = (1/2)*1.7*104x2
824.04(1.2+x) = 8500x2
on solving for x
x = 0.393 m = 39.3 cm
(c) We know that spring in series
(1/k) = (1/k1)+(1/k2)
Sine K1 = K2 = K
k =K/2 = 1.7*104 /2 = 0.85*104 = 8.5*103 N/m
Hence frequency , f = (1/2Pi)*(K/m)1/2
f = 1.6 Hz
(d) We will follow the same as we did in part b
mg(1.2+x) = (1/2)Kx2
84*9.81(1.2+x) = 4250x2
x = 0.589 cm = 58.9 cm

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