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 1.54 104 N/m.

(a) What is the frequency (in Hz) at which he bounces, given that his mass plus the mass of his equipment is 78.0 kg?

2.24 Hz

(b) How much would this rope stretch (in cm) to break the climber's fall if he free-falls 2.00 m before the rope runs out of slack? Hint: Use conservation of energy.

49.8 cm

(c) Repeat both parts of this problem in the situation where twice this length of nylon rope is used.

frequency (in Hz) ____________ Hz

stretch length (in cm) ____________ cm

I have part a & b, but cannot figure out c.

Explanation / Answer

c)

keq = k1 + k2 = 2*k = 3.08 x 10^4 N/m

f = (1/2)(k/m)

f = (1/2)(3.08 x 10^4 / 78)

f = 3.1642 Hz from part a)

PE gets converted into spring energy U. If the rope stretches x, then

mg(2m + x) = 1/2kx^2

(78kg x 9.81 m/s^2 x 2m) + (78kg x 9.81 m/s^2 x (x)) = 1/2 x 30800 N/m x x^2

15400x^2 - 765.18x - 1530.36 = 0

solve quadratic equation

quadratic with roots at x = -0.2914 m not possible

and x = 0.34 m = 34 cm answer from b)

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