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 1.30 ? 104 N/m.

(a) 1.83 Hz is the answer.

What is the frequency (in Hz) at which he bounces, given that his mass plus the mass of his equipment is 98.0 kg?

  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.

62.28 cm

(c)

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

frequency (in Hz)  Hzstretch length (in cm)  cm. in Hz it is 1.294. How do you solve for for Part C in cm?

Explanation / Answer

(a)Frequency is given by f = (1/2?)?(k/m) where k = force constant = 1.3*104 N/m

m = mass = 98.0 kg

Substituting these values, we get

f = (1/2?)?(1.3e4/ 98kg) = 1.83 Hz

(b) PE gets converted into spring energy U. If the rope stretches "x," then
mg(2 + x) = ½kx²
98kg * 9.8m/s² * 2m + 98kg * 9.8m/s² * x = ½ * 13000N/m * x²
Dropping units for ease (x is in meters),
6500x² - 960x - 1921 = 0
quadratic with roots at x = -0.475 m ? not possible
and x = 0.6225 m = 62.25 cm

(c) Twice the length of rope is used. This means that now the rope behaves like two springs in series combination.

keq = k/2 = 6.5*103 N/m
Then part (a) becomes f = 1.296 Hz after substituting keq instead of k.
and part (b) leads to
98kg * 9.8m/s² * 2m + 98kg * 9.8m/s² * x = ½ * 6500N/m * x²

3250x² - 960x - 1921 = 0

quadratic with roots x = -0.635 m (not possible)

and x = 0.9305 m = 93.05 cm

Hope this helps!

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