the length of nylon rope from which a mountain climber is suspended has a force
ID: 1556307 • Letter: T
Question
the length of nylon rope from which a mountain climber is suspended has a force constant of 1.15 x 10^4 N/m.
a. what is the frequency, in Hz, at which he bounces, given his mass and the mass of his equipment is 98kg.
b. how much would this Rope stretch in centimeters to break the climbers fall if he free Falls 2.8 meters before the Rope starts to stretch
c. what is the frequency in Hz at which he bounces given his mask and the mass of his equipment is 98 kilograms if the Rope is twice as long
d. how much would this Rope stretch in centimeters to break the climbers fall if he free Falls 2.8 meters before the Rope runs out of slack if the Rope was twice the length
Explanation / Answer
As we know that
(a)
f = (1/2)(k/m) = (1/2)(1.15e4kg/s² / 98kg) = 1.72494 Hz
(b) PE gets converted into spring energy U. If the rope stretches "x," then
mg(2m + x) = ½kx²
98kg * 9.8m/s² * 2.8m + 98kg * 9.8m/s² * x = ½ * 11500N/m * x²
Dropping units for ease (x is in meters),
5750x² - 960.4x - 2689.12 = 0
quadratic with roots at x = -0.6054335958990299 not possible
and x = 0.7724596828555516 m
(c) If we double the length the rope then
Frequency remains same
(d) part b also same because energy of the system remains conserved
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