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 1.75 104 N/m. (a) What is the frequency at which he bounces, given his mass plus equipment to be 88.0 kg? in Hz (b) How far in cm would this rope stretch to break the climber's fall, if he free-falls 1.2 m before the rope runs out of slack? in m (c) Repeat both parts of this problem in the situation where twice this length of nylon rope is used. (d) 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

given that

k = 1.75*10^4 N/m

m = 88 kg

(a)

we know that

f = 1/2*pi * sqrt(1.75*10^4/88)

f = 2.24 Hz

(b)

from conservation of energy

PE is converted into spring energy U

m*g(h+x) = 1/2*k*x^2

88*9.8(1.2 + x) = 1/2*1.75*10^4*x^2

1034.88 + 862.4*x = 8750*x^2

8750*x^2 - 862.4*x - 1034.88 = 0

solving by quadratic equation

x = 0.396 m

(c)

length of rope is double means

k' = k/2 = 17500/2 = 8750 N/m

so f = 1/2*pi * sqrt(.875*10^4/88)

f = 1.58 Hz

m*g(h+x) = 1/2*k*x^2

88*9.8(1.2 + x) = 1/2*.875*10^4*x^2

1034.88 + 862.4*x = 4375*x^2

4375*x^2 - 862.4*x - 1034.88 = 0

solving by quadratic equation

x = 0.594 m

(d)

x = 0.594 m = 59.4 cm

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