Question

The length of nylon rope from which a mountain climber is suspended has a force constant of 1.40  10⁴ N/m. (Hz)
(a) What is the frequency at which he bounces, given his mass plus equipment to be 70.0 kg? (m)
(b) How much would this rope stretch to break the climber's fall, if he free-falls 2.00 m before the rope runs out of slack?
(c) Repeat both parts of this problem in the situation where twice this length of nylon rope is used. Bounce frequency and distance stretched.

Answer 1

Given,

a) The frequency is given by,

b) Let the rope stretches x m to break the climber’s fall. Given he free-falls 2.00 m before the rope runs out of slack. Considering the conservation of energy we can say that the PE stored in the stretched nylon rope will be equal to the loss of PE due to the decrease in the position of height (x+2) m.

Hence,

On solving the above quadratic equation, we get,

x = 0.4944m (Ans.)

c) In the case where twice the length of nylon rope is used, it may be treated as two parallel ropes of the same length. In this case the equivalent force constant will be just twice that of the first

The mass remains the constant, ie, m = 70kg

Substituting this value of k we can easily repeat the same calculation using above equations.

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a) The frequency is given by,

b) Let the rope stretches x m to break the climber’s fall. Given he free-falls 2.00 m before the rope runs out of slack. Considering the conservation of energy we can say that the PE stored in the stretched nylon rope will be equal to the loss of PE due to the decrease in the position of height (x+2) m.

Hence,

On solving the above quadratic equation, we get,

x = 0.3385m (Ans.)