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 1.1 times 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 94 kg? A numeric value is expected and not an expression. How much would this rope stretch, in centimeters, to break the climber's fall if he free-falls 1.2 m before the rope starts to stretch? A numeric value is expected and not an expression. What is the frequency, in Hz, at which he bounces, given his mass and the mass of his equipment is 94 kg if the rope is twice as long? A numeric value is expected and not an expression. (d) How much would this rope stretch, in centimeters, to break the climber's fall if he free-falls 1.2 m before the rope runs out of slack if the rope was twice the length? A numeric value is expected and not an expression.

Explanation / Answer

k = 1.1 x 10^4 N/m = 11000 N/m

a)m = 94 kg

The frequency at which he bounces will be given by

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

f = 1/2 x 3.14 x sqrt (11000/94) = 1.72 Hz

Hence, f = 1.72 Hz

b)from conservation of energy

m g (1.2 + x) = 1/2 k x^2

94 x 9.8 x 1.2 + 94 x 9.8 x = 1/2 x 11000 x^2

5500x^2 - 921.2 x - 1105.44 = 0

the above quadratic eqn gives us

x = 0.54 m

Hence, x = 0.54 m

c)The stiffness constant will become

k' = 2k = 22000

f' = 1/2pi sqrt (k'/m)

f' = 1/2 pi sqrt(22000/94) = 2.44 Hz

Hence, f' = 2.44 Hz

d)from conservation of energy

m g (1.2 + x) = 1/2 k' x^2

94 x 9.8 x 1.2 + 94 x 9.8 x = 1/2 x 22000 x^2

11000x^2 - 921.2 x - 1105.44 = 0

the above quadratic eqn gives us

x = 0.36 m

Hence, x = 0.36 m

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