Question

The length of nylon rope from which a mountain climber is suspended has a force constant of 1.40 × 10^4 N/m .

(a) What is the frequency at which he bounces, given his mass plus and the mass of his equipment are 90.0 kg?
(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? (Hint: Use conservation of energy.)
(c) Repeat both parts of this problem in the situation where twice this length of nylon rope is used.

Answer 1

w = 12.5 rad/s and A = 50 10⁻² m

Step-by-step explanation:

Let's treat this problem as a case of oscillatory movement

a) The angular velocity is given by

w² = k / m

w = √ 1.4 10⁴/90.0

w = 12.5 rad / s

b) It is climber falls by two meters, its mechanical energy at the highest point is

Em = U = mgh

Em = 90 9.8 2

Em = 1764 J

The energy of the harmonic oscillator can be calculated

E = ½ k A²

Where A is the amplitude of the movement. This is going to be the stretch for the mechanical energy of the fall

A² = 2E / k

A = √(2 1764 /1.40 10⁴)

A = 50 10⁻² m

c) To realize this part, you must know how the force constant changes with the length, in the case of two springs joined along its axis

1 /k equivalent = 1 / k1 + 1 / k2