Question

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

Answer 1

Answer:

w = 12.5 rad/s   and A = 50 10⁻² m

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