Answer to Question #106347 in Classical Mechanics for Christopher Bach

Question

Johan climbs a cliff wall. He is attached to an elastic safety rope that is 13.0 m long and has a spring constant of 1.2 kN / m. The safety line is attached to Johan's center of gravity, which is 3.1 m above the safety line's fastening point in the rock wall.
He loses his grip and falls straight down. When the safety line is stretched, the case is braked. During the case, the safety line is extended by 5.5 m.
Johan swings up and down a few times before stopping. He weighs 86 kg.

a) Calculate the oscillation time at the end of the oscillation process.
b) How far below the line's attachment point A does Johan hang when the halt has stopped?
c) Determine Johan's maximum speed during the case.



Answer 1

As per the given question,

Length of the elastic rope =13m

Spring constant (k) =1.2kN/m

The height of the safety line =3.1m above the ground

The extended safety line =5.5m

The weighting of the john =86 kg

a)

We know that time period of the oscillation of the spring,

$T=2\pi\sqrt{\dfrac{m}{k}}$

$\Rightarrow T=2\pi \sqrt{\dfrac{86}{1.2}}$

$\Rightarrow T=2\pi \times8.46$ sec

$T=53.15 sec$

b) Now after the stretching of the spring,

$\dfrac{kx^2}{2}=mgh$

$\Rightarrow h=\dfrac{1.2\times (5.5-3.1)^2}{2\times 86\times 9.8}$

$h=$ 0.004m

c) Now, applying the conservation of energy,

$\dfrac{kx^2}{2}=\dfrac{mv^2}{2}$

$\Rightarrow v=\sqrt{\dfrac{kx^2}{m}}$

$\Rightarrow v=\sqrt{\dfrac{1.2\times 2.4\times 2.4}{86}}$

$v=0.28m/sec$