Question

A 55.0 kg bungee jumper steps off a bridge with a light bungee cord tied to her and to the bridge. The unstretched length of the cord is 15.0 m. She reaches the bottom of her motion 40.0 m below the bridge before bouncing back. Her motion can be separated into an 15.0 m free-fall and a 25.0 m section of simple harmonic oscillation. Use the principle of conservation of energy to find the spring constant of the bungee cord.

Answer 1

`69.06\ \text{N/m}`

Step-by-step explanation:

m = Mass of bungee jumper = 55 kg

g = Acceleration due to gravity =
`9.81\ \text{m/s}^2`

h = Height at the bottom = 40 m

k = Spring constant

x = Displacement of bungee cord = 25 m

Applying the principle of conservation of energy to the system we have

`mgh=(1)/(2)kx^2\\\Rightarrow k=(2mgh)/(x^2)\\\Rightarrow k=(2* 55* 9.81* 40)/(25^2)\\\Rightarrow k=69.06\ \text{N/m}`

The spring constant of the bungee cord is
`69.06\ \text{N/m}`.