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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.

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Answer:

[tex]69.06\ \text{N/m}[/tex]

Explanation:

m = Mass of bungee jumper = 55 kg

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

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

[tex]mgh=\dfrac{1}{2}kx^2\\\Rightarrow k=\dfrac{2mgh}{x^2}\\\Rightarrow k=\dfrac{2\times 55\times 9.81\times 40}{25^2}\\\Rightarrow k=69.06\ \text{N/m}[/tex]

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

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