Question

A bungee jumper with mass 58.5 kg jumps from a high bridge. After arriving at his lowest point, he oscillates up and down, reaching a low point seven more times in 45.0 s . He finally comes to rest 20.5 m below the level of the bridge.Calculate the spring stiffness constant AND the unstretched length of the bungee cord.

Answer 1

10.23077 m

55.88391 N/m

Step-by-step explanation:

m = Mass of person = 58.5 kg

T = Time period =
`(45)/(7)=6.42857\ s`

k = Spring constant

g = Acceleration due to gravity = 9.81 m/s²

Time period is given by

`T=2\pi\sqrt{(m)/(k)}\\\Rightarrow T^2=4\pi^2(m)/(k)\\\Rightarrow k=(4\pi^2m)/(T^2)\\\Rightarrow k=(4\pi^2 58.5)/(6.42857^2)\\\Rightarrow k=55.88391\ N/m`

The spring stiffness constant is 55.88391 N/m

The force on spring is given by

`mg=kx\\\Rightarrow x=(mg)/(k)\\\Rightarrow x=(58.5* 9.81)/(55.88391)\\\Rightarrow x=10.26923\ m`

The unstreched length of the cord is 20.5-10.26923 = 10.23077 m