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.

Respuesta :

Answer:

10.23077 m

55.88391 N/m

Explanation:

m = Mass of person = 58.5 kg

T = Time period = [tex]\frac{45}{7}=6.42857\ s[/tex]

k = Spring constant

g = Acceleration due to gravity = 9.81 m/s²

Time period is given by

[tex]T=2\pi\sqrt{\frac{m}{k}}\\\Rightarrow T^2=4\pi^2\frac{m}{k}\\\Rightarrow k=\frac{4\pi^2m}{T^2}\\\Rightarrow k=\frac{4\pi^2 58.5}{6.42857^2}\\\Rightarrow k=55.88391\ N/m[/tex]

The spring stiffness constant is 55.88391 N/m

The force on spring is given by

[tex]mg=kx\\\Rightarrow x=\frac{mg}{k}\\\Rightarrow x=\frac{58.5\times 9.81}{55.88391}\\\Rightarrow x=10.26923\ m[/tex]

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