Question

A 53 kg bungee jumper jumps from a bridge. She is tied to a bungee cord whose unstretched length is 13 m, and falls a total of 35 m. Calculate the spring stiffness constant k of the bungee cord, assuming Hooke's law applies.

Answer 1

Let X be the deflection from the zero-stretch length (13).

When motion is considered, max spring potential energy (at max stretch) equals the gravitational potential energy loss at maximum extension.

(1/2)k Xmax^2 = M g *35 m

Xmax = 22 m

k = 2 M g *35m/(22)^2
= 75.1 Newton/meter

Answer 2

ghey

Answer 3

Its elastic potential energy