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
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.
3 answers
ghey
Its elastic potential energy