Question

A diver with mass 55kg jumps 0.8m into the air. She lands on a diving board that bends down a deflection of 0.4m. What is the spring constant for the diving board?

Answer 1

The diver falls from a position .8 m up to a position .4 meters down from the straight board
so
the potential energy lost by the diver was
m g h = 55 * 9.81 * 1.2 = 647 Joules
int a perfect world with no losses of energy that 647 Joules is stored in the fibers of our perfect diving board
(1/2) k x^2 = 647
k * 0.4^2 = 1294
k = 8093 Newtons/meter