Posted by Shanelle on .
A diver (mass = 75.0 kg) walks to the end of a flexible diving board. The end of the board bends down a distance of 0.55 m. The diver's weight on the diving board makes it behave like a spring.
a) Determine the spring constant of the diving board.
b) Calculate the diver's maximum speed as the board springs back to its equilibrium position.
c) What is the diver's maximum height above the board's equilibrium position?
Please explain! :)
My teacher told me that I could use Hooke's law for a) (f=kx), Elastic potential energy (Ee=1/2kx^2) for b), and Ee=Eg (Eg=mg(change in height)) for c). I worked a) out to be 1337.72 by using F=mg to find force to use in the F=kx equation.
b) 1.04 m/s
c) o.o6 m