Question

A bungee jumper of 68 kg is attached to a bungee cord with an unstretched length of 20 m. He jumps off a bridge and when he finally stops, the cord has a stretched length of 50.0 m. Treat the stuntman as a point mass, and disregard the weight of the bungee cord. Assuming the spring constant of the bungee cord is 82.6 N/m, what is the total potential energy relative to the water when the man stops falling if the distance between the man and the water is 6m?​

Answer 1

Below

Step-by-step explanation:

The man will have Gravitational POTENTIAL energy due to Earth's gravity of mgh

mgh = 68 kg * 9.81 m/s^2 * 6m = 4002 J

The bungee has ELASTIC potential energy of 1/2 k x^2

where x = stretch distance and k = spring constant

= 1/2 (82.6 N/m) ( 50-20 m)^2 = 37170 J

Total = 4002 + 37170 = 41172 J