In the dangerous "sport" of bungee-jumping, a daring student jumps from a balloon with a specially designed elastic cord attached to his waist. The unstretched length of the cord is 23.4 m, the student weighs 818 N, and the balloon is 31.3 m above the surface of a river below. Calculate the required force constant of the cord if the student is to stop safely 2.74 m above the river. Answer in units of N/m.

k = 1755 N/m

Explanation:

Given:

- The length of the cord L = 23.4 m

- Weight of the student W = 818 N

- The elevation of balloon H = 31.3 m

Find:

Calculate the required force constant of the cord if the student is to stop safely 2.74 m above the river.

Solution:

- We know the potential energy of the student changes by  

                      ΔP.E = m*g*( H - 2.74 )

                      mg*(31.3 - 2.74) = 818*28.56 = 23362.08 J

- When he stops at 2.74 m above ground his KE = 0 so ALL his lost potential energy must be stored in the extended or stretched bungee cord.

- He falls 23.4 m before the bungee cord starts to stretch. That means it doesn't start stretching until he is 31.3 - 23.4 = 7.9 m above the ground.  

- It has to stop  stretching at 2.74 m above the ground so the

                                total stretch = 7.9 - 2.74 = 5.16 m

- Therefore his PE from 31.3 m to 2.74 m is stored in a 5.16 m stretch of the bungee cord.                      

                                ½kx² =  ΔP.E

                                k = 2*ΔP.E / x^2

                                k = 2*23362.08 / 5.16^2

                                k = 1755 N/m

magnitude and direction

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