A 85 kg student stands on a bridge with a 12-m-long bungee cord tied to her feet. You can assume that the bungee cord is massless and has a spring constant of 250 N/m. The student jumps off the bridge and falls until the bungee cord is fully stretched, where she comes to a stop. How far below the bridge does the student fall before coming to a stop

The student fall  below the bridge(12+3.33 = 15.33 m ) before coming to a stop. Where 12 is the length of bungee cord.

According to Hooke's law in the elastic limit of a material the the strain in the spring is directly proportional to the applied force or stress.  

Formula

Where,

=spring force

k=spring constant =  250 N/m

x=spring stretch or compression = ?

The force,

F = mg

So,

Therefore, the student fall  below the bridge(12+3.33 = 15.33 m ) before coming to a stop. Where 12 is the length of bungee cord.

To know more about Hook's Law,

link

Explanation:

According to Hooke's law "provided the elastic limit of a material is not exceeded the extension e is directly proportional to the applied force"

Given data

mass m= 85 kg

spring constant k= 250 N/m

assuming g= 9.81 m/s^2

extension e= ?

We know that

solving for the extension e

to get how far bellow the bridge the student falls before coming to stop, we have to add the extension e to the length of the bungee cord.

=12+3.33= 15.33 m

does anyone know or what?