Question

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

Answer 1

Answer:

  15.33 m

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

$F= ke \\mg= ke$

solving for the extension e

$85*9.81= 250*e\\\e= \frac{833.85}{250} \\\e= 3.33m$

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

Answer 2

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

$\bold {F_s = -kx}$

Where,

= spring force

k = spring constant =  250 N/m

x = spring stretch or compression = ?

The force,

F = mg

$\bold {F = 85 \times 9.18}\\\\\bold {F = 833.85 }$

So,

$\bold {x = \dfrac {F_s }{k} }\\\\\bold {x = \dfrac {833.85} {250} }\\\\\bold {x = 3.33}$  

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,

https://brainly.com/question/10991960