Question

A 75 kg person jumps from a window to a fire net 21 m below, which stretches the net 1.1 m . Assume that the net behaves like a simple spring

A. Calculate how much it would stretch if the same person were lying in it.

B. How much would it stretch if the person jumped from 33 m ?

Answer 1

first find the spring constant by using conservation of energy
mgh=1/2kx^2
m=75kg
g=9.81 N/kg
h=21m
x=1.1m

k = 2mgh/x^2 = 2*75*9.81*21/1.21 = 25538.429 N/m

if the same person were lying in it

mg = kx

=> x = mg/k = 0.02881 m it would stretch if the same person were lying in it

B)

mgh=1/2kx^2

=> x = sqrt(2mgh/k) = sqrt(2*75*9.81*33/25538.429) = 1.3789 m