Question

A 65 kg person jumps from a window to a fire net 18 m below, which stretches the net 1.1 m. Assume the net behaves as a simple spring and

A) Calculate how much it would stretch if the same person was lying on it.
B) How much it would stretch if the person jumped from 35 m

Answer 1

0.03167 m

1.52 m

Step-by-step explanation:

x = Compression of net

h = Height of jump

g = Acceleration due to gravity = 9.81 m/s²

The potential energy and the kinetic energy of the system is conserved

`P_i=P_f+K_s\\\Rightarrow mgh_i=-mgx+(1)/(2)kx^2\\\Rightarrow k=2mg(h_i+x)/(x^2)\\\Rightarrow k=2* 65* 9.81(18+1.1)/(1.1^2)\\\Rightarrow k=20130.76\ N/m`

The spring constant of the net is 20130.76 N

From Hooke's Law

`F=kx\\\Rightarrow x=(F)/(k)\\\Rightarrow x=(65* 9.81)/(20130.76)\\\Rightarrow x=0.03167\ m`

The net would strech 0.03167 m

If h = 35 m

From energy conservation

`65* 9.81* (35+x)=(1)/(2)20130.76x^2\\\Rightarrow 10065.38x^2=637.65(35+x)\\\Rightarrow 35+x=15.785x^2\\\Rightarrow 15.785x^2-x-35=0\\\Rightarrow x^2-(200x)/(3157)-(1000)/(451)=0`

Solving the above equation we get

`x=\frac{-\left(-(200)/(3157)\right)+\sqrt{\left(-(200)/(3157)\right)^2-4\cdot \:1\left(-(1000)/(451)\right)}}{2\cdot \:1}, \frac{-\left(-(200)/(3157)\right)-\sqrt{\left(-(200)/(3157)\right)^2-4\cdot \:1\left(-(1000)/(451)\right)}}{2\cdot \:1}\\\Rightarrow x=1.52, -1.45`

The compression of the net is 1.52 m