Question

A 70kg person jumps from a window into a fire net 20 m

below,which stretches the net 1.1m. Assume that the net behaveslike
a simple sprng, and calculate how much it would stretch if thesame
person was lying in it. How much would the net stretchif the person
jumped from 50m?

Answer 1

0.02867 m

1.722 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* 70* 9.81(20+1.1)/(1.1^2)\\\Rightarrow k=23949.3719\ N/m`

The spring constant of the net is 23949.3719 N

From Hooke's Law

`F=kx\\\Rightarrow x=(F)/(k)\\\Rightarrow x=(70* 9.81)/(23949.3719)\\\Rightarrow x=0.02867\ m`

The net would stretch 0.02867 m

If h = 50 m

From energy conservation

`70* 9.81* (50+x)=(1)/(2)23949.3719x^2\\\Rightarrow 11974.68595x^2=686.7(50+x)\\\Rightarrow 50+x=17.43801x^2\\\Rightarrow 17.43801x^2-x-50=0`

Solving the above equation we get

`x=(-\left(-1\right)+√(\left(-1\right)^2-4\cdot \:17.43801\left(-50\right)))/(2\cdot \:17.43801), (-\left(-1\right)-√(\left(-1\right)^2-4\cdot \:17.43801\left(-50\right)))/(2\cdot \:17.43801)\\\Rightarrow x=1.722, -1.6`

The compression of the net is 1.722 m