- #1
Furby
- 21
- 0
Bungee Jump--Springs and Energy
A person bungee jumps off a bridge.
The person's mass is 90kg.
The height of the bridge above the water is 80m.
The bungee chord has a length of 25m.
The bungee chord's spring constant is 30 N/m.
What is the minimum distance the person will be from the surface of the water?
PEs=1/2*k*x2
PEg=m*g*h
I assigned the point of jump at the height of the bridge as y=0, thus as the jumper falls his PEg increases, becoming more negative. Kinetic energy is growing at the same rate until 25m, the chord's equilibrium point, where the chord will begin to stretch and exert a force on the jumper opposite to gravity.
Afterwards the kinetic energy begins to convert to PEs always growing in equal value to the constantly increasing PEg. Thus at the end, or 'bottom' of the person's jump, he will no longer have any KE, and thus:
mgh=1/2*kx2
h=25+x
Use some algebra and you get a quadratic:
15x2-882x-22050=0
x=77.715m
Which add the initial 25m and you get 102m, obviously exceeding the 80m height. The bungee jumper would then easily hit the water. I solved the problem another way assigning y=0 at the point of greatest KE (25m) and set PEg to be 0 at this point as well, and received the same answer.
My first assumption is that there was a typo and the problem is intended to have a 300 N/m spring constant, but I'm curious to see if I did anything wrong?
Homework Statement
A person bungee jumps off a bridge.
The person's mass is 90kg.
The height of the bridge above the water is 80m.
The bungee chord has a length of 25m.
The bungee chord's spring constant is 30 N/m.
What is the minimum distance the person will be from the surface of the water?
Homework Equations
PEs=1/2*k*x2
PEg=m*g*h
The Attempt at a Solution
I assigned the point of jump at the height of the bridge as y=0, thus as the jumper falls his PEg increases, becoming more negative. Kinetic energy is growing at the same rate until 25m, the chord's equilibrium point, where the chord will begin to stretch and exert a force on the jumper opposite to gravity.
Afterwards the kinetic energy begins to convert to PEs always growing in equal value to the constantly increasing PEg. Thus at the end, or 'bottom' of the person's jump, he will no longer have any KE, and thus:
mgh=1/2*kx2
h=25+x
Use some algebra and you get a quadratic:
15x2-882x-22050=0
x=77.715m
Which add the initial 25m and you get 102m, obviously exceeding the 80m height. The bungee jumper would then easily hit the water. I solved the problem another way assigning y=0 at the point of greatest KE (25m) and set PEg to be 0 at this point as well, and received the same answer.
My first assumption is that there was a typo and the problem is intended to have a 300 N/m spring constant, but I'm curious to see if I did anything wrong?