# Bungee jumper problem

• paulimerci
In summary, the bungee jumper starts with gravitational potential energy (GPE) due to their height above the ground. As they jump off the bridge, this energy is converted into kinetic energy and eventually into elastic potential energy as the bungee cord is stretched. Given the values of m = 55kg, h = 10m, k = 25 N/m, and a stretched distance of 15m, the total energy of the system can be calculated using the formula TE = 1/2 kx^2 + mgh. With the reference point for potential energy at the initial level, the total energy is 0. However, if a different reference is chosen, the total energy can be any
kuruman said:
. If we accept the purported answer of 8202 J as the mechanical energy, we can find the zero of the gravitational potential energy function
I think it is reasonably clear that it was the mechanical energy in "rest" position that is wanted, and not only is there no reason to suppose it is conserved, it clearly is not in bungee cords (or you would hit your head on the launch platform).

The bungee cord is characterized by a spring constant. To me this means that it obeys Hooke’s law and that the force it exerts is conservative. Ignoring air resistance, all the forces acting on the jumper are conservative and can be derived from potentials. I agree that a bungee cord that obeys Hooke’s is not practical, but this is what we are given here.

kuruman said:
The bungee cord is characterized by a spring constant. To me this means that it obeys Hooke’s law and that the force it exerts is conservative. Ignoring air resistance, all the forces acting on the jumper are conservative and can be derived from potentials. I agree that a bungee cord that obeys Hooke’s is not practical, but this is what we are given here.
That’s an interesting point about Hooke's law, but the corollary of work being conserved here is that the cord was already extended by 15m when the jumper jumped.
A more reasonable view of bungee cord behaviour is that there are two different spring constants, one for increasing stretch and a smaller one for decreasing. That merely adds another ambiguity to the list: which one are we given?

Last one:
Imagine the bungee jumper at top. They see no bungee cord after the attendant says ,”Ok...you are all hooked up and ready to go. Enjoy your jump!!””

haruspex said:
That’s an interesting point about Hooke's law, but the corollary of work being conserved here is that the cord was already extended by 15m when the jumper jumped.
A more reasonable view of bungee cord behaviour is that there are two different spring constants, one for increasing stretch and a smaller one for decreasing. That merely adds another ambiguity to the list: which one are we given?
It is interesting to see how different people prioritize the information in this and other ambiguous problems when trying to provide an interpretation that will possibly salvage them. I considered as most important the idea that only conservative forces act on the jumper and interpreted the statement "and comes to rest 10m above the water" to mean "instantaneously at rest". I justified my view because a spring constant implies a Hooke's law force that depends only on position, hence can be derived from a potential and is conservative.

You assign higher priority to the view that the jumper comes to rest and remains at rest after stopping. I confess to the misdeed of bending "and comes to rest 10 m above the water" to mean an instantaneous "rest" to salvage energy conservation. My question at this point is, can we have our cake and eat it too? Can we blend a Hooke's law spring with the jumper coming to a complete stop when the cord is extended by 15 m?

Your idea of two spring constants is a good start but with the second one being dissipative, i.e. depend on the velocity and not the position. We can model the jump as a vertical mass-spring system with a disembodied hand (second spring constant) providing an additional upward force such that the mass comes to rest at the equilibrium point, ##x_{\text{eq}}=\frac{mg}{k}##. Then ##W_{\text{el.}}+W_{\text{hand}}+W_{\text{grav.}}=0.## This can be used to find the work done by the dissipative force $$W_{\text{hand}}=\frac{1}{2}(25~\text{N/m})\times (15~\text{m})^2-(55~\text{kg})\times \left(9.8~\text{m/s}^2\right)\times (15~\text{m})=-5273~\text{J}.$$How this helps find the total energy is beyond me.

You can make a limitless number of assumptions. How would you know which one to select if you don't know the answer to the question? Based on just the text as is given and the values given, and with no other context, the only reasonable assumption is that the author did not think it true. If the behaviour of the cord were to be described by more than a simple spring constant, this would be given in any reasonably formulated problem. In introductory physics problems the bungee cords are just springs following Hooke's law. This were context would help.

I get the feeling trying to "make sense" of this blunder of problem isn't helping @paulimerci solve these types of problems? How about someone just propose a similar "sound problem" like this one instead, in which COE is undoubtedly meant to be involved( Introductory Physics). Its hard to tell with all the bum information surrounding this problem and the diagram posted in #17 by the OP what issues they would actually have solving this type of problem.

paulimerci and kuruman
The question "what is the energy" with no other qualification and conservative forces does not make sense no matter what numbers are given. There is no wrong answer to this. This may be the most important thing for the OP to understand from all this

erobz
kuruman said:
You assign higher priority to the view that the jumper comes to rest and remains at rest after stopping.
No, I saw it as quite ambiguous, at first, as to whether that is complete rest or instantaneous rest. But in post #34 I showed how to deduce it means instantaneous.
My objections to taking it as work conserving are what I wrote in post #38: it would follow that the instantaneous rest is at the launch position, so the cord was already extended 15m before the jump, which would be extremely unusual.

Since it seems clear from the answer that the author intended the solution in post #28, data such as the spring constant are deliberate red herrings. Given the general sloppiness, it would not be surprising that the need to specify which phase the constant applies to was overlooked.
kuruman said:
i.e. depend on the velocity and not the position
No, each depends only on extension, but which applies depends on whether it is in stretching phase or relaxing phase.

@paulimerci

You could try this one instead.

##m= 55~\rm{kg}## bungee jumpers' mass
##g = 9.8~\rm{\frac{m}{s^2}}##

The bungee jumper leaves a bridge of height ##H##, with a bungee cord of free length ##l_o ##. At the point where they instantaneously come to rest, they are ##63~\rm{m}## below the bridge deck, and the cord has experienced a ##110 \% ~\rm{el.}##

What is the ##k## value of the bungee cord (ideal linear spring)?

(I think I concocted an ok problem...If not...I'm sure we'll hear about it.)

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