Understanding the Bungee Jumper Problem: Energy Transformation and Calculations

[paulimerci]

Homework Statement

Find the total energy of a 55kg bungee jumper who jumps off a bridge and comes to rest 10m above the water while the bungee with a spring constant of 25N/m is stretched 15m.

Relevant Equations

Total energy = 1/2 kx^2 + mgh

There is a three-step energy transformation. Gravitational potential energy(GPE) is stored energy associated with an object's height above the ground. So here, the bungee jumper who is on a bridge 10m above the water has GPE. As the jumper jumps off the bridge, this stored energy is converted into K.E soon this energy will begin to be transformed into Elastic potential energy as the bungee cord is flexed to its maximum extent.

Given values are m = 55kg , h = 10m, $k = 25 \frac {N}{m}$, stretched distance = 15m, displacement (x) = stretched distance - unstretched distance = 15-10m
The total energy of the system, TE, is the sum of all the different energies present in the system:
TE = Elastic PE + mgh
$$ TE = \frac {1}{2} kx^2 + mgh$$
$$ TE = \frac {1}{2} 25 \cdot 5^2 + 55 \cdot 9.8 \cdot 10$$
$$ TE = 5702J$$
Have I done it right?

 

[erobz]

Homework Statement:: Find the total energy of a 55kg bungee jumper who jumps off a bridge and comes to rest 10m above the water while the bungee with a spring constant of 25N/m is stretched 15m.
Relevant Equations:: Total energy = 1/2 kx^2 + mgh

There is a three-step energy transformation. Gravitational potential energy(GPE) is stored energy associated with an object's height above the ground. So here, the bungee jumper who is on a bridge 10m above the water has GPE. As the jumper jumps off the bridge, this stored energy is converted into K.E soon this energy will begin to be transformed into Elastic potential energy as the bungee cord is flexed to its maximum extent.

Given values are m = 55kg , h = 10m, $k = 25 \frac {N}{m}$, stretched distance = 15m, displacement (x) = stretched distance - unstretched distance = 15-10m
The total energy of the system, TE, is the sum of all the different energies present in the system:
TE = Elastic PE + mgh
$$ TE = \frac {1}{2} kx^2 + mgh$$
$$ TE = \frac {1}{2} 25 \cdot 5^2 + 55 \cdot 9.8 \cdot 10$$
$$ TE = 5702J$$
Have I done it right?

Check your initial height $h$. Also, the problem states the spring was stretched $15~\rm{m}$.

 

[erobz]

Is it zero?

I think you are getting a bit mixed up. Set the PE datum to be zero where the jumper instantaneously comes to rest ( 10 m above the water). How far did they fall, how far did the spring stretch?

 

[erobz]

Now that I'm looking at it a bit closer, what is free(unstretched) length of the bungee cord?

 

[paulimerci]

I think you are getting a bit mixed up. Set the PE datum to be zero where the jumper instantaneously comes to rest ( 10 m above the water). How far did they fall, how far did the spring stretch?

The jumper falls 15 m above the water with the added spring stretch (10+5=15m)?

 

[nasu]

@paulimerci As you wrote it, the questions do not make much sense. The energy can be anything you want, depending on where you pick the reference. Why not take the reference for potential energy at the initial level so the total is zero (no KE, the elastic potential is zero)? As you have only conservative forces (gravitational and elastic) the total stays the same: zero. If you don't like zero, move the reference and you get whatever you want. The other details are irrelevant if the requirement is just to "find the total energy".

 

[kuruman]

In my opinion the statement of the problem is faulty because it does not specify the zero of energy which is arbitrary. Although it is safe to assume that the zero of kinetic energy is when the person is at rest relative to the ground and that the elastic potential energy is zero when the bungee cord is relaxed, where do we take the zero of gravitational potential energy? In other words, $y$ in $mgy$ is measured relative to what point?

To @paulimerci : Do you have a numerical answer to this problem?

Ah, @nasu beat me to it but I will post anyway.

 

[erobz]

The jumper falls 15 m above the water with the added spring stretch (10+5=15m)?

Draw a diagram and let's see what you are proposing.

 

[erobz]

Even though the problem statement turned out to be faulty. I still feel like you might be getting a bit tripped up on what distances are what.

 

[haruspex]

@paulimerci As you wrote it, the questions do not make much sense. The energy can be anything you want, depending on where you pick the reference. Why not take the reference for potential energy at the initial level so the total is zero (no KE, the elastic potential is zero)? As you have only conservative forces (gravitational and elastic) the total stays the same: zero. If you don't like zero, move the reference and you get whatever you want. The other details are irrelevant if the requirement is just to "find the total energy".

There is also ambiguity regarding the reference time. Is the required energy before the jump or at rest at the end?
And I cannot make sense of a 55kg mass reaching equilibrium with 25N/m cord stretched only 15m. What planet is this on?
@paulimerci, please check that you have quoted the question exactly.

 

[paulimerci]

The jumper falls 15 m above the water with the added spring stretch (10+5=15m)?

But why I have to add the length of the bungee cord and

In my opinion the statement of the problem is faulty because it does not specify the zero of energy which is arbitrary. Although it is safe to assume that the zero of kinetic energy is when the person is at rest relative to the ground and that the elastic potential energy is zero when the bungee cord is relaxed, where do we take the zero of gravitational potential energy? In other words, $y$ in $mgy$ is measured relative to what point?

To @paulimerci : Do you have a numerical answer to this problem?

Ah, @nasu beat me to it but I will post anyway.

Yes, its 8202.5J

 

[paulimerci]

There is also ambiguity regarding the reference time. Is the required energy before the jump or at rest at the end?
And I cannot make sense of a 55kg mass reaching equilibrium with 25N/m cord stretched only 15m. What planet is this on?
@paulimerci, please check that you have quoted the question exactly.

That's what the problem stated. I didn't make it on my own.

 

[nasu]

The two energies should be the same in absence of non-conservative forces. Time is not relevant. If there are non-conservative forces you can't do much anyway.

 

[erobz]

Yes, its 8202.5J

They took PE = 0 at water level(as far as I can tell).

 

[haruspex]

The two energies should be the same in absence of non-conservative forces. Time is not relevant. If there are non-conservative forces you can't do much anyway.

If the jumper comes to rest work is not conserved. Unless it means instantaneously at rest (for the second time)?

 

[haruspex]

They took PE = 0 at water level(as far as I can tell).

Since you have found an interpretation that matches the answer, would you please post it?

 

[paulimerci]

Draw a diagram and let's see what you are proposing.

 

[erobz]

As far as I can tell this is what was done.

Using the elevation where the jumper instantaneously comes to rest as PE = 0

They find the height $H$ of the bridge ( which is irrelevant) above PE = 0 to be given by:

$$mgH = \frac{1}{2}kx^2 = \frac{1}{2} \cdot 25 \cdot 15^2 = 2812.5~\rm{J}$$

Then they must be assuming from that point they have additional 10 m gravitational potential to the water.

$$ mg10 + \frac{1}{2} \cdot 25 \cdot 15^2 = 55 \cdot 9.81 \cdot 10 + \frac{1}{2}\cdot 25 \cdot 15^2 = 5395.5 ~\rm{J} + 2812.5~\rm{J} = 8208~\rm{J} $$

Not a perfect match?

Actually, it is a perfect match if you use 9.8 as the value for $g$

 

[paulimerci]

As far as I can tell this is what was done.

Using the elevation where the jumper instantaneously comes to rest as PE = 0

They find the height $H$ of the bridge ( which is irrelevant) above PE = 0 to be given by:

$$mgH = \frac{1}{2}kx^2 = \frac{1}{2} \cdot 25 \cdot 15^2 = 2812.5~\rm{J}$$

Then they must be assuming from that point they have additional 10 m gravitational potential to the water.

$$ mg10 + \frac{1}{2} \cdot 25 \cdot 15^2 = 55 \cdot 9.81 \cdot 10 + \frac{1}{2}\cdot 25 \cdot 15^2 = 5395.5 ~\rm{J} + 2812.5~\rm{J} = 8208~\rm{J} $$

Not a perfect match?View attachment 318962

Actually, it is a perfect match if you use 9.8 as the value for $g$

Thank you, I'm confused where $x$ is the displacement of the end of the spring from its equilibrium position and so shouldn’t we write as 15-10 = 5m for $x$ and why we are adding them?

 

[erobz]

Thank you, I'm confused where $x$ is the displacement of the end of the spring from its equilibrium position and so shouldn’t we write as 15-10 = 5m for $x$ and why we are adding them?

It says the jumper comes to rest 10 m above the water. That is the position where the spring is fully stretched.

 

[paulimerci]

It says the jumper comes to rest 10 m above the water. That is the position where the spring is fully stretched.

okay, so what's the stretched and unstretched length here?

 

[erobz]

okay, so what's the stretched and unstretched length here?

The unstretched length is irrelevant. But you could figure it out from finding $H$ in the diagram I provided.

 

[kuruman]

Since you have found an interpretation that matches the answer, would you please post it?

Here is how one gets the quoted number $$\frac{1}{2}(25~\text{N/m})\times (15~\text{m})^2+(55~\text{kg})\times \left(9.8~\text{m/s}^2\right)\times (10~\text{m})=8202~\text{J}$$The interpretation is that the cord starts exerting a force on the jumper immediately after the jump, the gravitational potential energy is measured from the level of the water and the bridge is 25.0 m above the water.

 

[erobz]

Here is how one gets the quoted number $$\frac{1}{2}(25~\text{N/m})\times (15~\text{m})^2+(55~\text{kg})\times \left(9.8~\text{m/s}^2\right)\times (10~\text{m})=8202~\text{J}$$The interpretation is that the cord starts exerting a force on the jumper immediately after the jump, the gravitational potential energy is measured from the level of the water and the bridge is 25.0 m above the water.

No free length of the bungee cord... This problem clearly has issues! I like yours better, but it's still not well thought out problem statement IMO.

 

[nasu]

With the values given for spring constant and "maximum" extension, the elastic force (375 N) is less than the jumper's weight ( about 550 N), isn't it? How can this be the maximum extension? Why not go lower? Maybe a typo in the spring constant?

 

[erobz]

With the values given for spring constant and "maximum" extension, the elastic force (375 N) is less than the jumper's weight ( about 550 N), isn't it? How can this be the maximum extension? Why not go lower? Maybe a typo in the spring constant?

Riddled with physical oddities.

 

[erobz]

Here is how one gets the quoted number $$\frac{1}{2}(25~\text{N/m})\times (15~\text{m})^2+(55~\text{kg})\times \left(9.8~\text{m/s}^2\right)\times (10~\text{m})=8202~\text{J}$$The interpretation is that the cord starts exerting a force on the jumper immediately after the jump, the gravitational potential energy is measured from the level of the water and the bridge is 25.0 m above the water.

I think you can analyze it as though it does have a free length.

Set PE = 0 at water(like you've done)

Let the height of the bridge above the water be $H$.

Then you can analyze it from the point where the spring starts to stretch to where they momentarily stop:

$$ mg ( 15 ~\rm{m} + 10 ~\rm{m} ) + mg ( H - (15 ~\rm{m} +10 ~\rm{m} )) = mg (10 ~\rm{m}) + \frac{1}{2} 25 \rm{ \frac{N}{m}} (15 \rm{m})^2 $$

$$ \implies mgH = mg (10 ~\rm{m}) + \frac{1}{2} 25 \rm{ \frac{N}{m}} (15 \rm{m})^2 $$

but it seems to be a very small free length...never mind...negative free length. SMH...

 

[PeroK]

But why I have to add the length of the bungee cord and
Yes, its 8202.5J

We often tell students off for simply putting the numbers they are given into the first equation that comes to mind. But, in this case:
$$m = 55kg, g = 9.8 m/s^2, h = 10m \ \Rightarrow \ PE_1 = mgh = 5390J$$$$k = 25 N/m, x = 15m \ \Rightarrow \ PE_2 = \frac 1 2 kx^2 = 2812.5J$$$$PE = PE_1 + PE_2 = 8202.5J$$And don't worry about having too many significant digits in the answer. Just plug and chug and everyone is happy.

 

[haruspex]

everyone is happy.

Not everyone.
With that mass and elastic constant the equilibrium extension (using 10m/s² for g) is 22m. If the jumper is at rest with an extension of 15m then she must have been down to at least 7m below equilibrium, 14m below the current rest position… 4m into the water.
The jumper is distinctly unhappy.

 

[erobz]

Not everyone.
With that mass and elastic constant the equilibrium extension (using 10m/s² for g) is 22m. If the jumper is at rest with an extension of 15m then she must have been down to at least 7m below equilibrium, 14m below the current rest position… 4m into the water.
The jumper is distinctly unhappy.

The inventor/manufacturer of the bungee cord is happy. They just appear out of thin air!

 

[paulimerci]

Here is how one gets the quoted number $$\frac{1}{2}(25~\text{N/m})\times (15~\text{m})^2+(55~\text{kg})\times \left(9.8~\text{m/s}^2\right)\times (10~\text{m})=8202~\text{J}$$The interpretation is that the cord starts exerting a force on the jumper immediately after the jump, the gravitational potential energy is measured from the level of the water and the bridge is 25.0 m above the water.

If 25.0m is the height of the bridge, why we don't plug in GPE (mgh)?

 

[paulimerci]

There are many interpretations that you discuss here I'm finding challenging to understand. Although I am aware that the question is flawed, I would greatly appreciate it if someone could elaborate.

 

[erobz]

There are many interpretations that you discuss here I'm finding challenging to understand. Although I am aware that the question is flawed, I would greatly appreciate it if someone could elaborate.

The interpretation that "makes sense" the bungee has no free length so that it starts applying the force immediately after the jump...unlike an actual bungee that would allow for some amount of free fall. The other issue is that the bungee has somehow stopped the jumper while applying much less force the jumper's weight. If you just hung the man (without them jumping off the bridge and gaining kinetic energy) the stretch of the spring would be larger than the quoted 15 m. Others are talking of work not being conserved to make sense of it... I think the question is just a blunder.

 

[haruspex]

If 25.0m is the height of the bridge, why we don't plug in GPE (mgh)?

As mentioned in posts #10 and #15, there are two ambiguities in the question:
- Does coming to rest mean at rest at equilibrium or just momentarily stationary?
- Is it asking for the energy at the beginning or at the "rest" position?

Since 15m extension with a 25N/m cord is not enough to hold a 55kg mass at equilibrium, we can deduce it means momentarily stationary. That means it is at the top or bottom of a bounce.

If work is conserved, the top of a bounce would be back at the jumper's starting height, so presumably the cord is not under tension there. And if at the bottom, the extension would be 22m. So we can deduce work is not conserved.
Therefore the starting height, and the total energy at that height, are unknowable.

It follows that the question must mean the energy at the momentarily stationary position.

 

[kuruman]

There are many interpretations that you discuss here I'm finding challenging to understand. Although I am aware that the question is flawed, I would greatly appreciate it if someone could elaborate.

Here is the problem with the problem as I see it.

At the top of the bridge the kinetic energy is zero and so is the elastic potential energy $U_{\text{el}}$. The mechanical energy is only in the form of gravitational potential energy. If we accept the purported answer of 8202 J as the mechanical energy, we can find the zero of the gravitational potential energy function $U_{\text{g}}(y)=mgy$. $$ME_i=U_{\text{g}}(y_i)\implies y_i =\frac{U_{\text{g}}(y_i)}{mg}=\frac{8202~\text{J}}{(55~\text{kg})\times(9.8~\text{m/s}^2)}=15~\text{m}.$$ Thus, the gravitational potential energy is zero 15 m below the initial position. Now the problem tells us that the tethered jumper stops instantaneously after dropping 15 m, i.e. she stops right where the potential energy is zero. Therefore at that point both the kinetic energy and gravitational potential energy are zero. Assuming mechanical energy conservation, this means that all the energy is in the elastic potential form. Hmm . . . let's see if it is. $$U_{\text{el}}=\frac{1}{2}(25~\text{N/m})\times (15~\text{m})^2=2812~\text{J}.$$ Nope. I don't think there is an interpretation that can fix this inconsistency while preserving both the given answer, the given quantities and conservation of mechanical energy.

 

[haruspex]

. 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).

 

[kuruman]

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.

 

[haruspex]

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?

 

[erobz]

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!!””

 

[kuruman]

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.

 

[nasu]

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.

 

[erobz]

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.

 

[nasu]

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

 

[haruspex]

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.

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.

 

[erobz]

@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.)