# A Different Spring/Work Question

1. Jun 7, 2010

### Mechanic

I've been struggling with the following concept for years now - maybe somebody here can help.
Imagine two identical rockets facing each other in deep space with their engines turned off with a spring between them. Turn their eingines on and they exert a force such that they compress the spring. Eventually the compressing force of the rockets exactly equals the restoring force of the spring - the compressing stops and the spring is kept in a compressed steady state by the force of the rockets.

Now, the act of compressing a spring requires work to be done...and energy consumed. The equation for the work done - FdotD or (1/2)kx(squared) - makes sense for the process of compressing the spring. Got it. But what about the energy required to keep the spring compressed? I can see the fuel guage of the rockets go lower over time while the compression is maintained so I know energy is being consumed. So, very specifically, is there anything in the literature describing the energy consumed in the act of MAINTAINING the steady state compression of the spring similar to the well known (1/2)kx(squared) description of the act of PRODUCING the compression? Please note the request for applicable references in the established literature. Thanks.

2. Jun 7, 2010

### Academic

Energy isnt required to keep a spring compressed, only force. If the spring is fully compressed, the energy spent by the rocket goes into the exhaust.

3. Jun 8, 2010

### Mechanic

OK – is there anything in the literature on the energy going into the exhaust? I’d really like to see that.

But…clearly energy is being consumed, and it is related to whether the spring is compressed or not. I can see the fuel gauge on the rockets move down over time – so I know energy is being consumed. If I throttle back on the rocket engines, thereby reducing the amount of energy consumed per unit time, the springs expand. The spring will remain compressed only when energy is being consumed. Isn’t there anything in the textbooks on this?

4. Jun 8, 2010

### GRDixon

I haven't come across this in any texts, possibly because it's a non-problem. Post #2 is essentially correct. While the spring is in the process of being compressed, part of the energy released by fuel combustion goes into compressing the spring, and part goes into accelerating the exhaust gas. Once the spring is fully compressed, all of the energy released by fuel combustion goes into accelerating the exhaust gas. If you throttle back, then the spring will expand and the released potential energy will be manifest as additional exhaust gas kinetic energy, until the spring settles to its new compressed state.

5. Jun 8, 2010

### vlado_skopsko

Since there is no movement there is no energy transfer between the spring and the rockets, only force is being applied on the spring to keep it compressed. And the energy is coming through the exhaust in the form of heat.
Here on that system ( rockets- spring ) you don't have conversation of energy because heat energy is leaking out of it.
To use the conversation of energy you have to see the bigger picture and watch the rockets spring and the space as a system and take into account the heat that is produced.
Those rockets are using fuel to make heat like your barbecue is, without going anywhere.

6. Jun 8, 2010

### dulrich

As other posters have stated, the spring is not moving so no work is being done on it. Clearly energy is being transferred, so energy must be going into the exhaust. When the rockets are lit, the source of the force is the momentum transferred from the exhaust (F = dp/dt).

Imagine instead of the rockets being lit you were to latch them together somehow. Then the latch is providing the force to keep the spring compressed. Ultimately this force is coming from the attractive forces at the atomic level in the latch. But the point is that the latch is providing a counter-balancing force to keep the spring compressed -- and does no work.

I think this comes down to a common misunderstanding of work. Work requires a displacement. Holding a block stationary in the air with your arms requires force, but no work.

This is like looking at a tree and asking how can it hold up its branches all day? Think of the amount of effort required to hold up your arms all day. But this is the difference between the physical definition of work (force times displacement which is related to the transfer of energy) and physiological "effort".

7. Jun 8, 2010

### Mechanic

(And now you will see where I really struggle with this.)

Yes - the rockets produce heat, and exhaust. However, looking at heat alone first, the amount of heat given off is related to how compressed the spring is - and this is the realation I'm looking for a definition of. Picking arbitrary numbers our of the air, it would be something along these lines...
At steady state, ie., after the changes due to throttle changes have settled out:
1. With the throttle set at 10 we see 100 watts of energy consumption giving rise to 2 meters of spring compression
2. With the throttle set at 20 we see 200 watts of energy consumption giving rise to 4 meters of spring compression
3, With the throttle set at 30 we see 300 watts of energy consumption giving rise to 6 meters of spring compression

Do you see the (fictional) relationship between the energy consumption and the amount of compression??

Also, please note that I understand very clearly that the definition work requires a displacement...yet I cannot reconcile the above, which is the root of this struggle.

8. Jun 8, 2010

### dulrich

I'll take a shot:

Suppose each rocket is 1000 kg and the exhaust is spewing out 1 kg per second at 100 m/s. (Obviously made up numbers -- forgive my ignorance).

The momentum in the exhaust is

$p = mv = (1 \text{ kg})(100 \text{ m/s}) = 100 \text{ kg-m/s}$

each second. That translates to a force of 100 N. Since the rocket is not actually moving its mass is irrelevant and that force is transferred to the spring. This causes a compression of -- whatever. Say the spring constant is 1000 N/m, so you get 10 cm of compression.

Now, the kinetic energy carried by the exhaust (each second) is

$KE = \frac{1}{2}mv^2 = (0.5)(1 \text{ kg})(100 \text{ m/s})^2 = 5000 \text{ J}$

So, the exhaust requires 5000 watts of power to create the 10 cm of compression.

Double-check me, but I think that's right.

9. Jun 8, 2010

### Academic

It seems to me, that this is simply how a spaceship was designed. It consumes energy to exert a force. You could design a different apparatus that would exert force on the spring without consuming energy, like the latch mentioned.

Maybe its like my table at home. It uses legs to support it, because that is the easiest way to support it. But it could be supported by some rockets pushing it up, balancing against the force of gravity.

You could in principle design a number of things that consume energy to exert a force.

Im not sure I answered your question, because Im still not very clear on what your confusion is.

10. Jun 8, 2010

### Mechanic

My confusion arises from the fact that I have a very well known equation for the energy required to perform the act of compressing a spring but no such well known equation for the energy required to keep the spring in a steady compressed state. I think dulrich above has a reasonable approach, but my real question is - Are there any references in the established literature on this question in general? In this case rockets were used to maintain the compression, but is there anything in the literature that says "For a spring with spring constant = 1000N/m, 5000Watts are required to maintained 10 cm of compression when no latching mechanism is used." It may seem like an obscure question, but it is of very special interest to me.
Thanks

11. Jun 8, 2010

### Academic

Thats because you dont need energy to keep a spring compressed. Like I said (and you agreed) energy is not needed to keep a spring compressed, only force is. The fact that you can contrive a situation where energy is consumed does not change this basic principle.

For example, in general energy is not needed to keep an object at rest. But I can contrive a situation where energy is needed. Consider a car on a icy hill... Its going to slide down, so I spin the tires at the right speed to keep me from sliding down. Energy is being consumed to keep me at rest. That does not change the principle that in general energy need not be consumed to stay at rest.

What the car needs to keep from sliding down is the same as what the rocket needs to keep from being pushed - force. It just happens that in these cases we are considering devices that consume energy to produce a force. But this is just a special case, there are also devices that do not consume energy and still apply a force. How much energy you need to produce a force is entirely dependent on the engineering of the device, its not a basic principle.

12. Jun 8, 2010

### Mechanic

Yes...very good points. I had been very deliberately avoiding the case where a latch can keep the spring in a compressed state...for brevity. But I abosolutely acknowledge that a latch can keep the spring compressed without the consumption of energy. And a table can keep a mass in place similarly. But (even if I didn't accurately state my desire) I'm looking for a generalized equation for the cases without any rigid restraints. Like for the car on the icy hill, and the rockets and the spring. In fact the car on the icy hill is a perfect example...is there anything in the literature on the energy required to keep the car stationary on the icy hill?? I do appreciate the insights provided.

13. Jun 9, 2010

### Academic

You would need to look up engineering and industry values for things like the coefficient of friction for the tires and ice, at the particular temperature etc. It would all be very specific to the situation and device you are considering. You could sum up some forces, see what the engine needs to provide and look at engine specs. If you want to know how your rocket would perform, then you need to design a rocket. If you want to know how one could perform hypothetically, then realize that you can hypothetically configure it so that it consumes any amount of energy you want - from zero joules to all the joules in the universe. You can hypothetically configure a number of devices that use energy with no net work gained; like running a heater and air conditioner at the same time. :) Same effect as not running it at all. With a combination of space ships and latches you could consume as much or as little power as you want to keep the spring compressed.

14. Jun 10, 2010

### Mechanic

I think I get your point…and I can see that two identical rockets facing each other with the spring entirely compressed could then be throttled up to consume arbitrary mounts of energy. But not so while the spring is not completely compressed. Speaking exclusively to the example of the spaceships with no latches…Different amounts of compression are associated with different, specific rates of energy consumption. As I laid out in my Jun8-10, 12:21 PM post, different rates of energy consumption result in different amounts of spring compression. And, I believe, any given amount of compression would require a single, specific energy consumption rate – and that rate would be the same regardless of the engineering solution used to produce the compression. That is, if the rockets consume 10 Watts to maintain compression of the spring by 3 meters then any other method – whether biologic, or engine driven pistons, or what ever - will also require 10 Watts to maintain a compression of 3 meters. That is the generality I’m looking for – somewhere in the literature. Thanks.

15. Jun 10, 2010

### rcgldr

You could replace the rockets with very large permanent magnets oriented so they compress the spring. Once stabilized, the magnets are generating a compressive force but not consuming energy.

16. Jun 10, 2010

### Mechanic

I would consider use of the magnets to be equivalent to using a latch – which I am avoiding as I’ve discussed above. The magnets would have to latched together or somehow held in place or else they would push themselves away from the rocket/spring system. And we are in deep space with no way to hold them in place – other than latching them together.

17. Jun 10, 2010

### jack action

Look at it this way:

Remove the spring and face the two rockets, nose-to-nose. Turn their engines on such that they don't move. How much power is needed to do so?

Answer: Power is irrelevant. Even the force is irrelevant, as long as the force's magnitude for each rocket is equal and opposite to the other.

Now assume that the rockets are of different design. One rocket has a thrust of 1000 N, then the other needs also 1000 N of thrust. But one produces exhaust gas with a velocity of, say, 100 m/s and the other at 200 m/s. The first rocket will have a power of 100 kW and the second one a power of 200 kW. But because their thrusts are equal and opposite, they won't move. The excess energy of the rockets will be transfered to the surroundings (i.e. the exhaust gas).

Put the spring back into the equation and you have the same result. The only difference is that the spring will vary in length until an equilibrium is reached, but from there it's the same concept. Not only the rockets' power can vary from near zero to infinity, but they don't even have to be the same, as long as their thrusts are.

Read http://en.wikipedia.org/wiki/Thrust#Thrust_to_power" for more info.

Last edited by a moderator: Apr 25, 2017
18. Jun 11, 2010

### Mechanic

In the case with the rockets nose-to-nose with no spring between them, and in the case where the spring is completely compressed, I basically agree. However, I’m focused solely on the case where the spring IS present and not fully compressed. In that case I see a direct correlation between the rate of energy consumption and amount of steady state spring compression as I laid out in my Jun8-10, 12:21 PM post. And I’m looking for a description in the accepted literature that generalizes that correlation. I’m beginning to believe that no such description exists – which surprises me.

19. Jun 11, 2010

### jack action

If you agree with the rocket nose-to-nose, you must agree that it is the same thing with a spring between the rockets. Put a solid bar between the rockets and you will agree that it is the same thing as the case where the rockets are nose-to-nose. But solid bars don't exist in real life: They will all go through some deformations under a given amount of force, thus they can all be modeled as a very stiff spring. Go with a softer spring, everything stays the same.

What is the difference between 2 rockets nose-to-nose with each having a thrust of 1000 N and the case where the same rockets, with the same thrust, but with a spring of 100 N/cm between them which will be compressed 10 cm? Or even the case where instead of a spring you have a solid bar which would compress 0.001 mm under that same thrust of a 1000 N?

The only difference would be the distance between the rocket. Force wise, everything is the same.

As for your Jun8-10, 12:21 PM post, it is not the correct way to look at it. The correct way would be:

1. With the throttle set at 10 we see 100 N of force giving rise to 2 meters of spring compression
2. With the throttle set at 20 we see 200 N of force giving rise to 4 meters of spring compression
3, With the throttle set at 30 we see 300 N of force giving rise to 6 meters of spring compression

Let's come back on Earth with the following example:

There are two tractors, one with an engine of 10 kW (which is geared such that it develops 1000 N of tractive force at a possible forward speed of 10 m/s) and the other with an engine of 100 kW (which is geared such that it develops 1000 N of tractive force at a possible forward speed of 100 m/s). If the tractor is chained to a spring 1000 N/m, which is itself chained to a wall, which tractor will you need to elongate the spring 1 m?

Answer: Either one will do the job. Why? Because they can both pull a 1000 N; no less, no more. All the energy will be dissipated in heat and throwing dirt with the spinning tires. The more powerful tractor will just do a bigger mess.

20. Jun 11, 2010

### Mechanic

Some how I believe we are talking past each other…Let me ask some Yes or No questions referring to my Jun8-10, 12:21 PM post:
1. I believe that the fuel gauges of the rockets will fall more rapidly when the throttles are set at 30 than when the throttles are set at 10. Is this true?
2. I believe that the rate at which the fuel gauges move down indicates the rate at which energy is consumed. Is this true?
3. I believe the steady state spring compression will be greater with the throttles set at 30 than when they are set at 10. Is this true?
4. Based on the above I believe that the amount of steady state compression is related to the rate of energy consumption. Is this true?
Thanks

21. Jun 11, 2010

### jack action

Yes.

Yes.

Yes.

No.

What you are not taking into account is the energy spent to accelerate the exhaust gases and the heat that is wasted. Both of these will also increase with throttle opening. These outputs are the ones related to energy consumption. Opening the throttle will increase the force, but it will also increase the velocity of the exhaust gas. Power is equal to force times velocity. If velocity = 0 (for the rockets and springs), power = 0. But that is not the case for the exhaust gases: As the power will increase, the exhaust gas velocity will increase too. This is why I wrote in a previous post, that both rockets could be of different design and have different power output and still be in equilibrium (as long as both their thrusts are equal).

From http://en.wikipedia.org/wiki/Thrust#Thrust_to_power":

Say the throttle of the airplane in the drawing is at a certain level. Then double the throttle opening. For sure fuel consumption will be greater. For sure the force on the cable will be greater. For sure the airplane won't move. So where does the extra energy goes? I can assure you that the exhaut nozzle of the jet engine will become a much more effective hair dryer! The tree will loose his leaves more easily due to the air speed increase and it will burn up more easily from the extra heat.

Last edited by a moderator: Apr 25, 2017
22. Jun 12, 2010

### Mechanic

Nope...the plane WILL move. The cable (modeled as a very stiff spring) will stretch a little more due to the increased throttle and the plane will move to the new location of the end of the cable. Steady state will be achieved and the plane will then remain at the new location. Triple the throttle opening and the cable will stretch even more, and the plane will move to the end ever more streched cable. And...the hair drier will dry hair more rapidly!

23. Jun 12, 2010

### jack action

Again, it is irrelevant that the cable is a spring or not. If the airplane has an engine that produces 1000 N with a velocity of 100 m/s for the exhaust gas, it will burn half the fuel compare to the same airplane with an engine that produces 1000 N but with a velocity of 200 m/s for the exhaust gas. Still the spring would have the exact amount of stretch because it would be under 1000 N of force in both cases. Power is irrelevant once there is no displacement.

Let's go back to basics. How much energy does it take to compress a spring?

$$E=\frac{1}{2}kx^{2}$$

How much power does it take to compress a spring in a time t?

$$P=\frac{E}{t}=\frac{\frac{1}{2}kx^{2}}{t}$$

See, power is related to the amount of time you take to compress the spring. Once it is compressed, there is no more energy transfer, hence no more power either. the more basic definition of power is:

$$P=\frac{dE}{dt}$$

Which means "the power is proportional to the variation of energy divided by the variation of time". No variation of energy, no power needed.

The amount of power you will put to maintain a force necessary for keeping the spring compression is only dependent of the machine's ability and efficiency to produce that force, whether the machine is a jet engine, a car or the human body. The most efficient "machine" will be the latch, as it requires no energy transfer to maintain the force needed.

Another example: Take a vehicle that is driving along the highway. You calculate the drag force times the vehicle speed and you get the power needed to maintain the vehicle motion. This is the minimum power that needs to be produced to keep the motion. If the vehicle has an electric motor it will produce just slightly more power than the minimum required (a little heat loss due to bearing friction, for example). If the car has an IC engine, the power produced will be a lot greater (Just think of the heat loss through the cooling and exhaust systems; plus the friction losses will be a lot greater due to the need for gear ratios to match the engine rpm to the wheel rpm). Note that when we say "losses", it is "lost" in the sense that it doesn't go where we want, but it still creates an effect, either move or heat something else. We often recuperate those losses for other means like using the cooling system of the IC engine to wam up the passenger compartment, something that an electric motor cannot do (more power will have to be created some other way).

Going back to the rockets-spring arrangement, what is the minimum power required to keep that motion? Answer: 0 W, since there is no motion. How much power will produce a machine to maintain the necessary force? It depends on the machine's ability to create that force.

24. Jun 12, 2010

### Mechanic

But I'm not talking about two different plane engines with the same thrust. I'm talking about one plane engine operating at different thrusts. And while operating at different thrusts, it will consume different amounts of energy, and extend the cable by different amounts. Thanks for all the inputs.

25. Jun 12, 2010

### Tommo1

This problem is a combination of two things.
1. An explosion: exhaust gases being pushed in opposite directions by both rockets. This is where the energy is going. This follows all the traditional momentum and force stuff. The gases are pushed and moving so work has been done.
2. The spring in the middle transmits the forces to each rocket and is essentially static and doing no work after it reaches equilibrium. F=-Kx for both sides.

Also, some of the energy is converted to heat.

Is this not all traditional physics in most text books.

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