# Kinetic Energy vs Momentum vs Power

1. Apr 25, 2015

Imagine this object as a big tanker ship, moving to the right

weight 500 000 metric tones
linear velocity 0.01 m/s
momentum cca 5 000 000 kg m/s
kinetic energy 25 000 J, which is 25 kilowatt-seconds

how much power could i extract from this object in order to slow it down to complete stop ?

Obvious answer is amount if kinetic energy, but seems to be negligible compared to its momentum.

According to this, 100% efficient motor or piston or any device consuming 25 kw during 1 second should stop the object in 1 second, which is in pretty unreal

So what am i missing here? How do i calculate extractable power correctly from such straight moving object?

thanx

2. Apr 25, 2015

### Orodruin

Staff Emeritus
Power is energy/time, so this depends how long you want to extract the power for.

3. Apr 25, 2015

### Simon Bridge

... you can only compare like with like. The size of the number is only useful in connection with what it is the number of. Kinetic energy and momentum are properly related by $K=p^2/2m$ so it is not surprising to find the kinetic energy and the moementum of the supertanker should have such disparate-a set of digits.

... well the thing is only going at 0.04kmph ... it's barely moving. On the up-side, it only took 25kJ to accelerate it to 1cm/s in the first place.

You have done it correctly - 100% energy conversion at an average rate of K/T will stop the object with initial kinetic energy K in less time than T... you need only factor in the efficiency of the energy conversion.

4. Apr 25, 2015

### CWatters

but it is correct.

Have a look on youtube at some of the videos of ships beached for recycling. They are typically run aground at full speed and some stop in a second or two..

5. Apr 25, 2015

### Eddie Sines

Very impressive.. you would think these ships would go much further on to the shore... but the friction must be massive given their weight... thanks for sharing..-e

6. Apr 26, 2015

### Simon Bridge

OK - but what about the guiding questions in post #3?

7. Apr 26, 2015

what guiding questions you mean?

8. Apr 26, 2015

### Simon Bridge

Hmm - no questions huh? I think I was reading a different #3. Too many threads open at the same time.
You have responded to CWatters example to the effect that the friction must have been huge to stop the ship so quickly, rather than that the kinetic energy of the ship must have been so small-seeming as in your post #1. Or did I misread you?

Do you have a response to the observations in post #3 - i.e. comparing like with like, the correct relationship os kinetic energy and momentum giving rise to a big difference in the size of numbers that don't mean nothin, and how slow 1cm/s actually is?

Actually, for your example, you can work out the force in question to change the momentum of your ship to zero over 1 second...
The gripping hand is that you'll want to adjust your intuition ... that takes work (he says from experience).
Just tell yourself: you don't get energy from momentum, you get it from energy.
Yeah I know it sounds dumb...

9. Apr 26, 2015

yes i tried it, it takes force 5 000 000 Newton during 1 second to stop it, distance traveled 0.005 meter
W=Fxd

so work is 25000J , during 1 second thats Power 25kilowatt-seconds, so math is correct ,

just intuition cannot handle fact that you can stop half a billion kilograms of moving mass with 1 second burst of 25 kw

10. Apr 26, 2015

### Simon Bridge

Yeah - you tend to think of a slow moving huge mass as unstoppable don't you?
Either 25kW is not as small as you thought or half-a-billion kgs going slowly is not as hard to stop as you thought.
Certainly 5 million Newtons sounds like a big number - but it acts over a very short distance ... so that's why the energy feels small.

What may help you out though is that the force you just worked out goes both ways (equal and opposite reaction) so anything delivering that 25kW would have to be braced against 5MN force for 1 sec ... which is a decent thump. For instance - your car can probably put out more than 25kW, but if you try to use it to push the tanker? Well there's probably not enough traction...

The key is to figure what is behind your intuition.

recall: many people feel that force is needed to keep something moving - so we get a lot of questions about how much force for a particular speed (or torque for a particular rpm). This is due to their personal experience of getting exhausted pushing things fast.

Anyway - you'll get used to it.
Good luck.

11. Apr 26, 2015

my intuition is based on this source of comparable force

http://en.wikipedia.org/wiki/Rocketdyne_F-1

force: 6770 KN vs our 5000KN

power: 30-40 MW vs our 25KW

rocket engine thrust is independent of distance traveled, its just constant source of braking force in our case

if this engine will fire for 1 second and stop the tanker, that makes sense to my intution, but why such difference in power? order of magnitude

12. Apr 26, 2015

### Simon Bridge

For the 25kW engine to stop the ship in 1s it has to be used 100% efficiently. In general this will not be the case... some energy will go someplace else.
What you've actually worked out is that any device stopping the ship in 1s will have put 25kW into the job. (Better: the device has to remove 25kW from the ship.) It says nothing about how much energy went elsewhere.
If the ship were stopped by a big spring, that is how much energy would end up stored in the spring (provided the spring survived the impact) - if the ship rammed the dock, that is how much energy went into damaging the dock.

A rocket engine is extremely inefficient - especially at low speeds.
The propulsive efficiency of a rocket depends on the ratio of vehicle speed vs exhaust speed.
Here the vehicle speed is almost nothing while the exhaust speed is typically vast so the propulsive efficiency is very low.
i.e. an exhaust velocity is typically 1000's of meters per second, the speed here averages about 0.5cm/s so the efficiency should be of order 0.01%

For this example it is easier to work out because we know the energies:
Lets say a more modest 30MW rocket produces 5000kN thrust - that would be believable right?
If we used it to stop the tanker in 1s then it expands 30MJ to do 25kJ of work ... that efficiency is: 0.08%
... which is consistent with the order-of-magnitude calculated above.

Once more the number work out - but you have deepened your understanding of what the maths is telling you.
That help?

13. Apr 26, 2015

### Simon Bridge

You could probably work it out for using a single F-1A (Saturn V) rocket ... it's exhaust velocity was a bit under 3km/s according to fact sheet:
http://www.projectrho.com/rocket/supplement/dvNomogram01.pdf
... but boy that was tough to find.

How long would it actually take the F-1 to stop the tanker? How much energy did it use doing that? How efficient does that make it?
Compare with the rocket efficiency equation.

It's ages since I've done anything with rocket science.

14. Apr 27, 2015

I think rockets are pretty efficient and in their frame of reference dont care if rocket is moving realtive to something else, they produce its thrust constantly

however our definition of work and power mean that if rocket just hoovers in the air and fights gravity, its velicity is zero, therefore work and power is zero, so their efficiency is zero

so let find better example , how to achieve 5 000 000 N in real life

15. Apr 27, 2015

lets try to stop tanker with 100 efficient water pump and water jet so no energy is wasted on anything else
what force can we generate with 25 kw pump?

setup 1.

Force of jet on a plate = mass flow rate per second x water jet speed
= 86.11 kg/s x 24.26m/s = 2089 Newton.......using 100% efficient system

in order to get 5 000 000 Newton , water pump would have to be bigger x2400, again tens of megawatts

so what am i missing here again ?

16. Apr 27, 2015

### jbriggs444

You are putting a lot of energy into the spray that is splashing off of the front of the tanker. Assuming no energy losses in the splash, the spray carries away slightly more than 100% of the power that you are supplying with the pump.

17. Apr 28, 2015

### Simon Bridge

I think you have misunderstood.
Efficiency is work over enery in... so the number cepends on what sort of work you want to do. So when you say you think rockets are pretty efficient you also need to say at what. Here you are interested in the ability of a rocket to change the kinetic energy of a ship. Because of examples you have of rockets using lots of power to make a big force, you have gained an intuitive impression that big forces using lots of power make big energy changes. This intuition is incorrect bacause rockets are innefficient.

To press home this point, lets say that instead of bringing the ship to rest, you want to speed it up. The ship is already going as fast as you like. If you hooked the rocket in my simplfied example to the back, it would be able to add 25kJ to the kinetic energy, by expending 30MJ ... thats still very innefficient. If all the 30MJ went into change in kinetic energy, at constant acceleration, in 1sec, what would the force be?

...well, yes. And?

The rocket was your example as to how you ended up with the intuition you did. The innefficiency of the rocket... its not an efficient way of turning energy into force... is behind the faulty intuition. That does not change because you can think up some other engine. This is what you said was behind your intuition.

.. which also turns out to be inefficient. Im telling you, you intuition has been informed by real world examples of lossy processes. This is why you have the faulty impression you do. The reason it takes lots more energy than you normally calculate to change somethings kinetic energy is because you have to use an inefficient process to do it.

By the looks of your calculation, the rocket is more efficient than the ideal water pump... mind you, it seems you are pumping the water through a height of 30m for some reason.

But there is something else that is important here: you have to be willing to conceed that your intuition is wrong.
Understandably and reasonably wrong but still wrong. If you want to insist that you are somehow right and the definitions of work and power are somehow incorrect then I cannot help you.

Your intuition is wrong because you have failed to take into account, in your calculations, factors that have informed your intuition. Its about the question you ask. If you ask how much KE to remove from the tanker to get it to stop... thats 25kJ. If you ask how much energy it will cost you to get it to stop... thats a different answer.

18. Apr 29, 2015

ok i understand now why intuition misguides in this case so much

25kJ it is

30m water height thats some water pressure value which needs to be set , besides massflowrate, so i set to 30m, could be any number, but higher pressure and lower mass flow rate generates even less force for our purpose

19. Apr 30, 2015

### Simon Bridge

The differential head is the maximum height the pump can push the water.
But no worries - you have to watch for the times when your feel for things is taking stuff into account that you didn't think of.

20. May 2, 2015

### dean barry

foundation physics says average power (required) = energy change / time
so, if KE = 25,000 Joules
To stop this ship in say 10 seconds, would take:
25,000 / 10
= 2500 Watts