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Kinetic Energy Question 
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#1
Dec1604, 02:52 AM

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After about 3 years of learning physics, I still can't understand how the kinetic energy formula works.
If the velocity of an object is doubled, why does the energy increase by a factor of four? Here's an example where this theory breaks down for me: I have a stationary object in space. To accelerate this object I have some sort of propulsion method which is powered by a battery. Lets say I accelerate it to 10m/s. This would require energy from the battery over a period of time for acceleration. Ok, so I drain the battery to half empty accelerating it. Now I want to double its velocity to 20m/s. The remaining half of the battery should be enough energy because the object is accelerating in the same way over the same period of time. So why is it true that I need 4x the energy to accomplish this? Is my example not possible? Am I missing something? 


#2
Dec1604, 03:51 AM

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You need 4 times the energy because you are applying a force over a distance that is 4 times greater. Remember, work (energy) is force times displacement (in the direction of the force).



#3
Dec1604, 05:07 AM

P: 6

But the energy output of the battery is no greater than it was before. It makes sense that the distance is 4 times greater, but the energy from the battery doesn't match the energy required mathematically.



#4
Dec1604, 05:59 AM

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Kinetic Energy Question
If the battery has energy [itex]E_B[/itex] available and it expends half its energy to get the speed of an object up to [itex]v_1[/itex] then
[tex]\frac {1}{2} E_B = \frac {1}{2}m v_1^2[/tex] and the remaining half of the battery's energy can be expended to accelerate the object to [itex]v_2[/itex] so that [tex]\frac {1}{2} E_B = \frac {1}{2} m v_2^2  \frac {1}{2} m v_1^2[/tex] It follows that [tex]m v_1^2 = \frac {1}{2} m v_2^2[/tex] so that [itex]v_2 = \sqrt {2} v_1[/itex] and the speed of the object does not double! 


#5
Dec1604, 06:03 AM

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P.S. You can do the same analysis with the battery using up 1/4 of its energy for the first leg of the journey and 3/4 of its energy for the second leg. You will find in that case that the speed doubles  i.e. it takes for times as much energy to accelerate to twice the speed.



#6
Dec1604, 06:04 AM

P: 6

Ahh.. Well Done.
Thanks. 


#7
Dec1604, 06:06 AM

P: 6

Would it not be 1/5 of its energy followed by 4/5?



#8
Dec1604, 06:09 AM

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No. Work it out following the example I gave above!



#9
Dec1604, 06:12 AM

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Oh that's right. I could use a little brushing up on my math skills.



#10
Dec1604, 12:33 PM

P: 27

Very intresting. Now I understand why Kinetic energy is refered to as force over distance and impulse is Force over time.



#11
Dec1604, 05:30 PM

P: 501

When you say mass x velocity squared are we talking about momentum?



#12
Dec1604, 09:49 PM

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DB,
Mass times velocity squared/2 is the kinetic energy while mass times velocity is momentum. delton, Work or change in energy is force times distance. 


#13
Dec1704, 07:00 AM

P: 501

Ahh, k now i see it, KE = 1/2 x M x V2 Joules and W or change in E = F x distance, Thnx



#14
Dec1704, 08:07 AM

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#15
Dec2004, 12:44 AM

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Space is a bad example for kinetic energy issues. The issue is work is related to the point of application of a force. In space the point of application is the engine itself.
In the case of a rocket in space, part of the rocket's mass is expelled in order to accelerate the remainder of the rocket. In the simple case where thrust remains constant, speed relative to the rocket's initial speed increases, but the force, thrust and fuel consumption rate remain constant. Jet engines also produce thrust, and part of the forwards force also is due to the fact that the point of application is relative to the engine itself, and not the air being pushed against by the thrust of the engine. Jet engines produce more power at higher speeds and altitudes. By comparason, a prop driven aircraft generates virtually all of its force by applying this force to the air. With a fixed power engine, for example an electric motor not affected by altitude, you trade off speed for force. At higher altitudes, there's less drag, so you gain some speed, because less force is needed. There's a ceiling though. This is because the lift produced is ultimately relative to the force from the prop, and as altitude increases, the lower limit for prop force remains constant, but the speeds increase with reduction in drag (and lift), and reach the limit of the power of the motor. Kinetic energy is relative to some frame of reference. Getting back to the case of the rocket, use the rocket as the frame of reference. If you don't like an accelerating frame of reference, put 2 rockets nose to nose with equal thrust so that neither accelerates. Work is being done on the mass that is expelled on the rocket engines. Trying to measure the force times distance is a bit tricky, but essentially, there's an ultimate velocity for the mass being expelled, and to make things easy, asssume a fixe rate for the amount of mass expelled per unit of time. If a rocket engine expelled 2 slugs (1 slug = 32.174 lbs) of mass per second with a terminal velocity of 2345.2 ft / sec, then the work (energy) would be: [tex] 1/2\ 2\ slug\ (2345.2 ft\ /\ sec)^2 [/tex] defining some constants here: [tex] 1\ slug = (1\ lb\ sec^2\ / ft)[/tex] and [tex]1\ hp = 550\ ft\ lb\ /\ sec[/tex] you get [tex] 1/2\ \ 2\ (lb\ sec^2\ / ft)\ 5,500,000\ ft^2\ /\ sec^2 = 5,500,000\ ft\ lb[/tex] This amount of engery is produced every second so the power is: [tex]5,500,000\ ft\ lb\ / \ sec = 10,000 hp[/tex] So the rocket engine is a fixed power engine if the frame of reference is the rocket itself, which is accelerating. But if a fixed frame of reference is used, then the power of the rocket increases over time as the thrust remains constant while the speed relative to the initial starting point increases. The thrust is the rate of mass expelled times it's terminal velocity: [tex](2\ slug\ /\ sec)\ 2345.2\ ft\ /\ sec) = (2\ (lb\ sec^2\ /\ ft))/\ sec)\ 2345.2\ ft\ /\ sec) = 4690.4\ lb [/tex] 


#16
Dec2004, 01:47 AM

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Getting back to the orignal post, as I mentioned space was a bad example.
Instead, assume a vehicle traveling on a road but in a vacuum (and no other drag). Here the point of application of force is to the road so a point on the road can be used as a frame of reference. Assume a constant 1 hp motor and no losses in transmitting this power, so you have 550 ft lbs / sec of power available. You can think of power as being related to force times speed. With the proper gearing, the motor will produce 55 lbs of force at 10 ft / sec. In order to go twice as fast, the gearing needs to be twice as tall. This reduces the force by 1/2, so you end up with 27.5 lbs of force at 20 ft / sec. 


#17
Dec2004, 09:50 AM

P: 3

This is a little OT, sorry.
Hypothetical situation. :) An asteroid massing 50 trillion metric tons in the main belt. I apply 10^19 joules of energy, a push so to speak. Via rocket or whatever. At a vector down the gravity well of the sun, towards earth. If I want to calculate the resulting velocity, do I use F=ma or K_e = 0.5 m v^2 ; or neither? TIA cs 


#18
Dec2004, 01:49 PM

P: 6

With the example of the car on the road then, the car does use more than twice the energy to double the speed, right? Not only does the motor have to do the exact same work all over again, but it also has to do it at twice the output rate since the gearing is twice as tall. With taller gearing, the engine has to work longer in order to achieve a desired velocity, due to the reduction in torque and therefore rate of acceleration. So it does the same work all over again, which right there is a doubling, and then it has to do it for twice as long which is a doubling again.
And for the hypothetical situation with the asteroid, you would use the KE formula because the energy is being transferred from one form to another. The F=ma equation is useful when you apply a force. With that you can then find acceleration to determine velocity through another equation. I hope that all makes sense. If not, I still enjoyed all the knowledge on planes! 


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