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Query about kinetic energy equation e = ½mv^2

  1. Feb 19, 2014 #1
    Hi

    I don't understand why the relationship between kinetic energy and velocity is not linear. Consider the following situation:

    1. A rocket is traveling through space at constant velocity with it's thrusters switched off.

    2. The thrusters are fired at full blast for precisely one second and then stop.

    3. Again, the thrusters are fired at full blast for precisely one second and then stop.

    If we compare the two thrustings, surely the energy expenditure and increase in velocity will be the same in both thrustings, hence a linear relationship? What am I missing in my understanding of this situation?

    I would be very grateful to anyone who can answer this.

    Thank you very much.
     
  2. jcsd
  3. Feb 19, 2014 #2

    Dale

    Staff: Mentor

    Because we already have a different word for something with a linear relationship with velocity: momentum

    It is not helpful to use two words to describe the same relationship.

    You are missing the energy in the exhaust. The same amount of chemical energy* is converted to KE in each case. In one case more of the KE goes to the rocket and less of the KE goes to the exhaust than in the other case.

    Generally, momentum concepts are much more useful for rockets than energy. The momentum concepts force you to consider the exhaust.

    *speaking non-relativistically
     
  4. Feb 19, 2014 #3
    Hi DaleSpam

    Thank you very much for your reply.

    I think that what you seem to be saying is that we must take into account that the mass of the rocket fuel is being converted into the mass of the exhaust. Am I correct?
     
  5. Feb 19, 2014 #4

    stevendaryl

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    That's a very good question. The way to answer it is to think about what you're actually doing when you turn on thrusters. What you're doing is throwing some mass toward the rear, which pushes you forward in reaction. So in accounting for energy in an accelerating rocket, you can't simply look at the velocity of the rocket; you have to also take into account the energy put into whatever mass you throw out the back.

    Let's make the situation discrete by assuming that you're sitting on a railroad car, with a pile of baseballs. Once per second, you throw a baseball at 100 meters per second (relative to your speed). For simplicity, ignore energy loss due to friction, or due to heating up you, the car, or the baseball. Then most of the energy put into throwing the baseball will go into the kinetic energy of the baseball, not the kinetic energy of the railroad car. So as you continue to throw baseballs, the velocity of the train will increase approximately the same amount each time. But the kinetic energy of the baseballs (relative to the rest frame of the tracks) will slowly decrease. That's because each baseball is traveling at 100 meters per second relative to the train, so when the train is going pretty fast, the speed of the baseball relative to the tracks will be substantially less than 100 meters per second.
     
  6. Feb 19, 2014 #5

    stevendaryl

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    I see that Dale gave the answer about the exhaust while I was still writing.
     
  7. Feb 19, 2014 #6

    Dale

    Staff: Mentor

    Some mass of fuel is being converted into mass of exhaust, but more importantly, the exhaust itself has KE. Suppose that you burn 10 MJ of fuel and find that the rocket's KE increased by 1 MJ then you would also find that the exhaust's KE increased by 9 MJ. Now that the rocket is going faster if you burn the same 10 MJ of fuel then you may find that the rocket's KE increased by 2 MJ and then you would also find that the exhaust's KE increased by 8 MJ.
     
  8. Feb 19, 2014 #7

    Here is one profound formula: E = F * s

    Try to analyze the situation using that formula.
     
  9. Feb 20, 2014 #8
    Thank you Jartsa

    To be honest, I am not sure that I know how to apply the work equation in the frictionless environment of space. If you drag a box 1 Meter across the ground and you are applying 1 Newton in order to do so then it makes sense that you have done 1 Joule of work. However in space, the distance keeps increasing under momentum without applying any force at all. Do you know how the work equation applies to space?
     
  10. Feb 20, 2014 #9

    Nugatory

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    Staff: Mentor

    In outer space or on a near-frictionless surface like an air-hockey table, if the speed is constant (that is, there is no acceleration), there's no force involved, ##F=ma## holds because both sides of the equation are equal to zero, and no work being done because ##F## is zero in ##W=Fd##. If you do apply a force, ##F=ma## will continue to hold, the object will accelerate, and work will be done according to ##W=Fd##.

    The reason work is being done when you drag a a box one meter across the groundat a constant is that you're fighting friction. If the object is moving and you stop applying force, it won't coast, it will stop. Conversely, if you apply force to it, but not not enough to overcome friction and make it move, you'll calculate the work you're doing from ##W=Fd## with ##d## equal to zero, s zero work is being done - which is to say that you can lean against an immobile object all day without doing any work at all.
     
  11. Feb 20, 2014 #10
    Thank you Nugatory

    How do you measure ##d## properly in space for the purposes of the work equation? If you apply a specific force to an object in space then it will move the object further than on the ground but surely that does not mean that more work has been done?
     
  12. Feb 20, 2014 #11

    Dale

    Staff: Mentor

    I find it easier to use this form:
    $$W=\int \mathbf{F}\cdot d\mathbf{x} =\int \mathbf{F}\cdot \frac{d\mathbf{x}}{dt} dt = \int \mathbf{F}\cdot \mathbf{v} \; dt$$
     
    Last edited: Feb 20, 2014
  13. Feb 20, 2014 #12

    E =F *d

    where F is the pushing force, d is the distance that the pushing force pushes, in other words how much the pushed object travels during the time that the pushing operation takes, and that time is the time during which there exists a pushing force.

    d depends on velocity v
    large v --> large d --> large E



    Some additional deliberations:

    Here's a definition I copied from the net:
    The coefficient of performance or COP of a heat pump is a ratio of heating or cooling provided to electrical energy consumed.

    Here's a definition I made up:
    The coefficient of performance or COP of an ion thruster is a ratio of increase of kinetic energy provided to electrical energy consumed.

    When speed of ion drive is lower than the speed of the exhaust, the COP is less than 100%.
    When speed of ion drive equals the speed of exhaust, the COP is 100%.
    When speed of ion drive exceeds the speed of exhaust, the COP is larger than 100%.
     
    Last edited: Feb 20, 2014
  14. Feb 20, 2014 #13
    Yes, it does mean more work has been done.
     
  15. Feb 21, 2014 #14
    Well work = f dx and w=change energy. W = ma dx, w = m dv/dt dx, w = m integral v dv, w= 1/2mv^2 the mathematics don't lie


    Sent from my iPhone using Physics Forums
     
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