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Kinetic energy

  1. Jul 18, 2015 #1
    Dear PF Forum,
    I'd like to ask a question regarding a post in Relativity sub forum.
    The formula for kinetic energy is ##E_k = 0.5mv^2##
    Which makes sense. Because if we push/accelerate 1 kg object for 1 m/s2 for 8 meters, we'll spend ##E = N.m = 8 joules##
    How much time do we need to do that? ##D = \frac{1}{2}at^2; t = \frac{2D}{a} = 4 seconds##
    The speed? ##v = a.t = 4m/s##
    How much energy do we need to directly stop that object? 8 Joules of course.
    ##E_k = 0.5mv^2 = 0.5(1Kg) * 4^2 = 8##.
    This all make sense to me.
    But consider this.
    A 1 ton rocket with matter/anti-matter fuel engine is traveling.
    The rocket accelerate at g ≈ 10m/s2. And keeps traveling that way.
    The rocket is picking it's fuel along the way, https://en.wikipedia.org/wiki/Interstellar_medium
    Either matter or anti matter. I know, we just can't find anti matter scatterd everywhere on the ground (or in interstellar medium for that matter, sorry for that anti-matter). Just supposed if we can find them, and for years the rocket has been accelerating and it has been annihilating, say, 100 tons of anti matter and 100 tons of matter.
    Here is the fact:
    1. Mass of the rocket: 1 ton
    2. Mass of the energy spent 200 tons ##E = mc^2 = 2E5 * (1E8)^2 = 2 * 10^{21} joules##
    How much energy do we need to directly stop this thing according to kinetic energy formula. Assuming the rocket travels near c?
    ##E_k = \frac{1}{2}mv^2 = 500 * 1E16 = 5E18 joules## taking away any heat loss, friction generated by interstellar medium.
    Energy to accelerate: 2E21 joules
    Energy to stop: 5E18 joules
    Did I mistakenly calculate?
    Further more, if the law of conservation energy is correct (which I know IT IS!) the energy to stop must also 2E21 joules.
    So the object is carrying energy bigger than ##E=mc^2##?
     
  2. jcsd
  3. Jul 18, 2015 #2
  4. Jul 18, 2015 #3
    Ahh, the rest mast vs moving mass. I must have forgotten, even if I don't know the formula, about that.
    Thanks for giving me this link. But I'm still reading time dilation and simultaneity of event.
     
  5. Jul 18, 2015 #4
    First thing we need to consider we can use only realtivistic equation when speed of object is near c ,which here its near c.

    Now before the rocket moves the total energy will be 1000 kg+2E5=Total mass, so ##E=mc^2## , ##E=201E3*(3E8)^2=1809E19##

    Lets calculate the force to stop it.The All fuel will be gone but the energy is conserved so that energy gone to speed how much speed.That energy is kinetic energy which its ##(γ-1)mc^2## and rest energy ##mc^2## so If we collect them we get ##γmc^2##. If you look carefully you will see that ##mc^2=1E3*(3E8)## and the other ##(γ-1)mc^2## is the exactly the energy of matter +antimatter energy.So total energy is conserved.
     
  6. Jul 18, 2015 #5
    I made the right solution you can look
     
    Last edited: Jul 18, 2015
  7. Jul 18, 2015 #6
    Yes, thanks. I forgot about the relativistic mass. But I'm not going to study it. :smile:, at least not now. Still struggling with time dilation and doppler. It's just that reading russ_watters post, something just hit me. "What about the kinetic energy?". It turns out that I forgot to take relativistic mass in the equation.
     
  8. Jul 18, 2015 #7
    Relativistic mass is not so hard.m'=mγ the equation is simple.
    Your first wrong idea is you didnt calculate rocket rest mass or you did not add it in calculations.
    Secons mistake is you used 1/2mv^2 to calculate kinetic energy which its also wrong. As I told you before If v is near c you had to use relativistic formulas.
     
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