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Ive got some Joules.

  1. Aug 14, 2013 #1
    To those who could help me out,

    I'm pretty rusty with physics I thought maybe one of you energy specialists or physics prodigies could give me an estimate at what I'm looking at.

    ~50million jouhls
    could be used for what kind of push or power would I get from this energy, would it be enough to move an heavy object or even give it enough power to move from a no motion, how much would I need to move 3 ton object from standstill to six feet away or would the jouhls burn out and not get the object an inch.

    Would I get a better result by spreading out the jouhls around one side of my object or focusing the energy in the middle or mid low mid high separation? Whats the best way to get the most punch and power in my placement of this energy is my question.
    Would that much power be dangerously overkill?

    And just for anyone who might know, where would I find books on mechanical engineers, organic chemistry and controlling highly compressed gas safely.

    Thank you guys,
    Tyler P.
     
  2. jcsd
  3. Aug 14, 2013 #2

    mfb

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    Joule is a unit of energy. It is not an object you can move around (or throw away), and it is not power.

    The energy required to move a mass a certain horizontal distance depends on friction.


    This is certainly not particle physics, I moved your thread to classical physics.
     
  4. Aug 14, 2013 #3
    Well, a joule is a unit of energy, as mfb pointed out. The joule has the units kg*m2*s-2.

    The force required to move an object depends on the friction force that is opposing it. If you had a 10 kg mass in isolated space (no external forces), it would accelerate due to the smallest possible force, given by ##\frac{F}{m}=a##.

    Now a CHANGE in energy, such as ##\Delta KE## is the work. If you did 5.0 X 107 joules of work on a mass ##m##, starting at rest, it would have velocity ##v=10^4 \frac{1}{\sqrt{m}}## from the work integral ##\int^b _a \vec{F}(\vec{r}) \cdot d \vec{r} = \Delta KE##
     
  5. Aug 14, 2013 #4

    mfb

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    The units in that equation do not match. If you express m in kg and v in m/s, the numerical result is right.
     
  6. Aug 14, 2013 #5
    Entschuldigung, aber ich verstehe Sie nicht.

    The units should match up. ##\frac{kg \cdot m^2}{s^2} = \frac{kg \cdot m^2}{s^2} \frac{1}{kg}## cancel out the kg, and take the square root, and you're left with ##\frac{m}{s}##, unless I'm missing something?
     
  7. Aug 14, 2013 #6

    ZapperZ

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    What is this? That equation is faulty. You are basically doing a/b = ab/b, which is nonsense!

    Zz.
     
  8. Aug 14, 2013 #7
    Well now I'm all confused.

    $$KE = \frac{1}{2}m v ^2 \rightarrow v^2= 2KE \frac{1}{m}$$

    And if you just put the units in, you get

    $$(\frac{m}{s})^2 = \frac{kg \cdot m^2}{s^2} \frac{1}{kg}$$

    When you cross out the kg, and take the square root: $$\sqrt{(\frac{m}{s})^2} = \sqrt{\frac{m^2}{s^2}}$$

    And we get the m/s as required....
     
    Last edited: Aug 14, 2013
  9. Aug 14, 2013 #8

    ZapperZ

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    What??!! You are going in circles!! You started with m/s, and then you pat your self on the back for getting back m/s??!!

    Zz.
     
  10. Aug 14, 2013 #9
    Yeah, because it shows that you get the required units back out of the equation.

    The original problem was because mfb said the units didn't work out, but writing the units out shows that using ##v = \sqrt{2KE/M}## gives the correct units of ##m/s##

    Oh, and I accidentally put an extra kg on left side of the equation of post #5. Which may have led to more confusion.

    I've got no idea what's going on now, it would be nice if you explained what you're talking about. I may have made some weird mistake that I'm just not seeing, but I'm fairly confident the units work out just fine.
     
    Last edited: Aug 14, 2013
  11. Aug 15, 2013 #10

    mfb

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    Sure, but you can't just remove the units of the kinetic energy.

    The left side has units ##\frac{m}{s}##, the right side has units ##kg^{-1/2}##. They are not equal.
     
  12. Aug 15, 2013 #11
    ##\frac{m}{s}= \frac{m}{s} kg^{1/2} \frac{1}{kg^{1/2}}## I fail to see a problem.
     
    Last edited: Aug 15, 2013
  13. Aug 15, 2013 #12

    mfb

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    You are introducing a new factor here which is not present in your equation.

    It's like saying "a=b (for all a,b) because a=a/b*b"
     
  14. Aug 15, 2013 #13

    Doc Al

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    I don't see anything wrong with what you did. You obviously meant this to give a velocity in m/s when the mass is given in kg.

    Of course, as a stand-alone equation (without the caveats above) it would make no dimensional sense, since you left out the implied units of √(Joules) on the right hand side. Nonetheless, I think folks are busting your chops a bit. :smile:
     
  15. Aug 15, 2013 #14

    mfb

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    Sloppy usage of units is the dominant source of avoidable errors at this level. We should not encourage it.

    That's what I said in post 4, but it was not given in post 3.

    $$v[m/s]=10^4 \frac{1}{\sqrt{m[kg]}$$
     
  16. Aug 15, 2013 #15
    At least someone isn't picking on me :cry:

    Alright, my mistake.
     
  17. Aug 19, 2013 #16

    NascentOxygen

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    A rock of 3 metric tons travelling at about 182m/sec has 50 million Joules of energy. That is a speed of 655 km/hr, making it approximately half the speed of sound.

    When pushing blocks to build a pyramid at Giza it is best to push horizontally and low down on the block. That way more of your effort goes into sliding it along rather than ploughing up the desert with it.

    That amount of energy would be a big help in getting a new pyramid under way.
     
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