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Space, mass & energy

  1. Sep 22, 2008 #1
    Hi;
    disclaimer: i'm not a physicist.
    just wondering ... if i was in space and needed to move 1 ton of mass ... how much energy would it require?
    if everything in space is weightless then does it really require much energy?
    what about inertia? ... if a 1 ton mass was floating toward me could i simply stop it (like superman) with 1 finger??
    thanks
     
  2. jcsd
  3. Sep 22, 2008 #2

    Dale

    Staff: Mentor

    How fast do you want to move it?

    KE = 1/2 m v²
    p = m v

    Where KE is kinetic energy, p is momentum, m is mass, and v is velocity.
     
  4. Sep 22, 2008 #3
    well ... just for the example ... let's say 10km per hour and i was going to use 100 pounds of force (arm strength) to move the object

    and, then one more variable ... if one was to use a smaller force to move the object over a longer time period then is it the same formula ? ... just with a time variable added ?

    for example ... instead of all the force in one instance ... 1/10th the force over 10X the time ... would that be correct ??

    so, of your two equations, which equation would I use ?

    thanks
     
  5. Sep 23, 2008 #4

    Tyx

    User Avatar

    Of course not. The fact that the truck floating towards you is not being pulled down by gravity does not mean it has less energy when it's moving, quite the contrary.
    If the object has a certain amount of kinetic energy, you will need that amount to stop it. Otherwise you will be swept away. It depends on what you are picturing when you say 'floating'.
    You can't stop a comet with your index finger just because you're in outer space. Superman packs a mean punch, really.
     
  6. Sep 23, 2008 #5

    Dale

    Staff: Mentor

    So, with the formula above you have KE = 1/2 1 ton (10 km/h)² = 3858 J

    If we take 3858 J and divide by 100 lbf (W = f d) we get 8.6 m. So a 100 lbf push could slow a 1 ton mass from 10 km/h over a distance of 8.6 m. Since most people don't have arms 8.6 m long this would not work.

    I think what you are interested in is W = f d where W is work f is force and d is distance. So if you apply the same force over a longer distance then you have done more work. Also remember that work is equal to the change in energy.
     
    Last edited: Sep 23, 2008
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