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Space Platform Problem

  1. Oct 29, 2013 #1

    sph

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    1. The problem statement, all variables and given/known data
    A 4.50x10^5kg space platform is speeding up from 2.300x10^3m/s to 2.600x10^3mls because a rocket fires with a thrust of 138kN in space. How long must the rock fire to do this?


    2. Relevant equations
    F=ma
    F=ma+mg


    3. The attempt at a solution
    I have tried to find acceleration so I could find the time.
    I have attempted to use the equation F=ma+mg to isolate for a but could not.
     
  2. jcsd
  3. Oct 29, 2013 #2

    tiny-tim

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    hi sph! welcome to pf! :smile:

    (you don't need the acceleration)

    force x distance = change in … ?

    force x time = change in … ? :wink:
     
  4. Oct 29, 2013 #3

    sph

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    sorry, I am not familiar with this. I have learned to use the kinematics equations though.
     
  5. Oct 29, 2013 #4

    tiny-tim

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    have you done force = rate of change of … ?
     
  6. Oct 29, 2013 #5

    sph

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    no, I haven't learned that yet either.
     
  7. Oct 29, 2013 #6
    Even though I believe that tiny-tim's method would solve your problem quicker it seems that you haven't learned that way of looking at Force yet (which ironically enough is actually how Newton originally came up with the second law).

    The other method involves just using Newton's Second Law of Motion Sum of all F = ma. You have the mass, the force, so what must the acceleration be? Once you find the acceleration just try to remember the definition of average acceleration and see what you can do from there. Good luck!
     
  8. Oct 29, 2013 #7

    sph

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    Thanks! amazing answer by an amazing person. Also thanks for letting me know what I am missing out on.
     
  9. Oct 29, 2013 #8
    Thank you for such a kind answer! I was nervous to post since this was my first post.

    By the way the other method that Tiny Tim was referring to was the definition of Force that Newton originally identified.
    Force = change in momentum over change in time. So Force can also be looked at as the derivative of momentum with respect to time (if you have gotten to calculus yet).
     
  10. Oct 29, 2013 #9

    sph

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    I am definitely going to ask my teacher about this! thank you once again and good luck with your future posts!
     
  11. Oct 30, 2013 #10

    tiny-tim

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    (just got up :zzz:)

    Hi Polishstudent! Welcome to PF! :smile:
    sph, that's completely correct

    and if you haven't done calculus (yet), the following applies if the force is constant …

    force x time = impulse = total change in momentum

    force x distance (strictly, force "dot" displacement) = work done = total change in mechanical energy :wink:
     
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