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I'm not sure how to do this

  1. Dec 14, 2009 #1
    From a height of 6,00m from the floor a man drops a package weighing 1,3kg. Another man below at the height of 1,60m from the floor stops the package 0,30m from the floor. (We take for granted that the man uses a constant amount of newton on the package and that there is no air resistance)

    A) Determine the velocity of the package when the man stops the package
    B) Determine the acceleration of the package during the deceleration
    C) Determine the amount of newtons the man uses on the package during the deceleration

    Is there anyone who can help me out with this question showing every step, been sitting with it a while now and don't know anywhere else to turn =/

    Thank you in beforehand :)
     
  2. jcsd
  3. Dec 14, 2009 #2

    PhanthomJay

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    Hi, welcome to PF! Someone here would be glad to help, but Forum Rules first require that you list what you think are the relevant equations and then show some attempt at a solution. Note that the low guy stops the package over a distance of 1.6 -0.3 = 1.3 m, by moving his hands down over that distance as he catches the package. You can use energy methods if you are familiar with them.
     
  4. Dec 14, 2009 #3

    rock.freak667

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    Think of conservation of mechanical energy for the first part.

    You will also need the kinematic equations of motion for the second part.
     
  5. Dec 14, 2009 #4

    tiny-tim

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    Welcome to PF!

    Hi Unkownentity! Welcome to PF! :wink:

    Try A) first: you have the initial velocity, and the total distance, and you want the final velocity, so what formula or principle do you think you should use? :smile:

    EDIT: ooh, PhanthomJay and rockfreak667 both beat me to it!
     
  6. Dec 14, 2009 #5
    This is as far as i have come

    By using newtons second law F = M * A

    and the acceleratoion in this case is the gravity therefore i cna rewrite the formula F = M * G

    F = 1,3 * 9,82
    F = 12,76

    (i'm bad at using symbols over computer, what i mean with V0 is the starting velocity which should be zero in this case)

    what i was thinking is by using this formula s = V0 *t * 1/2at^2 shouldn't i be able to get the amount of time it takes for it to fall and then by using (After getting the time) v = V0 + at i can find out the velocity, does that sound reasonable?
     
  7. Dec 14, 2009 #6

    tiny-tim

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    Hi Unkownentity! :smile:

    (you don't need the force … the acceleration is enough :wink:)


    Yes, that's the right idea.

    But you only seem to know two of the standard constant acceleration equations …

    there is a third one, v2 = v02 + 2as :wink:
     
  8. Dec 14, 2009 #7
    Yeah isn't that the timless formula or something we haven't got to learn that yet since that goes more in depth in the next course :P but otherwise my other idea was right?
     
  9. Dec 14, 2009 #8

    tiny-tim

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    (it's not the tiny-timless formula :blushing:)

    Yes, those two equations together will work fine. :smile:

    ok, next try B) … you have the initial and final velocity, and the distance, and you want the acceleration. :wink:
     
  10. Dec 14, 2009 #9
    i got that the answer on A was 10,50 m/s

    V0 = 0
    V1 = 10,5 m/s

    since we only want to know the acceleration during the deceleration should i instead use the distance 0,3m? then i could use S = 1/2 (V0 + V1) *t to get the time and then use S = V0 *t + 1/2at^2 and leave the acceleration unknown does this work?
     
  11. Dec 14, 2009 #10

    tiny-tim

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    You're not reading the question properly :redface:

    the package does not stop at the floor, it stops at 0.30m (and the deceleration started at 1.60m).

    But your method is correct: you can use the same equations "backwards" as in A), because the same four variables are involved. :wink:
     
  12. Dec 14, 2009 #11
    Oh okey so i just need to switch the distance, so i use 1,6 instead of using 0,3?
     
  13. Dec 14, 2009 #12

    tiny-tim

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    In part A), yes. :smile:
     
  14. Dec 14, 2009 #13
    shouldn't the distance on part A be 5,70 meters since the man stops it at that height?
     
  15. Dec 14, 2009 #14

    tiny-tim

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    hmm … it's a badly-worded question :frown:
    Part A) can't mean what it says (ie at 0.30m), because the velocity when he stops it is obviously zero :rolleyes:, so the first part must mean that he starts stopping it at 1.60m, and stops stopping it at 0.30m … and A) is asking for the velocity when he starts stopping it. :smile:
     
  16. Dec 14, 2009 #15
    i think he wrote it wrong i think he means that

    A) Determine the velocity of the package when the man grabs the package
     
  17. Dec 14, 2009 #16

    tiny-tim

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    Yes, at 1.60m. :smile:
     
  18. Dec 14, 2009 #17
    Okey so i got a veloity of 0,59m/s on A) does that sound reasonable?

    But what i don't understand is what distance i'm supposed to use on B) =/
     
  19. Dec 14, 2009 #18

    tiny-tim

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    How do you get that?
    1.60m - 0.30m = 1.30m.
     
  20. Dec 14, 2009 #19
    V = X
    A = 9,82
    t = 0,06

    V = V0 + at by incorperating the numbers into the formula i got V = 0,59m/s
    ----------------------------------------------------------------
    S = 1,6m
    V0 = 0m/s
    t = X

    S = V0 *t + 1/2at^2

    Here i did the same thing but with "T" so i could ge the time. (T = 0,06s)

    thats how i did it
     
    Last edited: Dec 14, 2009
  21. Dec 15, 2009 #20

    tiny-tim

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    Hi Unkownentity :smile:

    (just got up :zzz: …)
    where does 0.06 come from?

    where does 1.6m come from? :confused:
     
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