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Forces of girl on a sled

  1. Oct 23, 2008 #1
    I'm not sure what the rules are about posting homework problems, but I just had 2 I couldn't get around too, so here's the second one...

    1. The problem statement, all variables and given/known data
    A 40 kg girl and a 8.4 kg sled are on the frictionless ice of a frozen lake, 15m apart but connected by a rope of negligible mass. The girl exerts a horizontal 5.2 N force on the rope.
    What are the acceleration magnitudes of
    (a) the sledge?
    (b) the girl?
    (c) How far from the girl's initial position do they meet?

    2. Relevant equations

    3. The attempt at a solution
    Finding the answers to part a and b were simple..
    (a) a = f/m = 5.2N/8.4kg = .62 m/s2
    (b) a = f/m = 5.2N/40kg = .13 m/s2
    (c) I'm having problems figuring out part C.
  2. jcsd
  3. Oct 23, 2008 #2


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    Re: Force

    Write equations for their positions
    XG + XS = 15
    XG = 1/2*ag*t2
    XS = 1/2*as*t2

    Solve noting that t2 is the same when they meet
  4. Oct 23, 2008 #3
    Re: Force

    1/2*ag*t2+1/2*as*t2= 15

    t2 = 15 / (1/2*ag +1/2*as)
    t2 = 15 / (1/2*.62 +1/2*.13)
    t2 = 40
    t = 6.3m

    That doesn't seem right, I'm lost.
  5. Oct 23, 2008 #4


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    Re: Force

    Not quite. That gives you seconds.

    XG = 1/2*ag*t2 = 1/2*(.13)*(6.3)2 = 2.58m
  6. Oct 23, 2008 #5
    Re: Force

    ahh. Gotcha, thanks.
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