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Finding a net force

  1. Dec 5, 2009 #1
    1. The problem statement, all variables and given/known data

    the questions says,
    "Person A who's mass is 40kg, is being pulled across the ground by a horse at a speed of 4 m/s [E]. All of a sudden the horse accelerates for a period of 5 seconds. During this time person A experiences a displacement of 40 m [E]. What net force was experienced by person A?"

    2. Relevant equations

    3. The attempt at a solution

    I have no idea how to even begin this question. I'm not sure what equations i even need to use.
  2. jcsd
  3. Dec 5, 2009 #2


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    Homework Helper

    Hi dois, welcome to PF.
    In the problem the initial velocity, duration pof acceleration and the displacement is given. Can find find a kinematic equation from the text book which relates these quantities with acceleration?
  4. Dec 5, 2009 #3


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  5. Dec 5, 2009 #4
    ok so first I would need to find the acceleration so I can sub that answer into F= m*a
    so i would use
    acceleration= v/t
  6. Dec 5, 2009 #5
    The problem gives you the initial speed (Vi), time (t), and displacement (d). If you take a look at the kinematic equations, you'll find that you can solve one of them for acceleration. The kinematic equation you're looking for is this one.

    d = (vi)(t) + (1/2)(a)(t)^2

    Go from there.
  7. Dec 6, 2009 #6
    Ok so if we sub all of our know variables into the equation
    d = (vi)(t) + (1/2)(a)(t)^2
    then we would get:
    40= 4x5 +1/2 (?) x5^2
    40=20 +25/2 x
    20=25/2 x
    20=12 1/2 x
    x= 1.6

    but as soon as i figure out the acceleration where do I go from there to find out the net force?
    Last edited: Dec 6, 2009
  8. Dec 6, 2009 #7


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    Okay, your algebra looks good, so a = 1.6 m/sec squared.

    Remember I said this was a two step question. Go back and read my first post. That indicates what you do next.
  9. Dec 6, 2009 #8
    So then we would sub a into F=ma
    so (40kg) (1.6 m/s)
    and then our answer would be Fnet= 64 kg m/s [E]
  10. Dec 6, 2009 #9


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    That should do it, except that the units you give are wrong. The seconds should be squared. That combination of units has its own name: Newtons.
  11. Dec 6, 2009 #10
    alright, thank you!!
  12. Dec 6, 2009 #11


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    You are quite welcome.
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