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Motion along an inclined plane

  1. Feb 5, 2006 #1
    how can i possibly find this out without the mass of the boy?

    Johnny jumps off a swing, lands sitting down on a grassy 20 degree slope, and slides 3.5m down the slope before stopping. The coefficient of kinetic friction between grass and the seat of Johnny's pants is 0.5.What was his initial speed on the grass?

    i set the problem up:

    [​IMG]

    i know i have to start by finding the acceleration, which i can get by finding out the force opposite of the frictional force (i forgot to include it)

    F(II)= mgsin(20)

    problem is, i dont know the mass of the boy
     
    Last edited: Feb 5, 2006
  2. jcsd
  3. Feb 5, 2006 #2

    Hootenanny

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    Try thinking about kinematic equations,find acceleration. Then think how this relates to force.
     
  4. Feb 5, 2006 #3

    Hootenanny

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    HINT: Use the right equations and the masses will cancel.
     
  5. Feb 5, 2006 #4
    thanks for trying but i still dont get it

    :cry:
     
  6. Feb 5, 2006 #5
    and about the kinematic equations, it wont work because i only have distance and vf
     
  7. Feb 5, 2006 #6

    Hootenanny

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    Did you not have any other information, such as the height of the swing etc?
     
  8. Feb 5, 2006 #7
    none whatsoever, just what i posted
     
  9. Feb 5, 2006 #8
    uk= .5
    vi=?
    vf=0 m/s
    g=9.8 m/s^2
    x= 3.5m
    theta= 20 degrees
     
  10. Feb 5, 2006 #9

    Hootenanny

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    I've figured it out. I'll start for you
    [tex] F = \mu R [/tex]
    [tex] ma = \mu mg \cos 20 [/tex]

    You take it from there.
     
  11. Feb 5, 2006 #10

    what does that R stand for (Fn??)

    and thanks alot man
     
  12. Feb 5, 2006 #11

    Astronuc

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    If one uses the conservation of energy, there are three things to consider, the initial kinetic energy, the gravitational potential energy, and the energy dissipated by friction. The mass is the same in all cases, so it divides out of the energy equation, and one is left with specfic energies, i.e. independent of mass. For friction, the energy dissipated is simply force * distance and force is related to [itex]\mu[/itex]mg*cos[itex]\theta[/itex] where [itex]\theta[/itex] is the angle between vertical and the normal to the slope.
     
  13. Feb 5, 2006 #12

    Hootenanny

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    Yes R is the normal reaction force.
     
  14. Feb 5, 2006 #13
    o i see now, thanks soooo much guys

    i still dont get why my teaacher gave this to us thoug
     
  15. Feb 5, 2006 #14
    ok so heres what i got

    [tex]a=\mu g \cos20 [/tex] (since the m's cancel out)

    a=.5(9.8)cos20
    a=4.604

    so, i then used a kinematic equation

    vf^2= vi^2 + 2ad
    0=vi^2 + (-32.23)
    vi= 5.677

    but this is wrong......
     
  16. Feb 5, 2006 #15

    Hootenanny

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    What is the correct answer?
     
  17. Feb 5, 2006 #16
    i dont know... the hw only tells me when i get a wrong answer (and deducts points for each try)
     
  18. Feb 5, 2006 #17

    Hootenanny

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    This is how I would have done it, but you might want to try astronuc's method.
     
  19. Feb 5, 2006 #18
    ok, thanks
     
  20. Feb 5, 2006 #19

    astronuc, can you please simplify this? i havent really learned the conservation of energy law yet
     
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