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Calculate the forces of a mover pushing a crate

  1. Apr 7, 2007 #1
    1. The problem statement, all variables and given/known data
    A mover pushes a 85kg crate along the floor at a constant speed through a displacement of 3,1m[E]. The coefficient of kinetic friction between the floor and crate is 0.22.
    A) Calculate Fn and Fapp.



    3. The attempt at a solution
    Fg=mg
    =85kg*9.8N/kg
    Fg=833N
    So Fn=Fg
    but how can i find Fapp
     
  2. jcsd
  3. Apr 7, 2007 #2
    You may want to review friction;

    Fn=weight since it is vertical. But frictional force is a fraction of this. Hint:There is a constant that you are not using.
     
  4. Apr 7, 2007 #3
    make a body diagram and it will be more clear. Net force of the forces in the direction of the displacement equal the product of mass and acceleration. Solve for Fapp..
     
  5. Apr 7, 2007 #4
    Net force= forces applied on the axis of the displacement
     
  6. Apr 7, 2007 #5
    i don't know net force
     
  7. Apr 7, 2007 #6

    arildno

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    Sure you do.
    You know the crate moves with constant velocity.

    What does that mean, in terms of net force acting upon the crate?
     
  8. Apr 7, 2007 #7
    i don't understand that
     
  9. Apr 7, 2007 #8

    arildno

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    If an object moves with constant velocity, what is its acceleration?
    And how is acceleration tied together with forces?
     
  10. Apr 7, 2007 #9
    but i need time to find acceleration
     
  11. Apr 7, 2007 #10

    arildno

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    Eeeh, why?
     
  12. Apr 7, 2007 #11
    i need help with AP Physics free response questions
    :frown:
     
  13. Apr 7, 2007 #12
    its the 2002 ap physics c queston
     
  14. Apr 7, 2007 #13

    arildno

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    Then make your own thread! :smile:
     
  15. Apr 7, 2007 #14
    could someone reply if they are willing to help
     
  16. Apr 7, 2007 #15
    [tex]Given[/tex]

    [tex]m=0.2kg[/tex]
    [tex]\Delta {h}_{2}=0m[/tex]
    [tex]{v}_{2}=3.96m/s[/tex]

    [tex]Required[/tex]

    [tex]{E}_{g} and {E}_{k}[/tex]

    [tex]Solution[/tex]

    [tex]{E}_{g}={m}g\Delta h[/tex]
    [tex]={0.2kg}*9.8N/kg*0m[/tex]
    [tex]{E}_{g}=0J[/tex]


    [tex]{E}_{k}=(1/2)m{v}_{2}^2[/tex]
    [tex] =(1/2)0.2Kg(3.96m/s)^2[/tex]
    [tex]{E}_{k}=1.56J[/tex]
     
    Last edited: Apr 8, 2007
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