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Physics hwk; applied force

  1. Nov 2, 2007 #1
    a curler exerts a force forward on a 19kg curling stone and gives it an acceleration of 1.8m/s(squared) [forward]. the coefficient of kinetic friction of the ice on the curling stone is 0.080 [back]. calculate the value of the applied force. thank you in advanced
     
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  3. Nov 2, 2007 #2

    hage567

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    What have you tried? You must show some work in order to get help. What's Newton's second law?
     
  4. Nov 2, 2007 #3
    Newton's Second law is an object will accelerate only wher there is a net external force acting

    So far I figured out:
    m=19kg
    a(acceleration)=1.8m/s^2[forward]
    (u)coefficient of Fk=0.080[back]
    F(applied force)=?

    Idon't know how to setup the equation
     
  5. Nov 2, 2007 #4

    hage567

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    Do you know the equation for the frictional force?
     
  6. Nov 2, 2007 #5
    No, I am not sure I think its F=Fnet-Ff
     
  7. Nov 2, 2007 #6

    hage567

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    There are two forces acting on the stone. One is the applied force and one is the frictional force. The sum of the two is the net force. Remember this [tex]\Sigma F=ma[/tex]

    The frictional force is given by [tex]f = \mu mg [/tex]

    See if you can try it now.

    Maybe this can help you out too

    http://hyperphysics.phy-astr.gsu.edu/hbase/fric.html#fri
     
  8. Nov 2, 2007 #7
    I think I'm getting it better now.
    Would it then be: 0.080[back] x 19kg x 9.8 + (19kg)(1.8m/s^2)
    F=49N[forward]
     
  9. Nov 2, 2007 #8

    hage567

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    Yes, that's right!
     
  10. Nov 2, 2007 #9
    YYYesssss THANKYOU
     
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