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Kinetic friction on an incline.

  1. Mar 13, 2012 #1
    1. The problem statement, all variables and given/known data


    A box is sliding down an incline tilted at a 13.0° angle above horizontal. The box is initially sliding down the incline at a speed of 1.40 m/s. The coefficient of kinetic friction between the box and the incline is 0.370. How far does the box slide down the incline before coming to rest?

    2. Relevant equations
    mk= Fnormal*ms


    3. The attempt at a solution
    1.4 m/s in x direction= m*cos(13)
    m=1.4368
    m*ms= 1.4368*.37=.51363= mk
    1.4m/s-.51368mk= .868m
    Not sure what I am doing wrong. Thanks.
     
  2. jcsd
  3. Mar 13, 2012 #2

    Doc Al

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    Staff: Mentor

    I don't quite understand what you've done or even what your equations mean.

    Try this: What forces act on the box? Figure out the resulting acceleration of the box as it slides down the incline.
     
  4. Mar 13, 2012 #3
    I found mass and then found the amount of kinetic friction. Then I subtracted that from the 1.4 m/s. Ill be honest. I know I am doing it wrong and I have no Idea what I am supposed to do. I guess I will keep googling it until I find something. Thanks.

    The force that act on the box are gravity, fnorm, and kinetic friction.
     
  5. Mar 13, 2012 #4

    Doc Al

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    Good. Now express the net force parallel to the incline. Then use Newton's 2nd law to find the acceleration.
     
  6. Mar 13, 2012 #5
    kinetic friction = not sure. I dont have a mass.
    fnormal = not sure dont have mass.
    fgravity = not sure dont have a mass

    Unless my mass from the first part was correct?
     
  7. Mar 13, 2012 #6

    Doc Al

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    You won't need a numerical value for the mass. Just call the mass 'm' and continue. Write expressions for these forces symbolically, not numerically.
    Nope.
     
  8. Mar 13, 2012 #7
    Ok thanks.
    kinetic= -.370*m
    fnorm= cos(13)*9.8
    kinetic +fnorm= 9.17
    9.17m=m*a
    9.17=a
    Is this right^^?
    Then how do I find the distance?

    Thanks so much.
     
  9. Mar 13, 2012 #8

    Doc Al

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    No. The force of kinetic friction equals μFnorm.
    Not exactly. (But you're close.) First, what does the weight equal? Then how does Fnorm relate to the weight?

    Once you find the correction acceleration, you'll use kinematics to find the distance.
     
  10. Mar 13, 2012 #9
    This is incorrect. Force of kinetic friction = μF[itex]_{Normal}[/itex]

    Good.

    Kinetic friction and the normal force are not on the same plane, you can't add them together. Think about it. What is the only force that is causing the block to deccelerate? After you find the acceleration just use v[itex]^{2}[/itex]=v[itex]^{2}_{f}[/itex]+2ad and solve for distance.
     
  11. Mar 13, 2012 #10
    kinetic= -.370* cos(13)*9.8*m
    fnorm= cos(13)*9.8*m
    kinetic +fnorm= 6.0
    6.0m=m*a
    6.0 = a

    vf^2= vi^2 + 2ad
    0= 1.4^2+ 2*(6)*d
    d= -.1633
    Still not sure what I did wrong :(
     
  12. Mar 13, 2012 #11
    Again, don't add these two together. Think of it this way. What is your net force and what does friction have to do with it?
     
  13. Mar 13, 2012 #12
    Well net force is my overall force right? So I figure you just add them ( I added the negative sign in front of the friction, so It is really like subtracting).
     
  14. Mar 13, 2012 #13

    Doc Al

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    Good. That's one of the forces acting parallel to the incline.
    OK. But realize that this force acts perpendicular (normal) to the incline surface.
    Why are you adding those forces? (They don't act in the same direction, for one.) And why are you setting it equal to 6.0?

    What you need is the net force parallel to the incline. (What about gravity?)
     
  15. Mar 13, 2012 #14
    I thought the force normal was no longer perpendicular if I took the cos(13) of it. I thought the cos made it parallel. That is why I was adding them.

    Sorry I was unclear. i wasn't setting it equal to 6.0. That is what it equaled when added.

    ughh so confused! :(
     
  16. Mar 13, 2012 #15
    Thanks for your help thus far but, I am clearly not competent enough to do this problem. Can someone just show me the equations I need and I will try to figure it out.
     
  17. Mar 13, 2012 #16

    Doc Al

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    No. Multiplying the weight (mg) by cosθ gives you the normal force, which is perpendicular to the surface. But the friction force is parallel. To get the friction force, multiply the normal force by μ.

    But the friction force is just one of the forces. The other force is the component of the weight parallel to the surface. Hint: If multiplying the weight by cosθ gives you the perpendicular component of the weight (which is equal to the normal force), what would you need to multiply the weight by to get the parallel component?
     
  18. Mar 13, 2012 #17
    kinetic= -.370* cos(13)*9.8*m
    fgrav (weight)= m*9.8*sin(13)
    acceleration= -1.329
    0= 1.4^2+2(-1.329)d
    d=.737

    WE DID IT. I LOVE YOU FOREVER!
    Thanks.
     
  19. Mar 13, 2012 #18
    Nice work.
     
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