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Sliding Block

  1. Sep 29, 2009 #1
    1. The problem statement, all variables and given/known data
    A block with mass m = 5.0 kg slides down an inclined plane of slope angle 36.5o with a constant velocity. It is then projected up the same plane with an initial speed 1.05 m/s. How far up the incline will the block move before coming to rest?

    3. The attempt at a solution

    [tex] a_x = -g sin \theta [/tex]

    [tex] \Delta v^2 = 2a \Delta x [/tex]

    [tex] \frac{-v_o^2}{2a} = \Delta x [/tex]

    [tex] \frac{-1.05^2}{-2gsin \theta} = \Delta x [/tex]

    [tex] \frac{-1.05^2}{-11.658} = \Delta x [/tex]

    But this wasn't right.
     
  2. jcsd
  3. Sep 29, 2009 #2

    Doc Al

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    As it slides down it's moving at constant velocity. So what force must be acting on the block besides gravity?
     
  4. Sep 29, 2009 #3
    I would say friction but there is no mention of it in the problem.
     
  5. Sep 29, 2009 #4

    Doc Al

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    Good. They expect you to conclude that on your own, based on the fact that it's not accelerating as it slides down.

    So what must that friction force equal?
     
  6. Sep 29, 2009 #5
    I'm getting

    [tex] f = -mg sin \theta [/tex]
     
  7. Sep 29, 2009 #6

    Doc Al

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    Good. The magnitude is mgsinθ, but the direction depends on which way the block moves.
     
  8. Sep 29, 2009 #7
    I don't get how the friction helps any.
     
  9. Sep 30, 2009 #8

    Doc Al

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    To determine the acceleration, you need the net force. Friction is one of the forces acting on the block.
     
  10. Sep 30, 2009 #9
    You'd only do that when the block is projected back up the incline right? Then use the kinematic equation I had in OP?

    Edit: So I was thinking about this more. When the block is projected back up the incline, you'd have friction and gravity in the X direction. Since I found the magnitude of the friction above, the acceleration would be:

    [tex] +mg sin \theta + mg sin \theta = ma_x [/tex] masses would cancel, so:

    [tex] 2 g sin \theta = a_x [/tex]

    Does this look right so far? (I set the positive X direction going down the incline)

    Edit 2: Yes, I was right, thanks Doc :P
     
    Last edited: Sep 30, 2009
  11. Sep 30, 2009 #10

    Doc Al

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    Friction acts whether the block is sliding up or down the incline. The problem you need to solve involves the block sliding up, so you need to find the block's acceleration as it slides up.
    Yes.
     
  12. Sep 30, 2009 #11
    Okay I have a new question that is similar to this.

    A block is at rest on an inclined plane whose elevation can be varied. The coefficient of static friction is μs= 0.44, and the coefficient of kinetic friction is μk = 0.17. The angle of elevation θ is increased slowly from the horizontal. At what value of θ does the block begin to slide (in degrees)?

    So I started out with:
    Y-Direction
    [tex] F_N - mg cos \theta = ma_y [/tex]

    [tex] F_N = mg cos \theta [/tex]

    X-Direction

    [tex] -f + mg sin \theta = ma_x [/tex]

    [tex] -(\mu_s mg cos \theta) + mg sin \theta = ma_x [/tex]

    [tex] -\mu_s g cos \theta + g sin \theta = a_x [/tex]

    [tex] g(-\mu_s cos \theta + sin \theta) = a_x [/tex]

    But here is where I'm stuck and dunno what to do next.
     
  13. Sep 30, 2009 #12

    Doc Al

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    Two hints:
    - What does ax equal?
    - When does static friction = μN?
     
  14. Sep 30, 2009 #13
    I don't understand what you're asking, when there's no sliding?

    Since you're giving this as a hint, I'm guessing you mean it's equal to zero. But I thought that since the block is moving (question asks for kinetic friction) you don't know if ax is zero or not.
     
  15. Sep 30, 2009 #14

    Doc Al

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    No, my point was that static friction does not always equal μN. μN represents the maximum value of static friction between two surfaces--the actual friction may well be less than that, depending on the situation. (For example: A book rests on a table. What's the static friction acting on it?)

    Right.
    As the surface is tilted up, the acceleration remains zero until you get to the point where it just begins to start moving. That's the point you want, so it's only static friction that you care about. What's the maximum angle you can have and still not have the block slide? (Use the first hint, of course.)
     
  16. Sep 30, 2009 #15
    Yes, I knew that about static friction, however I think I'm stuck more mathematically, because if you simplify what I had up above when ax = 0 you get:
    [tex]
    -\mu_s cos \theta + sin \theta = \frac{1}{g}
    [/tex] What next?
     
  17. Sep 30, 2009 #16

    Doc Al

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    That's not quite correct. Take another look.
     
  18. Sep 30, 2009 #17
    Oops. I still don't see how to solve for the angle.

    [tex]

    -\mu_s cos \theta + sin \theta = 0

    [/tex]
     
  19. Sep 30, 2009 #18

    Doc Al

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    That's better. Play with it a bit. Move terms around.
     
  20. Sep 30, 2009 #19
    Just a question, is this going to end up being an identity, because I don't know those very well. It'd help if I knew there was an identity involved.
     
  21. Sep 30, 2009 #20

    Doc Al

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    No fancy trig identities involved. Hint: What trig function can be defined in terms of sine and cosine?
     
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