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Block on inclined plane

  1. Oct 11, 2007 #1
    1. The problem statement, all variables and given/known data

    A block with mass m = 23.4 kg slides down an inclined plane of slope angle 31.1 o with a constant velocity. It is then projected up the same plane with an initial speed 1.50 m/s. How far up the incline will the block move before coming to rest?



    2. Relevant equations
    KE=1/2mv^2
    W=F*displacement*cos(theta)


    3. The attempt at a solution
    I don't really know how to even start this problem. Can someone walk me through it?
     
  2. jcsd
  3. Oct 11, 2007 #2

    learningphysics

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    Use the coming downward part to find out the frictional force. There are different ways of doing this...

    one ways is: work by friction = change in mechanical energy.

    once you know the frictional force, you can do the next part.
     
  4. Oct 11, 2007 #3
    I haven't ever done anything with mechanical energy. How do you figure that out?
     
  5. Oct 11, 2007 #4

    learningphysics

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    hmmm... actually I was wrong.... that's probably not the best way to do the first part... probably best to just use forces...

    The net force acting on the block as it slides downwards is 0 (since acceleration is constant). Can you use this to figure out the frictional force?
     
  6. Oct 11, 2007 #5
    so then frictional force must be equal but opposite to mg??
     
  7. Oct 11, 2007 #6

    learningphysics

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    not mg... what is the component of mg along the plane?
     
  8. Oct 11, 2007 #7
    I'm not quite sure. Is it just g?
     
  9. Oct 11, 2007 #8

    learningphysics

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    no... mg divides into two components... one perpendicular to the plane... one parallel to the plane.

    use the angle 31 degrees.
     
  10. Oct 11, 2007 #9
    m*cos(theta)
     
  11. Oct 11, 2007 #10

    learningphysics

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    no. what you need is mgsin(theta).
     
  12. Oct 11, 2007 #11
    alright, my bad I meant to put mgcos(theta), but I see why it's mgsin(theta). So then I figure that out and divide by 1.5 m/s?
     
  13. Oct 11, 2007 #12

    learningphysics

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    well... mgsin(theta) is the frictional force... the magnitude doesn't change when going up or down... but the direction is different...

    So when the object is going up, what is the net force acting on it along the plane?

    Take the net force... divide by m... that's the acceleration.
     
  14. Oct 11, 2007 #13
    when going up would it be -mgsin(theta)?
     
  15. Oct 11, 2007 #14

    learningphysics

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    yes, that's the frictional force... but you've also got -mgsin(theta) due to gravity... so it's a total of -2mgsin(theta).
     
  16. Oct 11, 2007 #15
    so when I plug the numbers in I get -6.74 m. Is this right?
     
  17. Oct 11, 2007 #16

    learningphysics

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    I don't think so. How did you get that?
     
  18. Oct 11, 2007 #17
    -2mgsin(theta)/m
     
  19. Oct 11, 2007 #18

    learningphysics

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    so that's -2gsin(theta). I get -10.12m/s^2
     
  20. Oct 11, 2007 #19
    oh ok. So then I divide by the initial speed to get how far up it travels?
     
    Last edited: Oct 11, 2007
  21. Oct 11, 2007 #20

    learningphysics

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    No. You need to get the distance... you have v0, a and vf = 0. use kinematics to get the distance.
     
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