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Help on a problem with multiple parts, please!

  1. Sep 12, 2004 #1
    hey i was wondering if anybody could show me how to do a problem that contains multiple parts:

    55. A 3.00 kg block starts from rest at the top of a 30.0° incline and accelerates uniformly down the incline, moving 2.00 m in 1.50 s.

    a. Find the magnitude of the acceleration of the block.
    b. Find the coefficient of kinetic friction between the block and the incline.
    c. Find the magnitude of the frictional force acting on the block.
    d. Find the speed of the block after it has slid a distance of 2.00 m.

    I know that the initial velocity=0, Fg (weight)=29.4N, Fn (normal force)=14.7N, and I think that the falling force is 21.6N, but I'm not sure if i did all this right. Just show me what I need to do, and I'll do the rest! thanks for your help.
     
  2. jcsd
  3. Sep 12, 2004 #2

    Pyrrhus

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    The force making it slide is gravity, and there should be only two forces acting on it [normal and gravity] if there is no friction of course.

    Normal will be equal to [tex] mgcos\theta = n [/tex] because there is no movement on the y-axis.

    Now for [tex] mgsen\theta = ma [/tex] which will become [tex] gsen\theta = a [/tex] because there's movement on the x-axis.
     
    Last edited: Sep 12, 2004
  4. Sep 12, 2004 #3
    friction

    there is definetly friction involved, because one of the questions is asking for the coefficient of friction.
     
  5. Sep 12, 2004 #4

    Pyrrhus

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    oh i didn't read the questions :smile:

    Well another force :cool:

    [tex] F_{f} = \mu N [/tex]

    so changing the x-axis equation it will be if we take positive aceleration the way as the block slides it will be

    [tex] mgsen\theta - \mu N = ma [/tex]
    [tex] mgsen\theta - \mu mgcos\theta = ma [/tex]
    [tex] gsen\theta - \mu gcos\theta = a [/tex]
     
  6. Sep 12, 2004 #5
    I plugged everything in so that it was a=9.8sin30=4.9m/s/s. This is wrong, though. The back of the book says it should be 1.78m/s/s. What's wrong?
     
  7. Sep 12, 2004 #6

    Pyrrhus

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    I did the first analysis without friction...
     
  8. Sep 12, 2004 #7
    so how would i find mu?
     
  9. Sep 12, 2004 #8

    Pyrrhus

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    Kinematics! :smile:

    The problem states uniform acceleration which is
    [tex] gsen\theta - \mu gcos\theta = a [/tex]
     
  10. Sep 12, 2004 #9
    but mu is a value, isn't it (and one that I don't have, at that)? So how do I find accel. without mu? I'm sorry if I'm being difficult, but as you can see, this isn't my strongest subject :).
     
  11. Sep 12, 2004 #10

    Pyrrhus

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    Well the block starts from rest

    info:
    [tex] V_{o} = 0 m/s[/tex]
    [tex] X_{o} = 0 m[/tex]
    [tex] X = 2 m [/tex]
    [tex] t = 1.5 s [/tex]

    I think this equation will help

    [tex] X - X_{o} = V_{o}t + \frac{1}{2}at^2 [/tex]
    [tex] X = \frac{1}{2}at^2 [/tex]

    [tex] X = \frac{1}{2}(gsen\theta - \mu gcos\theta)t^2 [/tex]
     
  12. Sep 12, 2004 #11
    still not answering the mu thing, though. Sorry.....lol. d=1/2 (9.8sin30 - [mu]9.8cos30)
     
  13. Sep 12, 2004 #12

    Pyrrhus

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    Are you understanding? or should we go back?
     
  14. Sep 12, 2004 #13

    Pyrrhus

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    Of course it is!, just get [tex] \mu [/tex] alone

    [tex]\frac{\frac{2X}{gt^2} - sen\theta}{-cos\theta} = \mu [/tex]

    Plug in known values and voila!
     
  15. Sep 12, 2004 #14
    Understanding everything except for that darn mu thing. I'm really sorry I'm making this so hard.
     
  16. Sep 12, 2004 #15

    Pyrrhus

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    [tex] \mu [/tex] is just the greek letter given to friction quotient.
     
  17. Sep 12, 2004 #16
    OK. got that. so... 2(2)/9.8(1.5*1.5)- sin30 / -cos30 = mu. mu=-.48
     
  18. Sep 12, 2004 #17

    Pyrrhus

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    I guess you can finish it now...
     
  19. Sep 12, 2004 #18

    Pyrrhus

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    No, you're calculating it wrong. [tex] \mu = 0.3678 [/tex]
     
  20. Sep 12, 2004 #19
    don't know where i went wrong....
     
  21. Sep 12, 2004 #20
    i did it again and still got .48308
     
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