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Block on incline plane

  1. Dec 10, 2009 #1
    1. The problem statement, all variables and given/known data
    http://img687.imageshack.us/img687/8417/screenhunter002.jpg [Broken]

    2. Relevant equations
    I don't know.

    3. The attempt at a solution
    I don't even know where to begin...
    Last edited by a moderator: May 4, 2017
  2. jcsd
  3. Dec 10, 2009 #2
    This involves energy right? So kinetic and potential energies have to be considered.
    (I just finished learning this, so I might be wrong) :)
    Looks like a webwork problem to me... :D
  4. Dec 10, 2009 #3


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    A good starting point is to draw a free body diagram and identify the forces acting.

    Then a second good step is to introduce a suitable coordinate system One wit hte x axis parallel to the inclined plane would do. Then choose a suitable y axis.

    After you do that, you might want to write down Neton's second law of motion for those coordinate directions.
  5. Dec 10, 2009 #4
    Start with:

    Wnc = [tex]\Delta[/tex]K + [tex]\Delta[/tex]U

    for part 1, you are looking for Vf

    this gives you...
    -(0.21)(37.8)(7.96) = (1/2)(4.5)(Vf2) - (4.5)(9.8)(4.1)

    turns out that Vf is 7.23m/s .. tell me if it's right.

    for part 2, you are looking for s, the distance

    you use the same: Wnc = [tex]\Delta[/tex]K + [tex]\Delta[/tex]U
    - (0.21)(4.41)(s) = 0 - (1/2)(4.5)(7.232)
    s = 126.99m
  6. Dec 10, 2009 #5
    http://img693.imageshack.us/img693/9601/screenhunter002d.jpg [Broken]
    What do I do now?
    I found the Force of Friction by doing Ff=uFn. 21*37.8=7.938
    How do I find the speed?
    Do I use:

    But then to find the Vf, I need the distance and acceleration.
    How do I find the distance and acceleration?

    For the acceleration, I did.
    a=g*sin(31) to get 5.047. That's the acceleration if there was no friction, right?
    Then I used multiplied that by 4.5 to get the net force. I then subtracted the force of friction from the net force to get: 14.78
    I divided the new net force by 4.5 to get 3.28 as the new acceleration.
    Is that correct?
    Last edited by a moderator: May 4, 2017
  7. Dec 10, 2009 #6


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    That is certainly a good start. You can use

    [tex] V^2_f = V^2_i + 2ad [/tex]

    since the acceleration is a constant.

    Another way to do it is to note that the Potential energy at the start goes into kinetic energy at the bottom minus the work to overcome friction. The work to overcome friction is the distance traveled times the frictional force. You can solve this energy equation for the velocity at the bottom and easily check your answer from your other method.

    You can also use an energy equation to find out how far it travels. The kinetic energy at the bottom will go into work to overcome friction until the block stops.
    Last edited by a moderator: May 4, 2017
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