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Help a noobie please

  1. Sep 30, 2004 #1
    ok, I am really having trouble with Friction here, how exactly do u get the coefficient of kinetic friction? I have 2 problems here:

    1)A box slides down a 31.5° ramp with an acceleration of 1.28 m/s2. Determine the coefficient of kinetic friction between the box and the ramp.
    Answer: ??????

    2)A 3.20 kg block starts from rest at the top of a 30.0° incline and slides 2.20 m down the incline in 1.50 s.

    (a) Find the acceleration of the block.
    Answer: 1.96 m/s^2
    (b) Find the coefficient of kinetic friction between the block and the incline.
    Answer: ????????????
    (c) Find the frictional force acting on the block.
    Answer: ??????????
    (d) Find the speed of the block after it has slid a distance of 2.20 m.
    Answer: 2.94 m/s
    Now I also have a normal dynamics/kinematics problem which i though i knew, but webassign keeps marking it wrong. here it is:

    3) A block with a mass of 20 kg is held in equilibrium on an incline of angle = 30.0° by the horizontal force, F, as shown in Figure 4-31. Find the magnitude of F.
    Answer: ?????
    Find the normal force exerted by the incline on the block. (Disregard friction.)
    Answer: ?????

    Figure 4-31 is just a ramp (a right triangle) with a cube above it, and the F arrow is horizontally poiting to the right above the cube.

    This assigment is due today at 7am, if anybody can help me, please anwer this threat or IM me to aim(cronowarrior007) or yahoo (coolkegcomputer) or MSN (coolcomputer7@hotmail.com).

    Thanks so much :smile:
  2. jcsd
  3. Oct 1, 2004 #2
    Dont post twice please. Becomes anyoing reading same thing two times. Thank you.
  4. Oct 1, 2004 #3


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    Homework Helper

    We're not going to do your homework, so start showing us your work.


    -Draw Free Body Diagram

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

    Newton's 2nd Law
    [tex] \sum^{n}_{i=1} \vec{F}_{i} = m \vec{a} [/tex]
    Last edited: Oct 1, 2004
  5. Oct 2, 2004 #4

    [tex]D = V_it + \frac{1}{2}at^2[/tex]

    (Hint: rearrange the equation.)

    Hope that helps!
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