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Halfway done Force/Friction Problem

  1. Sep 14, 2008 #1
    [​IMG]

    A block of mass m1 = 290 g is at rest on a plane that makes an angle θ = 30° above the horizontal. The coefficient of kinetic friction between the block and the plane is µk = 0.10. The block is attached to a second block of mass m2 = 220 g that hangs freely by a string that passes over a frictionless and massless pulley.
    Find its speed when the second block has fallen 30.0 cm.
    cm/s

    What I have done:
    I turned the 2 masses into kg.

    M1 = .29kg and M2 = .22kg
    M1 Force of Gravity = .29kg * 9.81 m/s^2 = 2.8N
    M1 Normal Force = 28 * cos30 = 2.4N
    M1 Parallel Force = 28 * sin30 = 1.4N
    M1 Friction Force = .10 * 2.4 = .24N

    M2 Force of Gravity = 2.1N

    This where I am stuck, I don't know what I should do next to determine the speed of M2 when it is falling.


    Edit:

    The Net Force would be
    1.4N - .24N = 1.16N
    = 2.1N - 1.16N = .94N

    F = M * A
    .94N = .51kg * A
    A = 1.8 m/s^2

    Since there is 100 centimeters in a meter
    so A = 180 cm/S^2

    Speed = Distance * Time
    X = 30cm * T
    Where T = 30/180 = .17s
    30 * .17s
    Speed = 5.1cm/s

    Is this correct?
     
    Last edited: Sep 14, 2008
  2. jcsd
  3. Sep 14, 2008 #2
    The Net Force would be
    1.4N - .24N = 1.16N
    = 2.1N - 1.16N = .94N

    The friction force on m1 acts opposite the direction of motion of m1 with respect to the inclined plane and so it acts down the plane.

    Once a is found, you can use v = a t and d = (1/2) a t^2 to find v.
     
  4. Sep 14, 2008 #3
    I found
    F = M * A
    .94N = .51kg * A
    A = 1.8 m/s^2

    Since there is 100 centimeters in a meter
    so A = 180 cm/S^2

    I found Time to be .58s
    V = to be 104cm/s

    Is this correct?
     
  5. Sep 15, 2008 #4
    Can someone verify if this is the correct solution? Thanks!
     
  6. Sep 15, 2008 #5
    Anyone? I just need to verify this solution before I submit it online, please.
     
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