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Spring on inclined plane with a Block

  1. Apr 5, 2015 #1
    1. The problem statement, all variables and given/known data
    A block of mass m = 4 kg, is held at rest on a spring (point A), spring constant k =
    500 N / m, compressed by a distance AB = .DELTA.L as shown in Figure 3. When the block is freed ,
    it reaches the point B ( non-deformed position of the spring ) with a velocity of VB = 5 m / s. assume
    the coefficient of kinetic friction between the block and the incline is μk = 0.15 , determine :
    a) compression of the spring ;
    b) the distance traveled by the block to stop, point C (measured from point B);

    Problem original drawing:
    http://i.imgur.com/i98xq8J.png


    2. Relevant equations
    ∑Fy = m*ay
    Fk = μk * N
    Wi->f = F(cos(angle)*DELTA X)
    Ugi + Uei + (Wi->f) + 1/2 *m*vi2 = Ugf + Uef + 1/2*m*vf2


    3. The attempt at a solution
    My drawing of the problem:
    The spring become uncompressed at B.
    http://i.imgur.com/Db1HwA0.jpg

    v = speed (m/s)

    ∑Fy = m*ay
    ay = 0 // m = 4kg
    N - W*cos(35) = m*ay => N - (4*9.81)*cos(35) = 4*0
    N = 32.1435N

    Fk = μk * N
    Fk = 0.15 * 32.1435N
    Fk = 4.82153N

    Wi->f = F(cos(angle)*DELTA X)
    Wi->f = 4.82153N(cos(35)*DELTA X)

    Ugi + Uei + (Wi->f) + 1/2 *m*vi2 = Ugf + Uef + 1/2*m*vf2

    Ugi = 0J
    Uei = 1/2 * K * Xsf2
    Wi->f = 4.82153N(cos(35)*DELTA X)
    1/2 *m*vi2 = 0 since speed initial is 0m/s
    Ugf = m*g*hf = 4Kg*9.81* hf*sin(35)
    Uef = 0J
    1/2*m*vf2 = 1/2*4Kg*5m/s2

    I am stuck here.... :(
     
    Last edited: Apr 5, 2015
  2. jcsd
  3. Apr 5, 2015 #2

    SammyS

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    You can post those links as images if they're not too large, like I did with one of yours

    Otherwise, download them to your own file system, then upload to PF & show a thumbnail.
     
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