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Spring and mass problem, compression, avg force, displacemnt

  1. Mar 29, 2015 #1
    1. The problem statement, all variables and given/known data
    A 2kg box traveling at a velocity of 5 m/s hits a spring (k=100 N/cm). Find
    a. How much the spring will compress
    b. What is the average force on the spring
    c. If the surface has μs = 0.15 and μk = 0.12, how far will the box rebound from the spring
    d. If the surface is smooth, how far will the box rebound from the spring

    2. Relevant equations
    ½mv2 = ½kx2
    F=ma
    s=v0t+½at2

    3. The attempt at a solution
    a. ½mv2 = ½kx2
    ½*2*52 = ½*100*x2*100 (to change to m)
    0.005 = x2 --> x=7.07cm

    b. Favg = ½mv2
    ---this doesn't seem right---

    c. F=ma
    0.12*2*9.81*a --> 2.3a
    ---What is my F?--- Do I use Faverage ? ---
    Then s=v0t+½at2

    d. With no friction, v=5m/s
     
  2. jcsd
  3. Mar 29, 2015 #2
    It isn't. What happened to friction?
     
  4. Mar 29, 2015 #3
    You need to take into account friction when calculating the compression of the spring

    .5kx^2 represents the total force exerted on the block over the interval x. How do you do find the average value of a function over an interval?
    Also, make sure to once again take into account friction.
     
  5. Mar 29, 2015 #4
    A value for friction isn't introduced until "C". Therefore it is assumed as friction-less prior to this.
    I know "A" is correct. It is easy enough. However, the solution I was given for "B" is Favg = (0+7.07+*100)/2 = 353.3N

    This confuses me . I thought the equation should be Favg = ½mv2 But there isn't a sqr value.

    What am i missing.

    upload_2015-3-29_12-15-31.png
     
  6. Mar 29, 2015 #5
    For b) the force function is F=-kx. The integral of the function is (1/2)kx^2. They just used the force function instead of the work function to find the average force. Since the force function is linear the average force ends up being the fully compressed spring force divided by two.
    c) The acceleration of the block is constant since the frictional force on it is constant, so you can use the constant acceleration equations to solve for the distance the block travels before stopping.
    d) If there is no friction, then there is nothing to stop the block from just continuing to move at 5 m/s.
     
  7. Mar 29, 2015 #6
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