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Rate of spring compession from block on incline

  1. Nov 12, 2007 #1
    1. The problem statement, all variables and given/known data

    Block 1 slides down an slope to hit block 2, which compresses a spring. How much compression does the spring experience? Mass - block 1=10 kg, block 2 = 1 kg, spring constant = 500 N, angle = 35, acceleration due to gravity=9.8
    m/s, ukf = 0.1

    2. Relevant equations

    ws = (1/2)*kx^2 = w(net force) d cos 0
    ws= (1/2)*500*x^2 = (9.8*10) 3 (cos 35)

    3. The attempt at a solution

    without the coefficient of static, and the mass of block 2
    ws= (1/2)*500*.9815^2 = (9.8*10)3(cos35)
    not sure how to work in the coefficient of static on block 1, and uncertain how to account for the effect of block 2 (1 kg)
    1. The problem statement, all variables and given/known data

    2. Relevant equations

    3. The attempt at a solution
  2. jcsd
  3. Nov 13, 2007 #2
    Not enough information. What are the initial conditions for Block 1? How long before it hits Block 2?
  4. Nov 13, 2007 #3
    Block 1 is at rest, distance is 3M to Block 2.
    I am unsure if the the '(9.8*10) 3 (cos 35)'
    can be used, as well as how to integrate the
    coefficient of 0.1 for ufk. Finally, how to deal
    with this force when it meets block 2 (1 kg)
    which seperates Block 1 from the spring.
  5. Nov 13, 2007 #4
    Best way to get the Block 1 situation clear is to draw a free body diagram. That will give you the force accelerating it (the component of its weight acting down the slope less frictional force). That will lead to knowing how fast it is travelling when it hits Block 2 ...
  6. Nov 15, 2007 #5
    I have calculated 10*9.8*sin35 = 56.21 - 10*9.8*cos35*0.1 = 48.18 * 3M
    = 144.54N. This is the only way I have been able to find to do this. I'm not sure how to deal with the 1 kg block which sits at the end of the spring. The calculated mass will be exerted against the 1 kg block, which in turn will compress the spring X amount. It is this amount of compression that is sought. Based on PES = (1/2)KX^2. (with spring K=500) Forgetting about the 1 kg block for a moment, I thought I could possibly take 144.50/250 =0.578 = square root of 0.578 = 0.76M spring compression. Even if this is close to correct, it disregards the 1 kg block. Any input appreciated. Thanks.
  7. Nov 15, 2007 #6

    Shooting Star

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    Homework Helper

    You also have to know if the collision between the blocks is elastic or not. That way, you’ll know how much energy is transferred to block 2.

    The velo of block 1 just before hitting block 2 should be clear enough, since all the data are given.
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