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Spring question

  1. Jun 28, 2007 #1
    1. The problem statement, all variables and given/known data
    An object of mass 10kg is at rest at the top of a frictionless incline. The mass is 2m higher than the elastic spring (k=10 N/cm). What is the objects maximum speed? What is the maximum spring compression?

    I think I may have got the answers but I'm not sure. Can someone tell me if this is done correctly?


    2. Relevant equations

    Maximum speed:
    V = [2gh] ^1/2

    maximum spring compression
    1/2mv^2 + mgy + 1/2ky^2 = 1/2mv^2 + mgY + 1/2kY^2

    3. The attempt at a solution

    Maximum speed:
    V = [2gh] ^1/2
    V = [2 x 9.8 x 2]^1/2
    V = 6.3 m/s

    maximum spring compression:
    ball touches spring = spring compresses
    1/2mv^2 + mgy + 1/2ky^2 = 1/2mv^2 + mgY + 1/2kY^2
    1/2mv^2 + 0 + 0 = 0 + mgY + 1/2(10)Y^2
    1/2(10)(6.3)^2 = (10)(9.CoolY +1/2(0.1)Y^2
    198 = 98Y + 0.05Y^2
    0 = 0.05Y^2 + 98Y - 198

    solving using the quadratic equation gives answers of -1962 or 2, where 2 sounds like the correct answer
     
  2. jcsd
  3. Jun 28, 2007 #2

    G01

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    1/2mv^2 + 0 + 0 = 0 + mgY + 1/2(10)Y^2

    This line has a mistake in it. The equation you have set up is the conservation of energy equation between the point of max speed and the point of maximum compression of the spring, both of with have the gravitational potential energy (they are at the same height). So, the potential energy term should be equal on both sides of this equation.
     
  4. Jun 28, 2007 #3
    Is this right?

    maximum spring compression:
    ball touches spring = spring compresses
    1/2mv^2 + mgy + 1/2ky^2 = 1/2mv^2 + mgY + 1/2kY^2
    1/2mv^2 + mgY + 0 = 0 + mgY + 1/2(10)Y^2
    1/2(10)(6.3)^2 +(10)(9.8Y) = (10)(9.8Y) +1/2(0.1)Y^2
    198 +98Y= 98Y + 0.05Y^2
    0 = 0.05Y^2 - 198
     
  5. Jun 28, 2007 #4

    nrqed

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    :confused: The max speed is when the block reaches the extremity of the (yet) uncompressed spring. When the spring has a maximum compression, the speed is zero. The two points are at different heights.
     
  6. Jun 28, 2007 #5

    nrqed

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    I see two problems with the last few steps. First, you should put the spring constant in Newton per meter! 10 N/cm = 1000 N/m.
    Then, the solution of the quadratic will be the one with the negative value of y (since you use y=0 on the initial position, y final will be negative).

    I am confused about the fact that there is an incline but no angle provided? You are solving theproblem as if the mass was dropped vertically on a vertical spring. That may be right but then there should be no mention of an incline.

    Hope this helps.
     
  7. Jun 28, 2007 #6
    The question does mention the incline - no angle. It could be wrong solving for the vertical spring, you're right.

    If I solve for horizontal, how would I change it? I know you still need KE, PE and the equation 1/2kx^2
     
  8. Jun 29, 2007 #7

    nrqed

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    No, I think that then you have to solve it the way you did because they mention the 2 meters higher. So I think you are doing it the right way, except fro the comments about k and the sign of y that I mentioned in my previous post.
     
  9. Jun 29, 2007 #8

    G01

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    Oh I'm sorry, I read the problem wrong. I thought the spring was located after the incline on flat ground, not on the incline. I have to stop reading these homework problems at 1AM.

    Again, Sorry for the confusion, I did mean well!:redface:
     
  10. Jun 29, 2007 #9

    nrqed

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    Oh, I see where your comment came from now. Indeed, that could have been a possible interpretation!
    hey, no problem!
     
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