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Spring and box on incline

  1. Dec 19, 2007 #1
    [SOLVED] spring and box on incline

    1. The problem statement, all variables and given/known data
    A 50 kg mass rests atop a 20m high incline. The spring constant of a 10m long spring below is 1000N/m. The 50 kg crate is sent down the incline. Assume all surfaces in this problem are frictionless.
    What is the velocity of the crate traveling along the horizontal surface?

    2. Relevant equations
    V_(f^2 )=V_(i^2 )+2aD
    Fspring=kx
    F=ma



    3. The attempt at a solution

    I am so confused by this problem. I tried to find the length of the incline, but couldn't figure out how. The only thing I could think to do was to find the potential energy of the crate, which is 9810 J. I know that only force gravity is acting of the crate, and that Fgy=FN, but that doesn't seem to help me at all.
    Please help me if you can because I have a physics final exam tomorrow. Thank you so much.
    1. The problem statement, all variables and given/known data



    2. Relevant equations



    3. The attempt at a solution
     

    Attached Files:

  2. jcsd
  3. Dec 19, 2007 #2
    Use energy...
     
  4. Dec 19, 2007 #3

    Doc Al

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    Staff: Mentor

    Since there's no friction, what's conserved?
     
  5. Dec 19, 2007 #4
    I don't want you to give me the answer, but is there anyway you could give me a little more information? I know that energy should be conserved, but the only formulas I know for the conservation of energy involve velocity, which I can't figure out a way to find.
     
  6. Dec 19, 2007 #5

    Doc Al

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    Staff: Mentor

    There are three types of energy involved here: kinetic energy, gravitational potential energy, and spring potential energy. Only kinetic energy involves velocity.
     
  7. Dec 19, 2007 #6
    Is this right? :

    1/2mv^2 + mgy = 1/2mv^2 + mgy
    0 + (50kg)(9.81m/s^2)(20m) = 1/2(50kg)v^2 + 0
    19.81m/s = v

    So then the spring would be compressed .9905m because:
    f = ma
    f = 50kg(19.81m/s)
    f = 990.5N

    fp = kx
    990.5N = 1000N/m * x
    .9905m = x

    Thanks for all your help

    -KAT444
     
  8. Dec 19, 2007 #7

    Doc Al

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    Staff: Mentor

    This is exactly correct. :smile:

    This is not correct.

    To find out how the spring compresses, again consider energy. What's the formula for the energy of a compressed spring?
     
  9. Dec 19, 2007 #8
    Is this the right way:

    elastic PE = 1/2kx^2
    energy is conserved so . . .
    9810J = 1/2(1000N/m)(x^2)
    4.43m = x

    Thanks - KAT444
     
  10. Dec 19, 2007 #9

    Doc Al

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    Staff: Mentor

    Perfecto!
     
  11. Dec 19, 2007 #10
    I agree with your profile, you're an all around good guy.

    Thanks again - KAT444
     
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