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Energy equilibrum question

  1. Jul 11, 2009 #1
    Hello,
    I need soe help please, i cant seem to get this one.

    1. The problem statement, all variables and given/known data

    A 0.20-kg mass is hung from a vertical spring of force constant 55 n/m. When the spring is released from its unstretched equilibrum position, the mass is allowed to fall. Use the law of conservation of energy to determine:
    a) the speed of the mass after it falls 1.5cm
    b) the distance the mass will fall before reversing direction

    Books answers:
    a) 0.48 m/s
    b) 0.071m

    2. Relevant equations

    Eg = mgh
    Ee =0.5 k x^2
    Ek= 0.5 m v^2


    3. The attempt at a solution

    I have tried several things, but have erased them. I came up with 0.24 m/s for a) several times.
     
  2. jcsd
  3. Jul 11, 2009 #2

    cepheid

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    The question asks you to use conservation of energy. You know that as the mass's height decreases, the system loses gravitational potential energy, (which is converted to kinetic energy). But by the same token, the system gains elastic potential energy because as the mass falls, the spring is elongated. Can you set up an equation for conservation of energy that takes both of these changes into account? It looks like you have all of the necessary equations.

    For part b, the mass has to stop falling and come to rest before it changes direction and starts moving upwards. What is the reason why the mass would stop (come to rest)? The answer to that question should help you solve this part of the problem.
     
  4. Jul 11, 2009 #3
    I believe i got part A

    Eg = Ee + Ek

    for part b

    The mass would come to rest because the springs energy (in the up direction) is greater thatn the kinetic energy and gravitational potential.

    I tried Ek = Etotal and that gave me the wrong answer so im a tad confused.
     
  5. Jul 11, 2009 #4

    cepheid

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    Rest means zero velocity. Which means zero kinetic energy. The mass will come to a stop because all of its kinetic energy has been converted into elastic potential energy. In your original equation:

    Eg = Ee + Ek,

    we have Ek = 0
     
  6. Jul 11, 2009 #5

    cepheid

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    Energy doesn't have a direction: it is a scalar, not a vector.
     
  7. Jul 11, 2009 #6
    it would be negative though wouldnt it? Since the springs energy is opposing that of gravity, cost 180 = -1?

    I ended up getting the answer i used the following:
    Et = Eg + Ee

    I left x as the unknowns and it ended up being quadratic, i then solved it for x to be 0.071 m, as the book says.

    Im having a real hard time with this stuff, not sure why.
     
  8. Jul 11, 2009 #7

    cepheid

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    It wasn't necessary to solve a quadratic. If you had done what I suggested in post #4 and set the kinetic energy to zero (because the mass has stopped moving), then you would have had:

    mgx = ½kx2

    2mg/k = x ​
     
  9. Jul 12, 2009 #8
    Thank you for your help.
     
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