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Spring potential energy

  1. May 9, 2015 #1
    1. The problem statement, all variables and given/known data

    A ball bearing of mass m = 50.0 g, is sitting on a vertical spring whose force constant is 120.0 N/m. The initial position of the spring is at y = 0 m


    a) The spring is compressed downward a distance x = 0.200 m. From the compressed position, how high will the ball bearing rise? How high does the ball bearing rise above the equilibrium position at y = 0 m?

    b) What is the kinetic energy of the ball at the moment it is released from the spring? (assume all energy converted into kinetic energy)

    c) What is the maximum speed of the ball?


    2. Relevant equations
    a) PEs = (0.5)(K)(x)2, h = PE/mg
    c) root[(2)(KE)/m]

    3. The attempt at a solution

    a)PEs = (0.5)(120.0 N/m)(0.200m)2
    = 2.4J
    h = PE/mg
    h = 2.4J/(0.05kg(9.8))
    h = 4.89m
    h = 4.89m - 0.200m (the distance of spring compression)
    h = 4.69 m

    b) KE = 2.4J - mgh
    KE = 2.4 J -(0.050kg)(9.8)(0.200m)
    KE = 2.3J

    c) v = root [2(2.3)/(0.050kg)]
    v = 9.60 m/s
     
  2. jcsd
  3. May 9, 2015 #2

    Simon Bridge

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    Another "lets use memorized equations" approach?
    Anyway - did you have a question? If you just want someone to check your work - how about a fellow student?
    Technically you are training to work on problems where nobody knows the answer ... who will you ask then?

    One way to troubleshoot your own work is to write down your reasoning at each stage.
     
  4. May 9, 2015 #3
    I'm sorry if you felt that way, I'm legitimately self-motivated and working hard to achieve decent knowledge in physics. I'm uncertain what made you feel bad and I apologize if you think I only used helpers to check my work which isn't true. Again, I might just leave this forum if you don't seem comfortable with me asking quetions. :) Thank you for all help you previously helped with though
     
  5. May 9, 2015 #4
    I see absolutely nothing wrong with your posts, Glen. Perhaps Simon is simply suggesting that using this forum to ask specific questions about specific things you are confused about might be the most beneficial use of your time? Anyways, your posts are clear and you show all your work! That's quite enough for me.
     
  6. May 9, 2015 #5
    Hiya Glenboro, first off the answer to part b you should be able to do without any calculations really. If the ball is at a stand still (as you are holding it) then the second you release it, it will also be at a stand still, so if it is not moving at all how much kinetic energy (i.e. energy due to motion) can it have? Also since the ball is at an equilibrium position at y=0, we can actually ignore the effects of gravity, as it is a constant force and has already been accounted for in the equilibrium. Then you only have the force of the spring to consider, and this is then a simple harmonic oscillator problem. So if you compress a spring by x, how much will it extend past its equilibrium position?
    Finally par c is where you actually have to do some calculations, but we know that the ball will have maximum velocity at the equilibrium position (experiencing a nonzero acceleration oposite to the direction of travel for all other points). Therefore, the amount of WORK released by the sping from y=-0.2 to y=0 is the kinetic energy of the ball.
    You should be able to do the problem now, hope this helps. Just remember in the future that just writing down equations won't help you. Always spend a bit of time first thinking about what the question is asking and how the system works (draw a graph if you need too).
     
  7. May 9, 2015 #6

    Simon Bridge

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    post #1 does not contain a question though. Without a question or some statement of what you need we have to guess.
    please evaluate post #1 against your undertakings in the last thread... nobody expects you to get it perfect right away but a good plan today...

    I do not feel "bad".
    I was not expressing any emotion at all... I was making observations as a way to help you improve yourself.
     
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