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Working Spring

  1. Apr 11, 2005 #1
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    The setup shown is at rest. If the spring is compressed 6in. how high (h) above Equi. will the 16lb. weight reach? Max Velocity?

    I'm using U_1,2=.5k(x1)^2-.5k(x2)^2 (Potential Energy)

    I know it is a simple problem but there is something i am not catching. so far I have tried this:

    16lb.(h+6)+[.5(15)1.06in.-.5(15)6in.]=0
    1.06in=distance weight alone compresses spring
    I know this is WRONG. Where am I WRONG?
     
  2. jcsd
  3. Apr 11, 2005 #2
    Conservation of energy is your answer :

    kx²/2 is the potential energy when a spring is compressed over a distance x. Then, the spring 'fires off' the object. Ofcourse this object will fly upwards but gravity will eventually stop this motion. So, the initial energy will be needed in order to 'fight' against gravity. In the conservation law, you will need to evaluate both kinetic and potential ebergy at the biginning and at the end of the motion, so you will also need to bring in gravity. Do you know the gravitational potential energy ? If so, your work is done...


    marlon
     
  4. Apr 11, 2005 #3
    T1+V1=T2+V2
    T1=.5mv^2=.5(.497)0=0
    T2=.5(.497)*?
    V1=mgh=.497(32.2)0=0
    V2=.497(32.2)h

    This is what you are saying right?
    If so what goes in the "?" spot?
    If not, Why not?
     
  5. Apr 11, 2005 #4
    No, you are wrong.

    beginning : the spring is compressed 6 inches and the kinetic energy is ZERO

    end : the object is a distance h above the equilibrium point and the kinetic energy is zero.

    So you have : kx²/2 = mgh and solve this for h

    marlon
     
  6. Apr 11, 2005 #5
    (15lb/in.*6in)/2=16lb*h
    h=16.87in
    Actual answer is 45.2in

    We are both missing something.
     
  7. Apr 11, 2005 #6
    are the units of the spring constant newton/meter ??? because if so, you will need to convert your units into meters and kilograms...

    marlon
     
  8. Apr 11, 2005 #7

    dextercioby

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    He needn't,marlon,we do...:tongue2: Units are okay fixed,so the calculations should go easily.Maybe if he translated from US to SI we'd be able to follow his #-s


    Daniel.
     
  9. Apr 11, 2005 #8
    Indeed, :rofl:

    However, isn't the spring constant given in #/m

    what the hell is this # ??? :rofl:

    marlon
     
  10. Apr 11, 2005 #9
    You guys going to help or not? :mad:
     
    Last edited: Apr 11, 2005
  11. Apr 11, 2005 #10

    dextercioby

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    If i'm not mistaking,they use it as pounce avoirdupoids per inch.They should use a unit for force,but they use a unit for mass.I think that's incorrect,or even dumb.

    Daniel.
     
  12. Apr 11, 2005 #11
    k=2626.8N/m
    x=.1524m
    weight collar=71.04N
    g=9.81m/s^2
     
  13. Apr 11, 2005 #12
    Look; the solution is definitely the way i presented it to you. There must be something wrong with the units.

    marlon
     
  14. Apr 11, 2005 #13
    Never Mind. I Solved It With Out You All.
     
  15. Apr 12, 2005 #14
    yeah, whatever, :rolleyes:
     
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