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Stress / strain

  1. Jan 20, 2010 #1
    1. A mass of 100g is suspended from a light spring and extends the spring by 4cm when the mass in in equilibrium.
    a) Find the spring constant k of the spring?
    b) How much elastic energy is now stored in the spring?
    c) What would be the extension of the spring if a mass of 170g were suspended?
    d) How much extra energy would now be stored in the spring?
    e) Why is more work needed to stretch a spring by 1cm from its unextended length, than is required to stretch it by an extra 1cm when the spring is already extended?



    2. Let deltaX = e
    F=ke
    Ep=1/2Fe
    Stress=F/A
    Strain=e/L
    E=stress/strain
    F=ma

    100g = 0.1kg
    4cm = 0.04m
    Let gravity = -10m/s



    3. a) F=0.1 x 10 = 1N
    k=10N/0.04m = 250N/m

    b) E=1/2 x 10N x 0.04m = 0.2J
    c) F=0.170 x 10 = 1.7N
    e = 1.7N/250N/m = 0.006.8m
    d) E = 1/2 x 10N x 0.170m = 0.85J
    e) I know this holds true from my experience in reality, but when I consider the force extension graph and hook's law, it tells me that up until the elastic limit the same force is required to achieve the same extension. But considering E=1/2 x stretching force x extension, from this equation it is clear that an increase in extension will increase elastic potential energy.


    Could someone kindly check and guide, please?
    Thank you :bugeye:
     
  2. jcsd
  3. Feb 8, 2010 #2
    Is there something wrong with this thread. Many views but no reply. Is it cause it's such easy question?
     
  4. Feb 16, 2010 #3
    Ok. You'll better ignore the above post. I guess I really was under too much stress and strain.
    So, hopefully to put it right:

    mass=0.10kg
    Δx=0.04m

    a) Find the spring constant k of the spring?
    F=ma
    0.10x10=1.0N

    F=kΔx
    k=F/Δx
    1.0/0.04=25N/m
     
  5. Feb 16, 2010 #4
    b) How much elastic energy is now stored in the spring?
    Ep=1/2FΔx
    1/21.0x0.04=0.02J

    c) What would be the extension of the spring if a mass of 170g were suspended?
    F=ma
    0.170kgx10=1.70N
    Δx=F/k
    1.70/25=0.068m

    d) How much extra energy would now be stored in the spring?
    Ep=1/2FΔx
    1/2x1.70x0.68=0.578m
     
  6. Feb 16, 2010 #5
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