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Two methods for finding spring constant k

  1. Mar 4, 2014 #1
    Two methods for finding spring constant "k"

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

    A box of mass 5 kg is made to rest on a vertical spring. The box compresses the spring by 10 cm, find the spring constant.

    2. Relevant equations

    I felt this can be solved by two methods :

    1. Force of gravity = Outward force of spring

    2. The decrease in the potential energy = elastic potential energy stored in the spring.

    3. The attempt at a solution

    The solution in my notes has the first method. So,

    Solution by 1

    F = mg
    F = -kΔx ... Δx = 10 cm = 0.1 m

    => mg = kΔx
    => (5)(9.8) = k(0.1)

    => k = 490

    --- --- --------------

    Solution by 2

    Initial potential energy of box = mgh
    Potential energy at the compressed mg(h-0.1)
    Elastic potential energy of the spring = (0.5)k(Δx)2

    => mgh - mgh(h-0.1) = (0.5)k(Δx)2
    => mg(0.1) = (0.5)k(Δx)2
    => (5)(9.8)(0.1) = (0.5)k(0.1)2

    => k = 980

    ------

    So which of these methods is wrong and why?
     
    Last edited: Mar 4, 2014
  2. jcsd
  3. Mar 5, 2014 #2
    The second method is wrong, because "2. The decrease in the potential energy = elastic potential energy stored in the spring." The sum of the gravitational potential energy and the elastic potential energy is the total potential energy. Now your statement is equivalent to "total potential energy is constant". There is no such law.
     
  4. Mar 5, 2014 #3
    But even if you use conservation of energy (including the kinetic energy), the total energy of the system is conserved.

    Since the kinetic energy is zero in both instances, the potential energies are conserved right?

    Normally its

    K1 + U1 = K2 + U2

    Here, K1 and K2 are zero. Leaving

    U1 = U2.
     
  5. Mar 5, 2014 #4

    TSny

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    Suppose you are initially holding the box at rest at the top of the spring with the spring uncompressed. You then lower the box onto the spring until the spring compresses to the point where you can let go of the box and the spring supports the box at rest.

    How many forces act on the box while it is being lowered? Does your energy equation account for the work done by all of these forces?
     
  6. Mar 5, 2014 #5

    BvU

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    Nope. The exercise says "is made to rest on.." If you just let go of the box with the spring extended, it will oscillate up and down around the 10 cm.

    At the lowest point Ekin = 0 again and the compression is 20 cm. All potential energy from gravity has been converted into spring compression energy.

    At the highest point Ekin = 0 and all the initial energy is potential energy from gravity again.

    At the 10 cm point only half is spring compression, the other half is kinetic energy.

    [edit]T was faster - again!
     
  7. Mar 5, 2014 #6
    That is true.

    Others have commented on this, and I will from a slightly different perspective. Write the full equation for conservation of energy: ##K_1 + U_1 = K_2 + U_2##. Now, as you say, assume that ##K_1 = K_2 = 0##. This obviously means ##U_1 = U_2##. This is a quadratic equation, so it will have two roots. Does that really mean that the mass-spring system has two equilibrium points? Or does that mean something else?
     
  8. Mar 5, 2014 #7
  9. Mar 5, 2014 #8
    How is it a quadratic? We know the value of x. The unknown variable is 'k'
     
  10. Mar 5, 2014 #9
    Hi Chestermiller,

    The post by nasu in that link says that when you "place it gently" the energy of the of the weight-spring system is not conserved. What gives? I read the other posts but I thought in such a system energy is conserved at every point in time - only with translation from one form to another.
     
  11. Mar 5, 2014 #10
    To get the mass to the equilibrium position, you have to lower the weight gradually so it doesn't overshoot. If you just let it go at the top, it will develop kinetic energy and overshoot to twice the equilibrium compression. Then it will bounce back up.

    To lower it gradually, there is another force that has to be taken into account, and this force adds a work term to your energy balance. This is the force that you have to be applying as the mass descends gradually to its equilibrium position. As the mass descends, you are reducing the force you are exerting as the spring assumes more of the load. When the mass gets to its equilibrium position, the force you need to be applying will decrease to zero. At the time that the mass reaches the equilibrium position, the work you will have done will be -mgx. This is the missing energy in your energy balance. If you include this, you will get the same answer as the force equilibrium result.

    Chet
     
  12. Mar 6, 2014 #11
    Thank you Chestermiller! That clears it all. :-)
     
  13. Mar 6, 2014 #12
    Interpret that as if k were known, and x were unknown.
     
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