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When should i apply energy conservation?

  1. Sep 1, 2010 #1
    [Two ball of mass 60g are attached with mass less rubber thread and held in vertical position as show in figure.
    In this position length of rubber thread is 40cm and it is not stretched. The upper sphere is slowly raised vertically upward until the lower ball just becomes unsupported by ground. At this time length of thread is 1m. The rubber thread exerts a force which is proportional to its extension.

    Q. What is force constant of rubber thread ? Ans 1N/m


    Attempt 1

    On Applying work energy theorem
    x = 60 cm
    Mgx = 1/2kx^2
    K= 2N/m

    Attempt 2
    x = 60 cm
    On Applying Newtons Equatns
    Mg = kx
    k = 1N/m

    Doubt: Why is first Attempt wrong.
     

    Attached Files:

  2. jcsd
  3. Sep 1, 2010 #2
    Because energy is not conserved. While you raise the ball you exert force on it, which means you are giving the system energy. Unless you calculate that energy, it won't work. The

    Conservation of energy is NOT just the potential energies of the ball and the spring! There is also the speed the ball gains and the energy you use to stop that speed in the end.
     
  4. Sep 1, 2010 #3
    Then wouldn't you need to know that final=initial speed? Since if the ball's speed changes at all from it's initial vertical position it will have Gained energy.
     
  5. Sep 1, 2010 #4
    So if ball is not raised slowly and there is some non zero vertical velocity we can not apply energy conservation.
     
  6. Sep 1, 2010 #5

    vela

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    You can always apply conservation of energy if you can account for where all the work goes. That's not the same as saying whether that approach would be fruitful, however. In this case, it wouldn't tell you anything useful.

    In your attempt, you didn't apply the principle correctly which is why you got the wrong answer. You have to calculate the work performed by all forces on the ball and set that equal to the change in kinetic energy. When you lift the ball, you have the upward applied force, gravity, and the spring all acting on the ball. If you do it correctly, you'll end up with 0=0 or find you don't have the correct information to solve the problem.

    The key piece of information is that "the lower ball just becomes unsupported by ground." That's not easily interpreted in terms of energy, but it's straightforward to comprehend in terms of forces. That's why the force approach is the way to go for this problem.
     
  7. Sep 1, 2010 #6
    Vela , wow! I could never say it ins such a clear way (but then again I'm 18 and still have a lot to learn!)
     
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