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I Why is a spring a conservative force?

  1. May 22, 2016 #1
    Let's say you are holding a bowling ball in the air, and there is a spring on the ground directly below it. You drop the bowling ball and it lands on the spring, compressing it, and then the bowling ball rolls off the spring onto a point P right next to the spring. Wouldn't this do a different amount of work than if you were to just throw the bowling ball directly onto P, bypassing the spring?
     
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  3. May 22, 2016 #2

    Orodruin

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    You will have to be more specific. What would do a different work on what?
     
  4. May 22, 2016 #3
    i agree with the above comment -that more details are reqired
    but as it stands the amount of work will be different as the paths are different and we do not know what type of forces are operating in the next throw- as far as spring action is concerned if it is ideal spring the motion will be holding the conservation of energy.
     
  5. May 22, 2016 #4
    The work done on the bowling ball by gravity and the spring.

    Here are two ways to get to P with ball initially at height h:

    Method 1 (bypass the spring):
    Step 1) Move the ball horizontally so that it is positioned above P (0 work)
    Step 2) Drop the ball (Work = mgh)
    Total work on ball = mgh

    Method 2 (drop ball on spring):
    Step 1) Drop the ball, and the ball lands on the spring. If the uncompressed spring is y tall, the work from this step is: mg(h-y) plus the integral of (mg-ky) from 0 to y.
    Step 2) Move the ball horizontally onto P (0 work if floor is frictionless).
    Total work on the ball = mg(h-y)+mgy-(1/2)ky^2=mgh-(1/2)ky^2
     
  6. May 22, 2016 #5

    Orodruin

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    You are missing the work done by the floor on the ball in order to stop it in the case where it is not dropped on the spring. There is nothing strange going on here.
     
  7. May 22, 2016 #6
    What if the spring is tight enough that the ball never completely hits the ground, and then you push the ball off the spring right when the ball's velocity reaches 0 (i.e. right before the ball starts going back up)?
     
  8. May 22, 2016 #7

    Orodruin

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    This does not change the fact that the force from the floor will do work in the other case.
     
  9. May 23, 2016 #8
    Either way you are dealing with a generalized spring. Take the case where the ball is dropped directly on to the floor. If the floor is undamaged and left without plastic deformation, then it too acted as a spring that stopped the ball. Real materials, like a floor, have some level of elasticity. A floor simply has a different spring constant (elastic modulus) than the actual spring.
     
  10. May 23, 2016 #9
    In Method 1 the rebound of the spring must be accounted for. In Method 2 the floor never does any work unless it deflects (work = force X distance). For that calculation you would need the floor stiffness characteristics (another spring!). In the end, the available energy is in mgh. Without getting into the weeds and tiny details the end result has to add up to mgh for conservation of energy.
     
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