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Force and Energy of a block on a spring as it compresses.

  1. Mar 29, 2016 #1
    1. The problem statement, all variables and given/known data

    Suppose is a 10kg block sitting at rest on the ledge of a building 20 meters tall. The 10kg block applies a Force of F = ma = 10kg(10m/s^2) = 100N. The potential energy with reference to the ground would then be PE = mgh = 10kg(10m/s^2)(20m) = 2000J.

    If the 10kg block is then gently placed on a 20m tall spring scale of spring constant k = 10N/M

    How far does the spring compress?
    Determine the amount of energy in the fully compressed spring.
    How much force is the block applying to the spring when the spring is fully compressed?



    2. Relevant equations

    F*d = mgh
    F*d = (1/2)kx^2
    F = kx



    3. The attempt at a solution

    Answer to question 1)

    F = kx
    F/k = x
    mg/k = x
    (10kg)(10m/s^2) / (10N/m) = x
    10 meters = x

    So the spring compresses 10m when the 10kg block is gently placed on the spring.


    Answer to question 2)

    Gravitational PE = F*d = mgh = (10kg)(10m/s^2) * 10m = 100N*10m/s^2 = 1000J

    Spring PE = (1/2)kx^2 = (1/2)(10N/m)(10m)^2 = 500J



    Answer to Question 3

    1st guess:

    kx = F = 100N


    2nd guess:

    F*d = (1/2)kx^2
    F(10m) = 500J
    F = 500J/10m = 50N


    3rd guess:

    (1/2)(base)(height) = (1/2)(Δx)(Δy) = (1/2)(Distance)(Force) = (1/2)(10m)(F) = 1000J
    F = 2(1000J)/(10m) = 200N


    Don't know if any of this is correct. I don't fully understand the energy and force and whether or not potential energy even has any part in determining the energy and force. Or how spring potential energy determines the energy and force. Thank you for any reply.
     
  2. jcsd
  3. Mar 29, 2016 #2
    to attempt such 'events' draw a free body diagram of both situations and then decide if you wish to analyse the situation in detail or initial and final equilibrium states of the block and the spring.
    1. initially at the ledge the body is in equilibrium- one does transfer it gently- without disturbing the equilibrium but now the spring can not support it as it is not compressed - as time advances the body will move down with some velocity under the action of force (downward weight- K.x) if x is the displacement.
    2. naturally spring force will put it finally to zero velocity but at the point where mg=k.x , the body will have some K.E. left over so it will go further down and one can expect oscillations about a mean point of the spring and finally it will die out in real springs- so some part of energy is going to be dissipated .

    3. if one takes the final position of the spring and adds spring energy 1/2 . k.x^2 + P.E. of the body at the new height = total energy and gets a value which is not equal to the initial potential energy - one may wonder where the energy is gone?
    i think the above is a fair analysis of the physical situation but i may be corrected if something is overlooked.
     
  4. Mar 30, 2016 #3

    haruspex

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    I'm not certain how to interpret that. I can hold a block on top of an uncompressed spring, gently, then let it go, or I can let it down gently so that when I let go nothing moves. drvrm's interpretation (the former of those) is probably right, but it leads to the physically unrealistic situation that the spring can be compressed to nothing.
     
  5. Mar 31, 2016 #4
    one can try 'gently' pushing the spring down without generating any perceptible velocity- but it will be a cumbersome process taking a lot of time and the losses in the spring can not be avoided for 'real springs'.
    sorry, i could not follow the essence of the comment quoted below
    moreover as the problem has been set-up by the author -is there any experimental back -up /references?
     
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