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Work done by or on a system?

  1. Aug 28, 2014 #1
    Consider a system with a spring and a block. The block moves on a frictionless surface. At the equilibrium position, the speed of the block equals to 0. Now i pull the block by a force F with a displacement of x. Assuming that the speed at the final position is v.
    Can you calculate the energy of the system after the process ?

    If i apply the equation: ΔE= Q+W1 ( with W1 is the work donr on the system )
    Is the above equation equivalent with the eqution: ΔE=Q-W2( with W2 is the work done by the system)

    ( Q is the heat that is transfered to the system)
    If these equations is equivalent, please tell me how to calculate W2 ?
    thank
     
  2. jcsd
  3. Aug 28, 2014 #2

    Doc Al

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    Staff: Mentor

    Is the force constant over that displacement?

    (The work done by the system is just the negative of the work done on the system.)
     
  4. Aug 28, 2014 #3
    I think it is not important if F is constant or not. Because we can represent the work done on the system by W1=∫Fdx.

    In my opinion, W2 is the work done by the system, so it is done by the spring force ( This is only force created by the system) and equals to W2=-1/2kx2.

    If the two equation is equivalent, W1+W2=0. It is clearly wrong because W1+W2=ΔK=1/2mv2.

    This is my confusion
     
  5. Aug 29, 2014 #4

    Doc Al

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    Staff: Mentor

    This is where you are going wrong. If you pull the block with a force F, then the block is pulling you with a force -F. (Newton's 3rd law.) So the work done by the system W2= -∫Fdx.
     
  6. Aug 29, 2014 #5
    Thank you so much Doc Al. Now i know what is the work done by the system.
    Hope that i can learn more from you. Thank again :D
     
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