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Maximum extension in spring connecting two masses.

  1. Apr 15, 2015 #1
    1. The problem statement, all variables and given/known data
    In the figure shown below all surfaces are friction-less. Find the maximum extension in the spring(in meters) , if the blocks are initially at rest and the spring is initially in its natural length.

    Details and Assumptions:

    F=30N
    k=700N/m
    m=5kg





    2. Relevant equations


    3. The attempt at a solution


    51912951bc1a0bd9bbd05146be988a7122b7.jpg

    Let the displacement of ##2m## be ##x## and that of ##m## be ##y##

    Applying work energy theorem on block 1 (##2m##) and on block 2 we get

    ##Fx-\frac { 1 }{ 2 } k{ (x+y) }^{ 2 }=\frac { 1 }{ 2 } (2m){ v }_{ 1 }^{ 2 }##.......(1)

    ##3Fy-\frac { 1 }{ 2 } k{ (x+y) }^{ 2 }=\frac { 1 }{ 2 } (m){ v }_{ 2 }^{ 2 }##.............(2)

    ##{ X }_{ COM\quad }=\frac { ml }{ 3m } ## Position of CoM initially.

    ##{ X }_{ COM }=\frac { -2mx+m(L+y) }{ 3m } ## Position of CoM finally.

    ##\triangle { X }_{ COM }=\frac { my-2mx }{ 3m } ##

    ##{ a }_{ com }=\frac { 2F }{ 3m } ##

    ##{ v }_{ com }=\sqrt { 2\frac { 2F }{ 3m } \times \frac { my-2mx }{ 3m } } ##..............(3)

    Elongation in the string will be maximum when velocity of block 1 would be zero.

    Putting this in (1) I got

    ##3x=35(x+y)##.....(4)

    On solving eq(2) and substituting eq(4) I got

    ##V_{2}=\sqrt(36y-12x)##.....(5)

    As ##m_{1}v_{1}+m_{2}v_{2}=(m_{1}+m_{2})V_{com}##

    So ##3V_{com}=v_{2}##



    Using eq(5) and (4) I got

    ##-y=3x##



    Am I correct till here?




     

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    Last edited: Apr 15, 2015
  2. jcsd
  3. Apr 15, 2015 #2
    I am going to solve the equations. But I think I have done a mistake.:confused:
     
    Last edited: Apr 15, 2015
  4. Apr 15, 2015 #3

    haruspex

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    I don't understand your equations 1 and 2. Aren't you counting the energy in the spring twice, once for the distance F moves and once for the distance 3F moves? Just consider total energy.
    What is the relationship between the velocities at maximum extension?
     
  5. Apr 15, 2015 #4
    I separately applied work energy theorem for block1 and 2.

    At maximum compression velocities of the blocks along the line joining them should be equal but this is a case of maximum extension. I think the extension would be maximum when velocity of mass ##2m## is zero.
     
  6. Apr 15, 2015 #5

    haruspex

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    I don't see how you can do that. You have counted the total spring energy in each, so if you were to add them together you'd have too much. And there's no obvious way to apportion the spring energy between them.
    So? Why wouldn't it be the same result for the velocities?
     
  7. Apr 15, 2015 #6
    Work done by all forces is equal to the change in the kinetic energy of the system. Taking block of mass ##2m## as a system. So

    Work done by F is ##+Fx##
    Work done by spring ##-\frac{k(x+y)^{2}}{2}## -ve sign because force exerted by the spring is in opposite direction of displacement.

    Change in kinetic energy is ##mv_{1}^{2}## as the initial velocity was zero. Equating then I got eq(1) similarly I got eq(2).

    What am I doing wrong?
     
  8. Apr 15, 2015 #7
    And what is the work done by the force on the other block?
     
  9. Apr 15, 2015 #8
    Kinetic energy is ½mv2, right? So, according to your statement, the blocks should not have any kinetic energy when the spring has reached maximum extension.
     
  10. Apr 15, 2015 #9
    If consider the other block to be moving rightward then
    Work done by all the forces on that block would be ##+3Fy-\frac{k(x+y)^2}{2}##.
     
  11. Apr 15, 2015 #10
    I was making a mistake. I got that. The extension would be maximum at the moment when the relative velocity of the blocks along the line joining them would be zero.
     
  12. Apr 15, 2015 #11

    haruspex

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    k(x+y)2/2 is the total work done on the spring. You have both forces doing that much work on the spring.
     
  13. Apr 15, 2015 #12
    Oh! I think I got it.

    Let us consider block 1,block 2 and spring as a system.

    So work done by all the forces is equal to the change in the kinetic and potential energy of the system.

    ##Fx+3Fy=\frac{k(x+y)^2}{2}+\frac{3mv^2}{2}## (extension would be maximum if velocities of blocks are equal).

    Is work done by spring on block 1 ##-\int _{ 0 }^{ x }{ k(x+y)dx } ##?
     
  14. Apr 15, 2015 #13

    haruspex

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    I see no basis for claiming that, and I see no need to plug in any assumption about it.
    I feel there must be an easier way, but the only way I have been able to solve it so far is by writing out the ODEs for each block's motion and solving. There are some simplifications that can be made, e.g. by writing z = x+y.
     
  15. Apr 15, 2015 #14
    Why is it, exactly, that you cannot use F = -kx? Is it because of the masses? The masses don't diminish the forces being applied on them, they just decrease the acceleration, right?
     
  16. Apr 15, 2015 #15

    haruspex

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    Of course you can use F = -kx (in this case, tension = k(x+y)). Where has anyone said otherwise?
     
  17. Apr 15, 2015 #16
    Oh. Okay. I guess no one said otherwise. When you said ODE I thought everything I knew probably went out the window.
     
  18. Apr 16, 2015 #17

    ehild

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    I think, that is the easiest method. You can introduce new variables, the position of the CM and the difference between the position of the masses, u=y-x. The equations of the motion separate to two single variable ODE-s. The acceleration of the CM is constant, and the distance between the masses performs SHM between the relaxed length and maximum length of the spring.
     
  19. Apr 16, 2015 #18
    I revised the chapter of Center of Mass last night and I found that this can be solved even in a more easy way.

    Let us consider two blocks and the spring as a system.

    Net external force acting on the system is ##2F##

    So by Newton's second law acceleration of the CoM of the system

    ## a_{com}=\frac{2F}{3m}##


    Consider a reference frame which is moving with the CoM of the system as it is a non-inertial frame of reference so we have to include Pseudo force on the blocks.

    sss.jpg

    Let the displacement of block of mass ##m## be ##x## towards right and that of 2m be ##y## towards left as seen from the CoM of the system.

    Work done by all the external forces on the right block is ##(3F-\frac{2F}{3})x##


    Work done by all the external forces on the left block is ##(\frac{4F}{3}+F)y##

    This work done is equal to the change in the potential and kinetic energy of the system. If the extension has to be maximum then velocities of the block wrt CoM of the system should be zero. So the final kinetic energy of the system is zero.



    ##(3F-\frac{2F}{3})x+(\frac{4F}{3}+F)y= \frac{k(x+y)^2}{2}##


    Solving this equation I got ##x+y=0.2## which is the right answer.:smile:
     
  20. Apr 16, 2015 #19

    haruspex

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    Congratulations.
     
  21. Apr 16, 2015 #20
    Thank you!!!:-p
     
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