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Find Maximum extension in the spring-Work Energy Problem

  1. Oct 6, 2012 #1
    1. The problem statement, all variables and given/known data

    The spring block system lies on a smooth horizontal surface.The free end of the spring
    is being pulled towards right with constant speed [itex]v_{0}[/itex]=2m/s.At t=0 sec the spring of constant k=100 N/cm is unstretched and the block has a speed 1m/s to left.Find the maximum extension of the spring is

    2. Relevant equations



    3. The attempt at a solution

    Initial Energy of the block spring system is [itex]\frac{1}{2}m(1)^{2}[/itex]

    Since the spring is being pulled to the right,the spring stretches and a force towards right starts acting on the block.This force retards the block and speed of block decreases to zero and then the block speeds up towards right.The spring stretches to maximum when the block's speed becomes 2m/s .

    Energy of the system at the instant of maximum extension should be [itex]\frac{1}{2}m(2)^{2}[/itex] +[itex]\frac{1}{2}k(x)^{2}[/itex]

    How should I find value of x? Please help
     

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    Last edited: Oct 6, 2012
  2. jcsd
  3. Oct 6, 2012 #2
    Think how the system will evolve.
    1. At t = 0, the block is going towards left with 1 m/s.
    2. As time progresses, the spring pulls the block so the block's speed (still towards left) decreases.
    3. At some time t = trest, the block stops momentarily.
    4. After trest, the block starts moving towards right.
    5. Eventually, the block moves towards right with speed 2 m/s.

    When does the spring have maximum extension?
     
  4. Oct 6, 2012 #3
    The spring will have maximum extension when the block is moving towards right with speed 2 m/s . But work is being done on the system . So how should we calculate the value of x ?
     
  5. Oct 6, 2012 #4
    Nope. Consider the block. The only force on the block is the one due to the spring. We should apply the principle of conservation of energy for the block.

    Suppose the speed of the block is [itex]u[/itex] at some instant and the extension in the spring is [itex]x[/itex] What is the total energy of the system?
     
  6. Oct 6, 2012 #5
    Sourabh N ....If we consider Block as the system then the only external force acting on it is that due to spring .But if we consider Block+Spring as our system then the force is due to the external agent moving the spring with constant velocity 2m/s towards right.

    We should consider Spring+Block as the system ...The energy equations written pertain to block-spring system.
     
  7. Oct 6, 2012 #6

    Doc Al

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    View things from a frame moving at 2 m/s to the right, so that the free end of the spring is stationary.
     
  8. Oct 6, 2012 #7
    If things are viewed from a rightward moving reference frame with speed 2 m/s .Then
    the block is seen moving towards left with velocity 3 m/s.At the maximum extension the block moves with velocity 2m/s towards right hence is stationary from the moving frame .

    Initial Energy = [itex]\frac{1}{2}m(3)^{2}[/itex]

    Energy of the system at the instant of maximum extension
    = [itex] \frac{1}{2}k(x)^{2}[/itex]

    Solving this we get x=6 cm .

    Is this correct ?? Please reply
     
  9. Oct 6, 2012 #8

    ehild

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    Correct! Nice work.

    ehild
     
  10. Oct 6, 2012 #9
    Thanks to both Doc Al and ehild...

    How do we solve the problem from the reference frame of the ground ??
     
  11. Oct 6, 2012 #10

    Doc Al

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    Yes, that's the right idea.

    What's the mass of the block?
     
  12. Oct 6, 2012 #11
    mass of the block=4kg
     
  13. Oct 6, 2012 #12

    Doc Al

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    Ah, yes, it was specified in the diagram. Good!
     
  14. Oct 6, 2012 #13

    ehild

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    I show one way, seeing from the ground. I start from Newton's Law.
    Let be the position of the block x(t). x(0)=0.
    The other end of the spring is at y(t). y(0)=Lo, unstretched length, and y(t)=Lo+2t. The length of the spring is L(t)=y(t)-x(t)=Lo+2t-x(t), and the stretching is ΔL=2t-x(t)

    The force exerted on the block is F=kΔL=k(2t-x(t).
    ma=k(2t-x(t)) ma+kx=2kt, or md2x/dt2+kx=2t

    If you learn calculus you will know how to solve such equations. The general solution is the sum of an SHM and a motion with 2 m/s to the right. The SHM has zero displacement at t=0, so the solution is x(t)=Asin(wt)+2t.
    w=√(k/m)=50 1/s. The velocity is Awcos(wt)+2. As the velocity of the block is -1 m/s at t=0, Aw+2=-1, A=-3/50=0.06 m.

    At the end x(t)=-0.6 sin(50t)+2t. Substitute into the expression for ΔL and find when ΔL is maximum.

    edited!
    ehild
     
    Last edited: Oct 6, 2012
  15. Oct 6, 2012 #14
    I have understood most part except 2 things

    y(t)=Lo+2t

    What is the term 2t ?

    The SHM has zero displacement ..why?
     
  16. Oct 6, 2012 #15

    ehild

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    t is time and 2 is the velocity of the right end of the spring.
    The displacement of the SHM is zero at t=0. We can choose it. I edited the previous post.


    ehild
     
  17. Oct 6, 2012 #16
    ehild...Thank u very much
     
  18. Aug 1, 2013 #17

    ehild

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    The general solution can be written as the sum of the general solution of the homogeneous equation and a particular solution yp of the inhomogeneous equation: x(t)=Acos(wt)+Bsin(wt)+yp. You can find the attachment useful to find a particular solution.

    ehild
     

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