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Calculate the work done by tension

  1. Jul 9, 2013 #1
    1. The problem statement, all variables and given/known data

    Consider the following arrangement
    Calculate the work done by tension on 2kg block during its motion on circular track from point A to point B.

    2. Relevant equations



    3. The attempt at a solution

    We know that work done by a force is product of force and displacement.
    We know the displacement of point of application as 4 m. How to find the work done by the tension as it is not constant it is variable.
     

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    Last edited: Jul 9, 2013
  2. jcsd
  3. Jul 9, 2013 #2

    haruspex

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    Conservation of work will do the integration for you. Think about the changes in potential and kinetic energy of the two blocks. (Don't assume the system will be at rest when point B is reached.)
     
  4. Jul 9, 2013 #3
    how?
     
  5. Jul 9, 2013 #4
    You need a couple of steps to solve the problem

    1 Consider the two blocks and the string together as the system.Apply work energy theorem.Work done by tension will cancel out.This will give you the kinetic energy of the 2 kg block when it reaches point B.

    2 Now,consider the 2 kg block in isolation.Again apply work energy theorem.This time the work done by tension will be equal to the change in the mechanical energy of the 2 kg block .Note that the work of the normal force by the circular track on the 2 kg block will be zero.

    Hope this helps
     
  6. Jul 9, 2013 #5

    haruspex

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    Not directly. It will give the total kinetic energy. You then need to work out the ratio of the speeds of the two blocks when 2kg is at point B. From the diagram, I believe 2kg should moving horizontally at that point, so you can deduce this ratio from the geometry and a bit of calculus.
     
  7. Jul 9, 2013 #6
    can you show how to calculate the ratios of velocities with geometry and calculus..?
     
  8. Jul 9, 2013 #7
    Well,I meant the same :smile: .
     
  9. Jul 9, 2013 #8

    haruspex

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    Consider the movement after reaching point B. Let X be the ground-level point directly below point A. If x is the distance of the 2kg block from X, find an equation relating x to the height of the 1kg block that's valid for x>=3.
     
  10. Jul 10, 2013 #9
    haruspex,i beg to differ.Why do we need calculus to get the work done by tension ? Yes,the tension is variable but we are not required to find tension ,rather work done by tension .

    Work done by tension when the 2 kg block moves from A to B =[itex] \int_{0}^{\pi R/2} T dx [/itex],where T is variable.

    This work done by tension will cancel when we consider the two blocks and the string together as a system and apply work energy theorem.Now the speed of the two blocks will remain same.From this we can deduce the kinetic energy of 2kg block when it reaches B.There is no need to consider any ratio.

    Next,the work done by tension is simply equal to the change in mechanical energy of the 2kg block between points A and B .

    Please let me know if i am missing something obvious .
     
  11. Jul 10, 2013 #10
    how can u say that their velocities would be equal?
     
  12. Jul 10, 2013 #11

    ehild

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    The 2 kg block moves along the circle, so its speed is Rdθ/dt. It pulls the string, the length of the string between point O and the block can we obtained with simple geometry at any position θ (ignoring the size of the pulley). The total length of the string is unchanged, so the speed of the 1kg block is dL/dt.

    ehild
     

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  13. Jul 10, 2013 #12
    After that what to do?
     
  14. Jul 10, 2013 #13

    ehild

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    After what? Have you the expression of dL/dt in terms of theta?

    ehild
     
  15. Jul 10, 2013 #14
    YES ,dL/dt=Rdθ/dt.BUT THAT'S SAME AS V=RW HERE W DENOTES ANGULAR VELOCITY.
     
  16. Jul 10, 2013 #15

    haruspex

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    No. At B, the 2kg block is travelling horizontally. The string is, presumably, still taut, so makes a straight line diagonally up to the pulley. The speed of the 2kg block will not be the same as the speed of the string over the pulley.
     
  17. Jul 10, 2013 #16

    haruspex

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    No. See my reply just above to Tanya Sharma.
    I wasn't sure it was supposed to be an arc of a circle. It doesn't need to be. Just has to be sufficiently smooth, starting vertical and finishing horizontal.
     
  18. Jul 10, 2013 #17

    ehild

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    From the OP:
    ehild
     
  19. Jul 10, 2013 #18
    so what is the final equation to calculate the velocity?
     
  20. Jul 10, 2013 #19

    ehild

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    Try to figure it out. First find the length of string between the 2kg block and point O. Then take the derivative with respect to time.

    ehild
     
  21. Jul 10, 2013 #20
    first the length is 1m then 4m.
     
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