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Work Done by Tension of Mass on a string

  1. Dec 14, 2009 #1
    1.A 2.0 kg mass is spun in a horizontal circle at the end of a 0.66 m long string at a speed of 12 m/s. What is the work done by tension when the mass has gone halfway around the circle?

    A) 576 J B) 1810 J C) 904 J D) 0


    2. Relevant equations
    Fc=mv^2/r and w=f*d



    3. The attempt at a solution
    Well I first calculated the centripital force using the first equation. (2kg)(12m/s^2)/.66 and got 4.36N. The I used work=f*d and got 8.17. This is not even an answer. Any sugestions?
     
  2. jcsd
  3. Dec 14, 2009 #2
    The first of your results computes a force but doesn't include include a path length. Also the math is way off 288/0.66= 4.36?????
     
  4. Dec 14, 2009 #3
    Oh hahaha calculator error. I did calculate the path to be about 4.15 m so the answer would be B 1810 J. W=436 * 4.15=1810. Does this sound about right?
     
  5. Dec 14, 2009 #4
    The question states the displacement is 1/2 a rev.
     
  6. Dec 14, 2009 #5
    Remember, there is a dot product in the work equation.
    W=F dot D

    If you haven't had dot products yet, you've had...

    W=FD Cos(theta)

    Where theta is the angle between the force and displacement vector.

    With circular motion, what is the direction of the force vector?
    what is the direction of the displacement vector?
     
  7. Dec 14, 2009 #6
    I knew there was something that bothered me about this problem.
     
  8. Dec 14, 2009 #7
    Well the cos 180 is equal to -1. There are no negative choices.
     
  9. Dec 14, 2009 #8
    To the op, what flatmaster is saying in order for work to be done, the force has to be at least somewhat in the direction of displacement. This problem is like moving a mass horizontally in a gravitational field. We were gonna get there, but now that the cat has been sprung loose....
     
  10. Dec 14, 2009 #9
    So are we saying that the work would be zero since the force is in the opposite direction of displacement?
     
  11. Dec 14, 2009 #10
    Not opposite but perpendicular. Think about it: if work were done by tension the speed of a ball on a string would dramatically slow with every revolution. The reason I dragged you thru the calculations is that unless you can punch numbers correctly, this is an exercise in futility--you may have the right approach but screw up simple calcs, causing you to doubt your reasoning. I can't tell you how often I see this as a teacher. Many of my courses are calculator taboo for this very reason. Ball park estimates to test your results are important. You mucked up both the force and the displacement computations. Its like the old carpenters saw, measure twice, cut once.
     
  12. Dec 14, 2009 #11
    Thank you for your help!!
     
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