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Work/Kinetic Energy

  1. Nov 1, 2009 #1
    1. The problem statement, all variables and given/known data

    A ball of mass 0.805 kg is tied to the end of a string of length 1.61 m and swung in a vertical circle.

    Question 1: During one complete circle, starting anywhere, calculate the total work done on the ball by the tension in the string. Ans: 0J

    Question 2: During one complete circle, starting anywhere, calculate the total work done on the ball by gravity. Ans: 0J

    Question 3: Repeat part (a) for motion along the semicircle from the lowest to the highest point on the path. Note: Problem 1 is part a. Ans: 0J

    Question 4: Repeat part (b) for motion along the semicircle from the lowest to the highest point on the path. Ans: unknown




    2. Relevant equations

    W= Fnet * d * cosx




    3. The attempt at a solution

    Well the first three questions were kind of easy, since in one complete circle, you get back to the same point, so there is no work done really. The last one is a bit hard. I thought it was 0 also, but it isn't. My free body diagram is a verticle line, at the bottom is Force of gravity, and at the top is Force of tension. Can someone give me a hint on the 4th question?
     
  2. jcsd
  3. Nov 1, 2009 #2
    If the circle has a radius of r, then the difference between the highest point on the path and the lowest point on the path is 2r.

    What is the potential difference between those two points as far as gravity is concerned?
     
  4. Nov 1, 2009 #3
    So you are saying multiply the work by 2 because of semi circle?

    Because the only forces are Tension and gravity, the force, F = mg = 0.805(9.8)=7.889
    So W=F*D = 7.889(1.61)=12.7J

    Like that? So multiply that by 2?
     
  5. Nov 1, 2009 #4
    The ball, at the highest point on the circle, will have a certain potential energy. Call it [tex]U_{high}[/tex]

    The ball, at the lowest point on the circle, will have a certain different potential energy. Call it [tex]U_{low}[/tex]

    The difference between the two values is equal to the amount of work done by gravity. The path doesn't matter.
     
  6. Nov 1, 2009 #5
    So how do I actually figure that out?
     
  7. Nov 1, 2009 #6
    What is the potential energy at [tex]U_{high}[/tex]?
    What is the potential energy at [tex]U_{low}[/tex]?

    If you don't know, then do you know what the potential energy of an object of mass m is a distance h from the floor?
     
  8. Nov 1, 2009 #7
    [tex]U=mgh[/tex]
     
  9. Nov 1, 2009 #8
    I think we didn't touch on this topic yet, thats why I am having a hard time. We just covered Work/Kinetic Energy. Anyways, I don't think its difficult is it?

    Ok, so Uh = -7.889
    Ulow would be 0 correct, because its kinetic at the bottom, and pntential at the top.
     
  10. Nov 1, 2009 #9
    Did you cover that the work (W) done by a force [tex]\left(\textbf{F}\right)[/tex] is given as such?

    [tex]W=\textbf{F}\Delta\textbf{x}[/tex]

    Gravity is the force here, given by:

    [tex]\textbf{F}=-mg\hat{\textbf{y}}[/tex]

    So in this case, [tex]\textbf{x}=h\hat{\textbf{y}}[/tex]

    So substitute those into the work equation:

    [tex]W=-mg\hat{\textbf{y}}\Delta h\hat{\textbf{y}}[/tex]
    [tex]=-mg\Delta h[/tex]

    Now what is [tex]\Delta h[/tex]?
     
  11. Nov 1, 2009 #10
    Hmm... I worked it out, and got an answer of -25.4J, and its correct. According to what I know, I just doubled my answer. And since its moving up, its negative.
     
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