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Conservation of energy question

  1. Nov 12, 2007 #1
    1. The problem statement, all variables and given/known data


    A 5.0 meter long rope hangs vertically from a tree right at the edge of a ravine. The ravine is 3 meter wide, and a woman wants to use the rope to swing to the other side. She runs as fast as she can, grabs the rope, and swings over the ravine.
    a) As she swings what energy conservation takes place?
    b) When she is directly over the far edge of the ravine, how much higher is she than when she started?
    c)Given your answers in part a and part b, how fast must she be running when she grabs the rope in order to swing all the way across the ravine?

    2. Relevant equations
    ke= 1/2mv^2
    Ug= mgy

    3. The attempt at a solution

    a) kinetic to potential
    b) 1/2 mv^2= mgy
    y= 1/2mv^2/ mg
    y= mv/mg
    But we dont know V
  2. jcsd
  3. Nov 12, 2007 #2
    you do know the length of the rope and the X she has to travel. try using that to find your answer.


    change the numbers, but you have the general idea.
    Last edited: Nov 12, 2007
  4. Nov 12, 2007 #3
    yeah i know i know them but formula would i use for that
  5. Nov 12, 2007 #4
    i dont know the original speed that she grabs it
  6. Nov 12, 2007 #5
    yeah but how do i find theta for that
  7. Nov 12, 2007 #6
    you know two parts of the triangle.

    try the pythagorean theorum
    Last edited: Nov 12, 2007
  8. Nov 12, 2007 #7
    yeah but is it a triangle ?
  9. Nov 12, 2007 #8
    wont the 5 meters swinging change the length of the hypoteneuse
  10. Nov 12, 2007 #9
    what is the huypotenuse?
  11. Nov 12, 2007 #10
    would be c of pythagoran theorum hich would be the value of 5.83 meters
  12. Nov 12, 2007 #11
    you should look at the picture and apply it to your problem. the point of the triangle is where the lady ends up.
  13. Nov 12, 2007 #12
    yes i see what you mean by that but what is 2m, is x the 5 meters or is 2m
  14. Nov 12, 2007 #13
    2m in your problem is the rope
  15. Nov 12, 2007 #14
    okay so now im lost 2m is the long one and x is the shorter distance 3 meters, how do i find height
  16. Nov 12, 2007 #15
    no, x is your height. the width of the ravine is perpendicular to the height and the length of the rope when it's hanging straight down.
  17. Nov 12, 2007 #16
    yes i know so assuming i want to use the diagram you gave to solve for X i would take C^2= A^2 + B^2 where C is going to be height nad A is going to length of rope and B is going to be distance across ravine
  18. Nov 12, 2007 #17
    is the rope vertical when you get across?
  19. Nov 12, 2007 #18
    no i would say it is not because it is attached to one tree, like if i use pythagorean theorum i get a value of 5.83 m so would my height be 0.83 meters
  20. Nov 12, 2007 #19
    no. if you look closely, the horizontal line is the distance across the ravine, and the 2m is your rope before you go. what are the other two lines?

    hint: the hypotenuse in the picture is 2m
  21. Nov 12, 2007 #20
    well simplly is the rope is 5 meters then by what you siad 2m is equal to 5 meters.... so my height would be square root of rope^2- Distance ^2
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