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Need help figuring out this problem

  1. Jul 11, 2012 #1
    1. The problem statement, all variables and given/known data

    6. A stunned Batman finds himself in a vault on a ledge. In an attempt the trap him the Joker runs out of the vault while the door closes. Batman fires a grappling hook into the ceiling at an angle, θ, attached to a line of length, L , and manages to pull himself off the ledge swinging down to the floor below. At the very bottom of the swing he lets go of the line to the grappling hook in an attempt to slide the rest of the way out of the door. Find an equation for how far Batman slides, x, as a function of L, θ and μk, the coefficient of kinetic friction between Batman and the floor. If the door is 9.00 meters away and the length of the line is 6.00 meters, the angle 60 degrees and the coefficient of friction 0.250 did Batman make it out? If he did, how fast was he sliding when he went through the door?

    2. Relevant equations

    fk (x)=E1 -E2
    started out with that equation

    3. The attempt at a solution

    E1=mgL(1-cosθ)
    E2=1/2mv2
    fk=μkmg
    so,

    μkmg(x)=mgL(1-cosθ)-1/2mv2

    then,
    x=L(1-cosθ)/μk - v2 /2μkg


    thats what i got for the equation but im not sure its right because when i try to use it to find how far he slides i cant because i dont know what v is. im really lost.
     
  2. jcsd
  3. Jul 11, 2012 #2

    TSny

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    "How far Batman slides" means "How far Batman slides before coming to rest."
     
  4. Jul 11, 2012 #3
    yea still lost lol
     
  5. Jul 11, 2012 #4

    TSny

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    Batman slides until he comes to rest. What is the value of v when an object is at rest? So, what is the value of v that you should use in your equation?
     
  6. Jul 11, 2012 #5

    tiny-tim

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    Welcome to PF!

    Hi Karhitect! Welcome to PF! :smile:
    fine so far … that's the kinetic energy at the bottom of the swing :smile:
    this is rather confusing …

    what is your E2 supposed to be? :confused:

    if it's the energy at the start of the slide, then it's just E1, and v isn't relevant

    if it's the energy at the door, then v is the answer :wink:
     
  7. Jul 11, 2012 #6

    TSny

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    You should find that Batman slides more than 9.0 m before coming to rest. Therefore, he will make it out. If you want to know his speed as he goes through the door, what value would you use for x and what symbol would you solve for using your equation?
     
  8. Jul 11, 2012 #7
    ohh duh lol that makes sense...i found that he slides 12 m ???

    i think you would use 9m for x

    would i use my same equation only solving for v this time?
     
  9. Jul 11, 2012 #8

    TSny

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    Yes, 12 m. And you have the right idea for finding v at the door. :smile:
     
  10. Jul 11, 2012 #9
    ok so then i got vf= mgL(1-cosθ)-μkmg

    and i got 7.35 m/s ??
     
  11. Jul 11, 2012 #10

    TSny

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    You have x = L(1-cosθ)/μk - v2 /2μkg

    Rearrange: v2 /2μkg = L(1-cosθ)/μk - x

    You'll still need to multiply through by 2μkg and then take a square root.
     
  12. Jul 11, 2012 #11
    i think i did something wrong cuz i got a negative number...when i brought the 2μkg over i brought it next to the x...then i square rooted the whole thing. but because i got a negative number i couldnt square root...
     
  13. Jul 11, 2012 #12

    TSny

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    v2 /2μkg = L(1-cosθ)/μk - x

    When you multiply both sides by 2μkg, the whole right hand side gets multiplied.

    v2 = 2μkg [L(1-cosθ)/μk - x]
     
  14. Jul 11, 2012 #13
    ohhh i see i was thinking that you had to minus the whole thing by x...soooo i got 3.83m/s
     
  15. Jul 11, 2012 #14

    TSny

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    Looks Good! :smile:
     
  16. Jul 11, 2012 #15
    Thank you sooo much!!!
     
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