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Velocity down an incline

  1. Dec 25, 2014 #1
    1. The problem statement, all variables and given/known data
    A 75KG object, goes down a slide that is 10m tall. What is the speed of the object at the halfway mark and the speed at the bottom of the slide. The object starts at rest and there is no friction.

    2. Relevant equations
    v2f - v2i = 2as


    3. The attempt at a solution

    attached pic of where I am at on the problem
    incline:velocity.jpg
     
  2. jcsd
  3. Dec 25, 2014 #2

    ehild

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    You have the equation: v2f - v2i = 2as. Note that s is the displacement along the slope now. How long is it in terms of the angle if the height of the slope is 10 m?
     
  4. Dec 25, 2014 #3
    sinθ=(10/h)
    h equaling the hypotenuse = the slope
     
  5. Dec 25, 2014 #4

    ehild

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    No, h=10 m, and s is the hypotenuse. Look at the picture.
     
  6. Dec 25, 2014 #5
    ok so,

    sin(θ)=(10/s)
     
  7. Dec 25, 2014 #6

    ehild

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    Isolate s and use it in the equation v2f - v2i = 2as.
     
  8. Dec 25, 2014 #7
    so to isolate s, do you mean something like this?

    s=10sin(θ)
     
  9. Dec 25, 2014 #8

    ehild

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    Something like that, but not quite. Is it more clear if I say solve the equation sin(θ)=(10/s) for s?
     
  10. Dec 25, 2014 #9
    more like this?
    s=10/sin(θ)

    I'm sorry, this stuff has me so confused... These online classes aren't really meeting my expectations. :(
     
    Last edited: Dec 25, 2014
  11. Dec 25, 2014 #10

    ehild

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    Yes.
    Determine the acceleration a now.
     
  12. Dec 25, 2014 #11
    well, acceleration = gsin(θ)
    so, (9.81)sin(θ)=a
     
  13. Dec 25, 2014 #12

    ehild

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    Use a and s in the formula v2f - v2i = 2as.
     
  14. Dec 25, 2014 #13
    well v2i = 0 since the object starts at rest
    v2f = 2((9.81)sin(θ)) (10/sin(θ))
     
  15. Dec 25, 2014 #14

    ehild

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    Good! Evaluate.
     
  16. Dec 25, 2014 #15
    I'm not sure if this is right, but it's a shot in the dark...

    v2f =2 ((9.81)sin(θ) (10/sin(θ))
    umm??? sin(θ) cancels out on each side leaving, 2(9.81)(10) = 196.2
    v2f = 196.2
    vf = √196.2 = 14m/s

    is this even close?
     
  17. Dec 25, 2014 #16

    ehild

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    It is correct. That is the speed at the bottom of the slope when the whole length is covered. What is the speed, when the object is halfway down the slope?
     
  18. Dec 25, 2014 #17
    lol i would say half of that? Thats just me guessing...
     
  19. Dec 25, 2014 #18

    ehild

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    NO!!!!!!!!!!!!! what is the new distance s(half) if it is half the previous one?
     
  20. Dec 25, 2014 #19
    haha I knew it wouldn't be that easy ;)

    v2f = 2 * 9.81(10 * .5)
    vf = √98.1
    vf = 9.9m/s
     
  21. Dec 25, 2014 #20

    ehild

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    You did it!
     
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