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Bicycle on a hill Please help

  1. Feb 20, 2007 #1
    1. The problem statement, all variables and given/known data
    You ride your bicycle down a hill. Your total mass is 54 kg. At the top of the hill, your
    speed is 5 m/s. The hill is 1 m high and 101 m long. what is
    your speed at the bottom of the hill?


    2. Relevant equations



    3. The attempt at a solution

    PE= mgh
    (54)(9.8)(10)
    = 5292

    find the degree of the hill i think using tangent?
    tan (10/100) ??

    then solves for v

    KE = 1/2 mv^2


    Ik its something along those lines but im a bit confused with the hill (do i use tangent, etc.) thank you very much, any help as soon as they possibly can is very much appreciated. thanks :-)
     
  2. jcsd
  3. Feb 20, 2007 #2

    Kurdt

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    Well what happens is that at the bottom of the hill all the potential energy will be converted to kinetic energy (disregarding friction and air resistance). What you need to do is set the potential and kinetic energy equal to each other and solve for v. Remember to add on the initial 5m/s.

    To find the angle, it depends on if the 100m relates to the hypotenuse or the distance from the start of the hill to the bottom. model the hill as a right angled triangle and remember your trigonometric ratios.
     
  4. Feb 20, 2007 #3
    KE = 1/2 mv^2
    5292 = 1/2 (54)v^2
    and solve for V?
    and then just add on the 5m/s to that?

    For the length of the hill, i am guessing that it would be referring to the hypotenuse (because that would be the length that the bike travels).
     
  5. Feb 20, 2007 #4

    Kurdt

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    Ok so what trigonometric function do you know that involves the opposite side (10m) and the hypotenuse (100m).

    And yes just solve for v and add the 5m/s on.
     
  6. Feb 20, 2007 #5
    KE = 1/2 mv^2
    5292 = 1/2 (54)v^2
    and solve for V?
    V=196 + 5 = 201
    but what do i do with the hill? it would be Tan-1 (10/100)?
     
  7. Feb 20, 2007 #6

    Kurdt

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    Remember:

    [tex] \sin(\theta) = O/H
    \cos(\theta) = A/H
    \tan(\theta) = O/A [/tex]

    where O = opposite, A = Adjacent and H = hypotenuse.
     
  8. Feb 20, 2007 #7
    Sin = o/h
    10/100
    = .0017
    would that b right? but how do i figure it out now, the speed at the bottom of the hill, once ive got those numbers? thank you i really appreciate all of the help!
     
    Last edited: Feb 20, 2007
  9. Feb 20, 2007 #8

    Kurdt

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    You forgot to take the square root.

    You want to take sin-1 (10/100)
     
  10. Feb 20, 2007 #9
    okay, now when i get both of those numbers what do i do with the angle?
     
  11. Feb 20, 2007 #10
    thanks again
    for taking the time to help me
     
  12. Feb 20, 2007 #11

    Kurdt

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    Oh right. Nothing because you have used conservation of energy instead of kinematic equations so you don't really need the angle. I thought it was part of the question you'd neglected to post.
     
  13. Feb 20, 2007 #12
    so the answer would just be the square root of 201 which would be 14.177
     
  14. Feb 20, 2007 #13

    Kurdt

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    No the square root of 196 and then add the five.
     
  15. Feb 20, 2007 #14
    Okay, thanks very much. just one more quick question...Why don't I do anything with the hill (as in tangent) because it is going down an incline, wouldn't that have an effect on the speed (speed it up)? sorry for all of the questions
     
    Last edited: Feb 20, 2007
  16. Feb 20, 2007 #15

    Kurdt

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    If you were using the kinematic equations you'd have to find the component of force acting parallel to the hill and thus you'd need to use the trigonometric stuff. Since you've considered conservation of energy this doesn't matter because all you need to know is how far vertically the bike travels. this is because gravity is a conservative force.
     
  17. Feb 20, 2007 #16
    Wow! thank you, you have been very helpful!
     
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