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Homework Help: Conservation of Mechanical Energy - I can't see what I'm doing wrong

  1. Sep 20, 2010 #1
    1. The problem statement, all variables and given/known data
    A motorcyclist is trying to leap across a canyon. The first cliff is 70m high, she wants to jump to the cliff 35m high. She is travelling 38m/s. Ignoring air resistance, find the speed with which the cycle strikes the ground on the other side.

    I think (and am very worried by this) that my algebra is incorrect as I do not get the same answer as the textbook which is v = 46.2m/s

    2. Relevant equations

    1/2mv2+ mghf = 1/2mv02+mgh0

    3. The attempt at a solution

    1/2mv2+ mghf = 1/2mv02+mgh0

    v2+ ghf= u2 + gh0
    v2+ (9.8) (35) = 382 + 9.8 (70)
    v2= 1444 + 686 (-343)
    v2= 1787
    v = 42.3m/s

    I know the equation is incorrectly reduced. The textbook has

    (1) 1/2mv2+ mghf = 1/2mv02+mgh0

    to:

    (2) vf = √(v02 + 2g (h0 - hf)

    I'm worried because it might mean I'm having a problem with fundementals - I would have just forgotten I used to be great at this stuff.

    I have basically two questions:
    - Is the final equation (2) not from (1) but just a form of the kinematics equation - so (1) cannot be reduced to (2)
    - or what algebraic thing am I doing wrong? I cannot see how 2g happens as (1/2) / (1/2) = 1... I can't see what I can do to get 2g from equation 1.

    Thanks as always to this forum!
    1. The problem statement, all variables and given/known data



    2. Relevant equations



    3. The attempt at a solution
     
  2. jcsd
  3. Sep 20, 2010 #2
    where is your factor of 2 in your 2nd step of part 3?
     
  4. Sep 20, 2010 #3
    I'm not exactly sure with what you mean. How would I get a factor of 2? Perhaps there should be a factor of 2 but how do I get that? Once again, I am not sure how there is '2g' in the second equation...

    :/
     
  5. Sep 20, 2010 #4
    1/2mv2+ mghf = 1/2mv02+mgh0

    v2+ 2ghf= u2 + 2gh0
     
  6. Sep 20, 2010 #5
    I suppose I need the working from equation (1) to it's manipulation to be equation (2) - as I just don't conceptually understand how the g's don't cancel if the mass cancels....
     
  7. Sep 20, 2010 #6
    Okay. Thank you for showing me that. I still don't understand why that is the case... I don't want you to waste your time showing me as you've already taken the time :) but is there a url I can look at to see why the '1/2's don't cancel but instead = 2 ?? I'm just a bit stuck I'm afraid :)
     
  8. Sep 20, 2010 #7
    Divide the whole equation by 1/2, m is a common factor so we cancel it, but g is not

    I suggest you to improve your algebra first
     
  9. Sep 20, 2010 #8
    Oh yes, that was never in dispute - I believe I prefaced the problem with that statement. I'll take a look. Thanks for the insight tsw99 :)
     
  10. Sep 20, 2010 #9
    okay I worked out how we get to...

    v2 + 2ghf = u2 +2gh0
    v2 = u2 + 2gh0 - 2ghf

    Can anyone show me why +2gh doesn't cancel by showing me what does happen step by step? Alternatively can someone give me a url website/video that I can look at that will show me the princples as to how this equation ends up as that in my question above?
     
  11. Sep 20, 2010 #10
    Horizontally the bike moves with a constant 38m/s(ignoring air resistance).Vertically the bike gains speed due to gravitational attraction.You can find the speed gained by energy considerations (as tried above)or by using an equation of motion.Whatever method is used it must be remembered that the initial vertical velocity is zero(this is implied in the question).From this mg(H1-H2)=(mv^2)/2.From this v^2=2g*35.To find the speed on landing add the horizontal and vertical speeds vectorially by using Pythagoras.Is this the method you was using?
     
    Last edited: Sep 20, 2010
  12. Sep 20, 2010 #11
    I just looked more closely at your method and your mistake seems to be in line two of your attempt.To quote tsw 99 "where is your factor of 2?"
     
  13. Sep 20, 2010 #12
    Thanks so much for your reply. As I said above, I am drawing an algebraic blank from how one gets from the first equation in my first post, to the second equation which is:
    v2 = u2+2g (h0-hf)

    I know it's elementary, but I know how we get to v2 +2ghf =u2+ 2gh0 but not to v2 = u2+2g (h0-hf)

    So I guess I am asking someone to humour me and show me how to get to the final equation. I know it's basic algebra and I'm revising as we speak and have been for the last few hours - but it's just my brain has dumped it in an irretrievable place at this stage...

    So.. in conclusion - if someone could show step by step how I can manipulate the first equation to reach the final equation I would be most grateful.
     
  14. Sep 20, 2010 #13
    basically I cannot see why 2g would not cancel ... and if I knew what I needed I'd recognise how absurd that statement is - but if someone could condescend to show me how step by step I'd be so, so appreciative. Sorry, just having one of those self-confidence draining moments :)
     
  15. Sep 20, 2010 #14
    oh my gosh. I must have been in a haze of stupidity. I can't believe I couldn't do this. Fresh eyes makes a difference. Thanks to those hanging in there. Oh my - how embarrassing.
     
  16. Sep 20, 2010 #15
    You agree with tsw 99 that:
    v2+2ghf=u2 +2gho.
    To make v2 the subject of the equation get everything else to the other side of the equation by subtracting 2ghf from both sides.Now you have:
    v2=u2+2gh0-2ghf.
    Now you can put the numbers in and calculate it from there.Try it,it will give you the right answer.Now look again at the equation and you will see that 2g is common to to the last two terms on the right hand side so instead of writing 2gh0-2ghf you can write:
    2g(ho-hf).This makes the final equation a little bit neater and saves a small step in the calculation.
     
  17. Sep 20, 2010 #16
    Look at the times of our two posts above Hemingway.You just beat me to it.:wink:
     
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