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Conservation of energy proof

  1. Oct 26, 2004 #1
    This is driving me nuts.

    A stone with weight w is thrown vertically upward into the air from ground level with initial speed v(i). If a constant force f due to air drag acts on the stone throughout its flight, (a) show that the maximum height reached by the stone is

    h = v(i)^2 / 2g(1+f/w)

    and (b) show that the stone's speed just before impact with the ground is

    v = v(i) (w-f/w+f)^1/2

    I've been working through these for ages and I'm just lost. I know I need the conservation of energy equation 1/2mv^2 = mgh and probably the kinetic motion equation 2a(delta x) = v(f)^2 - v(i)^2, but I'm just not sure what to do with them.
  2. jcsd
  3. Oct 26, 2004 #2
    find the stone's acceleration using newton's 2nd law.
  4. Oct 26, 2004 #3
    Huh. I'm getting closer. But I still have no idea how I'm supposed to relate the conservation of energy and kinetic motion equations. (It's probably something really obvious, and I'll feel like an idiot for not realizing it sooner, but I've been staring at these for so long...)
  5. Oct 26, 2004 #4
    what is the acceleration that you have got ?
  6. Oct 26, 2004 #5
    Well, F = ma so a = F/m...and the force in question would be the initial velocity. But I'm not at all sure how that helps...
  7. Oct 26, 2004 #6
    If upward direction is taken to be positve than using newton's 2nd law you get
    -f+(-W)=ma but since W=mg than m=W/g
    then you get a = -g(1+f/w) and using this constant acceleration substistute into
    [tex]V^2=u^2+2as[/tex] to get the max height. becareful with the sign.
  8. Oct 26, 2004 #7
    ...right. I'm going to sound like an idiot, but could you go through that accelleration in a bit more detail? I'm still a bit confused.
  9. Oct 26, 2004 #8
    [tex]\Sigma\vec{F}=m\vec{a}[/tex] ,right /
    two downward forces which are the stone weight and the drag force act on the stone when it is moving upward ,right ?
    if the upward direction is chosen to be positive, then downward direction must be negative. then we have -W for the stone weight and -f for the drag force. negative signs indicate the direction and W and f are their magnitudes.
    you know that the magnitude of the stone weight is given by W= mg W, m and g are all positive because i am talking about magnitudes.
    then substitute into the equation above , you will get what i get.
  10. Oct 26, 2004 #9
    I think I've managed it now. Thanks a lot. :)
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