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Conservation of Energy/Angular Momentum

  1. Jan 10, 2007 #1
    1. The problem statement, all variables and given/known data
    The problem is "A wooden block of mass m slides with an initial speed of vi up a straight incline that makes an angle of theta with the horizontal. If the incline is rough with a coefficient of kinetic friction uk, determine the maximum height it rises in terms of h found in part a (which was H = vi^2/2g)"

    2. Relevant equations
    conservation of energy
    You can only use m (mass of block), vi (initial velocity), theta, and g (gravity)

    3. The attempt at a solution
    KE at the bottom = PE at the top + Work.
    mgh + Fd = 1/2mv^2
    h = [mvi^2 - ukmgcos(theta)d]/mg
    Problem is, I don't know what d is and it can't be part of my final answer. I know it's the distance that the block travels up the ramp, but how do I figure that out?
    Also, since it says in terms of h found in part a, does that mean I set everything equal to vi^2/2g?

    How would I do this with a hoop with radius R?
  2. jcsd
  3. Jan 10, 2007 #2
    So, you correctly applied the conservation of mechanical energy. :smile:
    Could you right d as a function of h and [itex]\theta[/itex]? :rolleyes:

    The hoop will have in the end not only linear velocity but also angular velocity. Apply again conservation of mechanical energy, but this time consider also the kinetic rotational energy. :tongue:
  4. Jan 10, 2007 #3
    Okay this is going to sound really confusing but am I on the right path for the block one?

    I got to the point where gh = 1/2(vi)^2 - ukg(cot(theta))h
    At this point, do I plug in what I got for H in part (a) since it wants it says "in terms of h found in part a"? What exactly are they asking me to do?
  5. Jan 11, 2007 #4
    Nevermind. I was being stupid and forgot they are two different H's
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