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Homework Help: Conservation of Energy and maximum height

  1. Mar 4, 2007 #1
    Hi, here's another problem I'm a little stuck on:

    A block of mass M slides up an incline with an initial speed Vi in the position shown. (The figure is of a block sliding up an incline at an angle "theta" with a final height of "H."

    Part A: If the incline in fritionless, determine the maximum height H to which the block will rise, in terms of the given quantities and appropriate constants.

    Part B: If the incline is rough with a coefficient of sliding friction "u" (mew), determine the maximum height to which the block will rise in terms of "H" and given quantities.

    For part A I just used Conservation of Energy and got the maximum height "H" to be H = (((Vi)^2)/(2g)), which is the right answer.

    For part B, I know I have to use Work = (Change in K + Change in U). So I have:

    (umg(cos (theta))) * distance = ((1/2)((Vf)^2) - (1/2)((Vi)^2)) + (mgHf - mgHi). The masses cancelled out and I assumed that the initial height and initial and final velocities equaled 0, so then my answer ended up being Hf = ucos(theta) *d, which isn't the right answer...

    Any suggestions?
    Last edited: Mar 4, 2007
  2. jcsd
  3. Mar 4, 2007 #2
    Why did you make the initial speed zero?
  4. Mar 4, 2007 #3
    When I tried to work the problem through the first time, I couldn't get the Vi to reduce to anything or cancel out, so then I tried to go back through with the assumption that Vi = 0 because there's no Vi in the final answer.
  5. Mar 4, 2007 #4
    It seems that Vi is a given in the problem. How do you reason that there is no Vi in the final answer?
  6. Mar 4, 2007 #5
    I have that the final answer is h = H/(1 + ((u)/tan(theta))) (given by teacher.) I'm having trouble getting to that answer. If I don't assume that Vi is zero, I have (using Work done): (u = coefficient of sliding friction.)

    W = delta K + delta U. (umg (cos(theta)) * d) = ((1/2)m((Vf)^2) - (1/2)m((Vi)^2)) + (mgHf - mgHi). Mass cancels, therefore: ug(cos(theta)) * d = ((1/2)((Vf)^2) - (1/2)((Vi)^2)) + (gHf - gHi). Initial height and final velocity are 0 (?) so then: ug(cos(theta)) * d = -(1/2)((Vi)^2) + gHf...

    And from there...I'm stuck. I don't know how to make "d" and "Vi" cancel.
  7. Mar 5, 2007 #6
    What is h?
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