1. Not finding help here? Sign up for a free 30min tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Maximum vertical height

  1. Dec 14, 2004 #1
    Determining maximum vertical height

    A block slides down a curved frictionless track and then up an inclined plane. The mass of the block is the point mass m. The coefficient of friction between the block and the incline is μk = 0.30 Use energy methods to find the maximum vertical height up the plane that the block reaches for θ = 45°, and h = 0.65 m.

    Mgymax = mgh-Wfk
    Mgymax = mgh- μkCos θ ∆x
    ymax = h- μkCos θ ∆x

    This is about how far I have come. Not sure if it is right though.

    I am not sure if ∆x should actually be ∆y or if either should be in the problem.

    I am seeking a direction on what to do next to complete this problem. I appreciate any help.
     
    Last edited: Dec 15, 2004
  2. jcsd
  3. Dec 15, 2004 #2

    Nereid

    User Avatar
    Staff Emeritus
    Science Advisor
    Gold Member

    It seems that something is missing from your description of the problem ... presumably [tex]\theta[/tex] is the angle of the inclined plane to the vertical (or is it horizontal?); but what's h?
     
  4. Dec 15, 2004 #3
    Well i suppose that theta is the angle of the incline with the horizontal. Let's suppose that h is the vertical height from which we start on the curved trajectory...

    At the point where the incline starts we have from energy-conservation :

    [tex]mgh = \frac{mv^2}{2}[/tex]

    Then we move up the plane until vertical distance [tex]h_{max}[/tex]

    energyconservation yields :
    [tex]\frac{mv^2}{2} = mgh_{max} - {\mu}N{\Delta p}[/tex] Where the delta p denotes the travelled distance of the object ON the inclined plane. ofcourse, using trigon. we know that [tex]h_{max} = \Delta p * sin(45)[/tex] so we can eliminate delta p and write all as a function of h_max...

    But personally i think that there should be a least one parameter that you can vary in order to solve the extremum-problem...you sure about this question...


    regards
    marlon
     
  5. Dec 15, 2004 #4

    Tide

    User Avatar
    Science Advisor
    Homework Helper

    Not to pick nits, but ...

    Is that vertical height as opposed to horizontal height? :wink:
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?



Similar Discussions: Maximum vertical height
  1. Maximum Height (Replies: 1)

  2. Maximum height (Replies: 9)

Loading...