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Dynamics 1: Motion Along a Line

  1. Oct 15, 2016 #1
    << Mentor Note -- poster has been reminded to fill out the Attempt at the Solution part of the Template in future HH threads >>

    1. The problem statement, all variables and given/known data

    An object of mass m is at rest at the top of a smooth slope of height h and length L. The coefficient of kinetic friction between the object and the surface, μk, is small enough that the object will slide down the slope if given a very small push to get it started.

    Find an expression for the object's speed at the bottom of the slope.

    Express your answer in terms of the variables m, μk, L, h, and appropriate constants.

    No clue how to do this.
     
    Last edited by a moderator: Oct 15, 2016
  2. jcsd
  3. Oct 15, 2016 #2

    PeroK

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    Unfortunately, you need to have some idea how to tackle the problem and show us how far you can get on your own before you get stuck. What do you know about motion, acceleration, gravity, kinetic energy etc.?
     
  4. Oct 15, 2016 #3
    This is what I get :
    final KE = initial PE - work done
    ½mv² = mgh - µk*mgLcos(arcsin(h/L)) → multiply by 2/m
    v² = 2gh - 2µk*mgLcos(arcsin(h/L)) → take square root

    v = √(2gh - 2µk*gLcos(arcsin(h/L))
     
  5. Oct 15, 2016 #4

    PeroK

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    You may want to do something about ##\cos(\arcsin(h/L))## but otherwise that looks right.
     
  6. Oct 15, 2016 #5
    What do I do with it ?
     
  7. Oct 15, 2016 #6

    PeroK

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    You could get rid of the trig functions.
     
  8. Oct 15, 2016 #7
    v = √(2gh - 2µk*gLcos(h/L)) ?
     
  9. Oct 15, 2016 #8
    or v = √(2gh - 2µk*gL(h/L))?
     
  10. Oct 15, 2016 #9

    PeroK

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    That makes no sense.
     
  11. Oct 15, 2016 #10
    Still gives me : The correct answer does not depend on: μκ.
     
  12. Oct 15, 2016 #11

    PeroK

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    ##\cos## and ##\sin## are related, which allows you to express the answer in terms of ##h## and ##L## without any trig functions. You don't have to do this.
     
  13. Oct 15, 2016 #12
    Don't really understand what you mean by expressing it in terms of h and L. This is the part im stuck on.
     
  14. Oct 15, 2016 #13

    PeroK

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    You can either use ##\cos^2 + \sin^2 = 1## or go back to your diagram and express ##\cos## in terms of ##h## and ##L## using Pythagoras.
     
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