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Motion of a block on a steep surface

  1. May 28, 2016 #1
    1. The problem statement, all variables and given/known data
    The block was given an initial velocity up the surface with an angle of 45 degrees to the ground. Calculate the ##\frac{t_1}{t_2}## with ##t_1## being the time it took to get to the highest point up the hill and ##t_2## the time it took to get down. In both cases the frictional force with coefficient m= 0.2 was acting on it.
    2. Relevant equations
    3. The attempt at a solution
    My tried solution is uploaded below in form of a picture.
    I have made free body diagrams but am stuck at getting the equation that involves time so that i can divide the two times. Any hints?
     

    Attached Files:

  2. jcsd
  3. May 28, 2016 #2

    PeroK

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    Think generally: if a body moves a distance ##d## how does the time depend on the acceleration?
     
  4. May 28, 2016 #3
    ##d=v_ot + \frac{1}{2}at^2##
    Im still kinda stuck...
     
  5. May 28, 2016 #4

    PeroK

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    The body goes the same distance up the slope as it does down the slope. Down the slope is easy, as ##v_o = 0##.

    You can use a trick for motion up the slope: the time taken to move a distance ##d## from an initial velocity to rest at deceleration ##a## is the same as the time to move a distance ##d## from rest at acceleration ##a##. This avoids needing to calculating ##v_0##.

    To see this, think about distance as the area under a velocity/time graph.
     
  6. Jun 18, 2016 #5
    I dont understand the trick, so in the new equation for the motion up the slope i would not have ##v_o## but would have some new ##a## while d stays the same right? How would i make that work?
     
    Last edited: Jun 18, 2016
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