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Box coming to a stop on a ramp

  1. Feb 13, 2015 #1
    1. The problem statement, all variables and given/known data

    You are helping your friend move, using a ramp to move boxes from the ground to the moving truck. You give a 26kg box a shove so it moves at 1.4m[PLAIN]https://s3.lite.msu.edu/adm/jsMath/fonts/cmmi10/alpha/144/char3D.pngs [Broken] at the bottom of the ramp. The angle that the ramp makes with the ground is 31[PLAIN]https://s3.lite.msu.edu/adm/jsMath/fonts/cmsy10/alpha/100/char0E.png. [Broken] The coefficients of static and sliding friction are 0.39 and 0.1, respectively.

    a) For how much time does the box slide up the ramp?

    [PLAIN]https://s3.lite.msu.edu/adm/jsMath/fonts/cmr10/alpha/144/char01.pngt= [Broken]


    b) How far up the ramp does the box slide?

    [PLAIN]https://s3.lite.msu.edu/adm/jsMath/fonts/cmr10/alpha/144/char01.pngx= [Broken]


    c) When the box stops sliding up the ramp, does it remain stopped or begin sliding back down?




    d) Explain how you determined your answer for part (c):



    2. Relevant equations

    u = a/g
    a = delta v / delta t
    other friction equations


    3. The attempt at a solution

    .39*9.8*26 = 99.372
    .1*9.8 = 25.48

    I believe you need to sin31 and set up an equation that compares the two frictional forces together to get an answer but I am having trouble setting it up.
     
    Last edited by a moderator: May 7, 2017
  2. jcsd
  3. Feb 13, 2015 #2

    Simon Bridge

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    You set up the equation from a free body diagram.
    Note: the numbers you supplied are meaningless without some words to go with them... you also need units.
     
  4. Feb 14, 2015 #3
    You have to think what makes it to stop.
    Add all the forces that bring it to halt.
    Use SUVAT equations for constant force/acceleratipn.
     
  5. Feb 14, 2015 #4

    Suraj M

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    You can also try another method to test your concept understanding.
    Consider the initial energy imparted and equate to work done by friction ans gravity; and change in potential.
     
  6. Feb 15, 2015 #5
    Second attempt: a = 9.81*sin(31)+.1*9.81*cos(31) = 5.89m/s^2

    (Vi)^2 = 2as

    s = ((1.4)^2)/(2*5.89) = .17m distance travled

    Still giving me an incorrect answer..
     
  7. Feb 15, 2015 #6

    haruspex

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    Looks right to me (.166 to 3 digits). Do you know what the answer is supposed to be?
     
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