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Homework Help: Newton's laws and incline surfaces

  1. May 22, 2009 #1
    1. The problem statement, all variables and given/known data

    A box is given an initial velocity of 5m/s up a smooth 20 incline surface . The distance the box travel before coming to rest is?

    2. Relevant equations

    I can't solve it correctly , I can't get the idea of this question

    3. The attempt at a solution

    x= ?

    F = ma
    mg cos ? = a

    tan-1 20 = 87.14
    180 = 90 + 87.14 + ?
    ?= 180 -90-87.14

    (mg cos 2.9 = ma ) /m
    g cos 2.9 = a
    a= 28.03

    x = 25 /( 28.03 )2
    x= 0.45 m

    My answer is so wrong !
    But i can't figure out how to solve this problem
    1. The problem statement, all variables and given/known data

    2. Relevant equations

    3. The attempt at a solution
  2. jcsd
  3. May 22, 2009 #2

    Doc Al

    User Avatar

    Staff: Mentor

    Presumably a 20 degree incline?

    What's the component of the weight acting down the incline?

    Not sure what you're doing here. I presume that the angle is given as 20 degrees.

    You have the wrong angle (and trig function).

    Sanity check: The acceleration due to gravity is only 9.8 m/s^2, so how can the acceleration parallel to the incline be greater than that?

    Find the correct acceleration and that kinematic equation will work.

    Read this to learn how to handle inclined planes: http://www.physicsclassroom.com/Class/vectors/u3l3e.cfm" [Broken]
    Last edited by a moderator: May 4, 2017
  4. May 22, 2009 #3
    Presumably a 20 degree incline?

    the question didn't mention the word "degree" or " ْ "
    it's exactly as I wrote it .
    So I toke 20 as the slop

    x=? vi=5 vf=0 a=?

    F = ma
    (m g cos20 = m a) / m
    9.8 cos20 = a

    x=1.37 m
    Which is also wrong

    180 = 90+20+?
    180 -90-20=70ْ

    F= m a
    (m g cos 70 = m a)/m
    9.8 cos 70 = a

    x=3.37 m
    I think this one is the right one,but not sure.

    thanks for the link was so useful
  5. May 22, 2009 #4

    Doc Al

    User Avatar

    Staff: Mentor

    I would take it to mean a 20 degree angle to the horizontal.

    It's wrong because you took cos20, which is the component perpendicular to the incline. You need the parallel component.

    This time you used the correct acceleration (which is g sin20), but you made a small error in your final calculation--probably a typo. (Redo the very last step.)

    Other than that minor problem, you got it right.
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