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Distance down an incline

  1. Apr 6, 2005 #1
    A 6 kg box slides down a long, frictionless
    incline of angle 34 degrees. It starts from rest at time
    t = 0 at the top of the incline at a height 17 m
    above the ground. Find the distance the box travels during
    the interval 0 < t < 0.6 seconds.
     
  2. jcsd
  3. Apr 6, 2005 #2

    dextercioby

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    What are your thoughts...?What kind of movement does the box have?

    Daniel.
     
  4. Apr 6, 2005 #3
    well the distance would be the speed x time but since i am only given time i do not know. the box is moving diogonally, is there some way i can use the height and the angle to find the distance?
     
  5. Apr 6, 2005 #4

    dextercioby

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    Make a FBD.See that the "driving" force is the tangential component of the gravity force.
    U got a linear accelerated motion with a given acceleration & given initial conditions.It's all done.

    Daniel.
     
  6. Apr 6, 2005 #5
    so i could just use X = Xo +Vot + (1/2) at^2?
     
  7. Apr 6, 2005 #6
    Since dexter gave you so much useful info, let me help you out here. Suppose the incline is downward to the right and the x-axis is along the incline. the y axis is perpendicular to the x-axis. There are two forces acting on the object : gravity and the normal force N.

    The clue is to calculate the components along both directions.

    [tex]ma_x = mgsin( 34°)[/tex]
    [tex]ma_y = -mgcos( 34°) + N = 0[/tex] this 0, because you stay on the incline

    You also know the initial position wtr to this frame of reference : 0 in the x-direction and 0 in the y-direction. The initial velocity is also 0.

    the formula that you will need is [tex]x = gsin(34) \frac{t^2}{2}[/tex]
    Keep in mind that x is NOT the horizontal distance but the distance travelled along the incline...

    regards
    marlon

    edit : to the OP : lose the m's... my mistake o:)
     
    Last edited: Apr 6, 2005
  8. Apr 6, 2005 #7
    thanx marlon, i had sumtin like that just didnt understand the (t^2/2) part
     
  9. Apr 6, 2005 #8

    dextercioby

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    It's really sad that a (probably) prolific future PhD stud in physics would make such an erroneous analysis in this simple problem...

    Daniel.
     
  10. Apr 6, 2005 #9

    dextercioby

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    In Marlon's last equation,the mass should have been absent...

    And,yes,[itex] \frac{t^{2}}{2} [/tex] times an acceleration is typical for uniformly accelerated motion.U should have known that.

    Daniel.
     
  11. Apr 6, 2005 #10
    WRONG correction dexter, the initial velocity IS ZERO in my chosen reference frame, so is the initial position and so is the force in the y direction

    marlon
     
    Last edited: Apr 6, 2005
  12. Apr 6, 2005 #11
    i knew something was wrong because i got an answer like 6 m, and the block couldnt have possibly gone that far in only a fraction of a second. thanx dex
     
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