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Sliding Block Down Ramp

  1. Sep 23, 2009 #1
    1. The problem statement, all variables and given/known data

    A block of mass m=1.62 kg slides down a frictionless surface (see below). The block
    is released at a height h=3.91 m from the bottom of the loop. Assume the bottom of the
    ramp indicated between the dashed lines in the diagram is a segment of a circle with R = 1.5m. Neglect air resistance.

    a) What is the force of the inclined track on the block at the bottom (point A)?
    b) What is the force of the track on the block at point B?
    c) At what speed does the block leave the track?
    d) How far away from point A does the block land on level ground?
    e) Sketch the potential energy U(x) of the block. Indicate total energy on the sketch.


    2. Relevant equations

    Are explained below

    3. The attempt at a solution

    A) I think, it's just the opposite force to the blocks weight at point a, and with no angles, it's simply mg = 15.89 N

    B) the normal force at point b now has to take into account the component of gravity in the y direction so would the force the track exert on the block be mgsin45? If so the answer is 10.32 N

    C) the speed at the end of the ramp corresponds to the kinetic energy it gained, from the potential energy at the top = mgh. however h in this case = (h - distanceAB). with a little bit of algebra I worked out the distance from the top to point B in the y direction (so new h) is 3.55 m. so PE mgh = 56.36 J = 1/2mv2. rearranging yields v = 8.34 m/s

    D) this is where i'm having problems. I know for a normal projectile question, if I'm given a projectile angle, distance from the ground and initial veloctiy I can calculate time in air, and eventually distance travelled. I have 2 of the 3, and cannot find a launch angle. Keep in mind this is not a straight ramp I'm dealing with, so I didn't think trig could help me. Do you think it is safe to just assume launching at a 45 degree angle? Please help if there is a way to do it.

    E) I think the potential energy graph looks a lot like the ramp itself, except it starts at the origin and jumps up immediately after. Am I right here?

    I only needed help on D) and E), but please even if I'm right and you know I am, confirm my other answers, it will help a lot for the learning process. Thanks in advance.
     

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  3. Sep 23, 2009 #2

    kuruman

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    (a) Is incorrect. What is the acceleration of the mass and in what direction is the net force?
    (b) Is incorrect. What you say would be the case if the acceleration were zero. It is not because the velocity is changing.
    (c) Appears correct. Method is OK - I did not check your numbers. I assume that when you say distanceAB you mean the vertical rise of the mass, not the arc length from A to B.
    (d) Draw a line tangent to the circle at point B. The angle this line makes with respect to the horizontal is the launch angle. Does it look like 45 degrees?
    (e) Where is your zero of potential energy? If it is at point A, there is no sudden jump. The potential energy is proportional to the height above A at any distance x so it does look like the ramp, pretty much, but no sudden jump.
     
  4. Sep 23, 2009 #3
    How then would you find acceleration? In drawing fbds and applying newtons laws, the Normal force only acts in the y direction (for point A) and so there would be no components in the other direction. How is the Normal force not opposite mg in this case?

    And what would be the correct method for b) at point B then?
     
  5. Sep 23, 2009 #4

    gabbagabbahey

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    As the name suggests, the Normal Force, is always normal (perpendicular) to the surface, whereas gravity is always downward...
     
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