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Question I set myself

  1. Aug 3, 2009 #1
    After reading a bit on classical mechanics, I decided to set myself a somewhat simple to most, but quite challenging to me question. I managed to come to a final answer. I think I may have gone a little wrong, but I'd appreciate any help or guidance. Can anyone confirm if I'm right or wrong? N.B: I didn't include the static coefficient of friction as I don't know how to use both in the same problem. However, I believe I would be right to say that it shouldn't affect the answer...?

    Okay, so the question I set myself is; If a box of mass 10 KG is put at the top of a ramp at 60 degrees and with a hypotenuse of length 6m, what will be the velocity of the box at the bottom of the ramp? (I included friction into the problem. I used a hypothetical kinetic coefficient of friction of 0.5).

    So the known variables are: Ramp angle = 60 degrees.
    Box mass = 10 KG
    Kinetic Coefficient of friction = 0.5
    Length of ramp (hypotenuse, not adjacent) = 6m

    So here's what I did...

    1) I found the force of gravity acting along the ramp. I found the force of gravity by doing the following.. 9.81*COS(90-60) = 8.5 m/s^2. Since F=ma, I found the component of gravity acting along the ramp to be 85N.
    2) Next I found the same for friction. Since friction is the normal force times by the coefficient of friction, i did 85 (normal force) times by 0.5 (coefficient). This comes to 42.5.
    3) To find the net force, I did the force of gravity (85N) minus the force of friction (42.5N). This comes to a total of 42.5N of force down the ramp.
    4) Since F=ma, I found the acceleration to be 4.25 m/s^2.
    5) With the formula v^2=2as, I found that 2*4.25*6=51. Finally, I found the square root of 51 to be 7.14.

    This shows that the velocity at the bottom of the ramp is 7.14 m/s.

    B.T.W, can anyone confirm whether this type of question is likely to come up in my GCSE next year? I want to get a solid idea on all of the topics while I can.
     
  2. jcsd
  3. Aug 3, 2009 #2

    Doc Al

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    Staff: Mentor

    OK. (The component of gravity parallel to the ramp = mg sinθ.)
    Recalculate that normal force.
     
  4. Aug 3, 2009 #3
    Oh, thanks. :)

    I'll do a re-run and post my answer in a bit. :)

    EDIT: Okay, I'm having a little trouble with the normal force. The way I did it before, was to assume that the normal force was opposing the force of gravity that I had already found (acting down the ramp). Now I realize that I need to find the force perpendicular to the ramp, not parallel to it. (Can anyone confirm this thought?).

    If this is the case, I'm stuck, because I'm not sure how to calculate this...
    Help appreciated. ;)
     
    Last edited: Aug 3, 2009
  5. Aug 3, 2009 #4

    Doc Al

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    Staff: Mentor

    Yes, the normal force opposes the component of gravity perpendicular to the ramp. You need to be able to find the components of gravity parallel and perpendicular to the ramp. (It's just a bit of trig.)

    Read this: http://www.physicsclassroom.com/Class/vectors/u3l3e.cfm" [Broken]
     
    Last edited by a moderator: May 4, 2017
  6. Aug 3, 2009 #5
    If you wanted to solve this question in an easier manner, i would propose to treat it as an energy problem as opposed to a forces problem. If you know the length of the ramp and its angle, you can find its height.
    From there, just use this equation: Ug=KE+W(f)
    where Ug is gravitational potential energy (mgh), KE is kinetic energy (.5mv^2), and W(f) is work done by friction, in this case, force of friction times distance of ramp.

    you may have already solved the problem by now, but its always good to see if problems can be viewed as energy problems so save time and to avoid working with unneeded vectors
     
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