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Annoying Efficiency Problem

  1. Jun 9, 2014 #1
    1. The problem statement, all variables and given/known data

    Objects A (2.0 kg), B (1.0 kg), and C (1.0 kg) are pushed up a ramp, which is 4.0 m and 1.2 m high. The applied forces are parallel to the ramp with magnitudes of 9.8 N on A, 4.9 N on B, and 3.1 N on C.


    2. Relevant equations

    Eff = W(output) ÷ E(Input) x 100%

    W = F x d

    E(potential) = mgh

    3. The attempt at a solution

    Efficiency of A:

    W(output) = 9.8 N x (4 + 1.2) = 50.96 J

    E(potential) = 2 x 9.8 x (4 + 1.2) = 101.92 J

    Eff = 50.96 ÷ 101.92 x 100 = 20% Efficiency

    I know I'm probably over-thinking this one tiny equation but the thing is, I've spent 1 and a half hours on this question alone. The answers I'm supposed to get are "A: 60%, B: 60%, C: 95%" but I can't get anywhere close to them no matter what I do. Maybe its because I forgot how to add the ramp components together, but I thought that was correct. I resorted to guessing as well, but literally no combination of numbers can get me to that answer; its inconceivable for me. I also have a formula sheet on me as well, but I don't know which one to use. This is also literally the only question on my review sheet that I cannot understand at all... Like I said, I think I'm definitely over-thinking it. May someone enlighten me on my mistakes?
     
  2. jcsd
  3. Jun 9, 2014 #2

    haruspex

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    What do you think the 4m distance is measuring? Why are you adding it to the 1.2m?
    The question is not very clear... do you think the objects are accelerating up the ramp? If not, what do you think all the forces are on the objects as they move?
     
  4. Jun 9, 2014 #3

    SteamKing

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    Your post is not clear on the dimensions of the ramp. Is the ramp supposed to be 4.0 m long measured horizontally or along the sloping surface?
     
  5. Jun 9, 2014 #4

    Nathanael

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    What you said is "work output" would actually be "work input" and the potential energy gained would be the "work output"

    Also, the object is pushed up the ramp, and the force is parallel to the ramp. The displacement is only 4 meters (you wouldn't add the 1.2 like you did). Input work would just be 9.8*4=39.2 J

    The change in potential energy is mgh. It would be 2*9.8*1.2=23.52 Joules (you don't add the the 4 meters, only the change in height matters)


    So you would get:

    Efficiency = 23.52 / 39.2 = 0.6 = 60%



    Try doing the other problems now and post if something went wrong.
     
  6. Jun 9, 2014 #5
    The 4m is going up the ramp, while the height is 1.2 m. Its hard to describe because I can't draw the picture here.

    Edit: Thanks Nathaneal for the quick response, and it was correct! You saved me, thank you very much.
     
    Last edited: Jun 9, 2014
  7. Jun 9, 2014 #6

    Nathanael

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    No problem :)
     
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