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MCAT Physics Question: Mechanics III

  1. Mar 12, 2012 #1
    1. The problem statement, all variables and given/known data

    Basically, Alice is pushing her brother Jeff on a toboggan at an ice skating rink that is flat except for a ramp that has an incline of 30 degrees

    Alice weighs: 60 kg
    Jeff weighs: 28 kg
    Toboggan weighs: 2 kg

    They decide to push Jeff and the toboggan up the ramp. They start from rest 10 m from the incline and she pushes him with a force that varies with the distance [the book has a figure of Force (N) vs. Distance (m)]. You can see the diagram if you google the question below.

    http://books.google.ca/books?id=HCU... she lets go just before the incline?&f=false

    Jeff goes speeding up the incline with a velocity of 2 m/s

    The Question is:

    "How much work does Alice do from the moment she begins to push Jeff 10 m from the incline until she lets go just before the incline?"

    A) 60J
    B) 75J
    C) 0J
    D) There is not enough information to answer the question


    2. Relevant equations

    I know how to solve the problem, I realize Figure 1: Force vs. Distance graph will give me the amount of work applied if I add the area under the graph. The problem is the graph doesn't show the Force and I am not sure how to solve for it.

    F=ma

    Vf^2=Vi^2 +2ad

    a= Vf^2/d



    3. The attempt at a solution


    I figure the mass is the Toboggan + Jeff.

    But solving for acceleration is the problem I am having. I know the initial velocity will equal to zero, but the velocity during the distance traveled I cannot figure out

    The answer they give for Force = 7.5 N


    Any help would be great,

    Thank you
     
  2. jcsd
  3. Mar 12, 2012 #2
    I'm not any expert but I think using the Work Energy Theorem will quickly give you an answer for work. I think your mass of Jeff + Toboggan was correct.

    Work = KE[itex]_{f}[/itex]-KE[itex]_{i}[/itex] = [itex]\frac{1}{2}[/itex]mv[itex]_{f}[/itex][itex]^{2}[/itex]-[itex]\frac{1}{2}[/itex]mv[itex]_{i}[/itex][itex]^{2}[/itex]

    Hope this helps. Good luck on the MCAT!

    Someone correct me if I'm wrong.
     
  4. Mar 12, 2012 #3

    PhanthomJay

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    You can't use your kinematic equation for constant acceleration when the force, and hence the acceleration, is not constant for the entire 10 m displacement. The problem is more easily solved using the work-energy theorem. Are you familiar with it?
     
  5. Mar 13, 2012 #4
    Yes,

    W total = ΔKE

    so would it end up being along the lines of this?


    W total = ΔKE

    Fdcosθ= 1/2mv2f - 1/2mv2i

    (10)F= 1/2(30)(2)2

    F= 6 N?
     
  6. Mar 13, 2012 #5
    I don't believe force has anything to do with this problem?

    It only asks for work, I got an answer of 60.0 J.
     
  7. Mar 13, 2012 #6

    PhanthomJay

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    You are incorrectly assuming that W = Fd cosθ. That is for constant force only. Go back to your original attempt where you said that
    Since the area under the graph is 1/2(30)(2)2 = 60 J, that's the work done by Alice, as whiskeySierra has also noted. Now apparently there is a part b to this problem that asks you to find the peak value of the Force. You should be able to find Fmax knowing that the area under the 'curve' is 60 J.
     
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