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Work and energy problem feedback

  1. Mar 25, 2012 #1
    I'm just starting to cover Work and Energy.

    I had a question which asked, "A 12 kg wagon is being pulled at an angle of 38° above horizontal. What force is applied to the wagon if it accelerates from rest to a speed of 2.2 m/s over a distance of 3.4 m?"

    For the record this is just a practice question and I am simply posting for feedback if I have made an error somewhere.

    Given we have m= 12kg F=? Angle= 38 degrees V1=0 V2=2.2 Δd=3.4

    I used the equation;

    Work = Change in energy

    W=ΔE

    FcosθΔd=(0.5)mv2^2-(0.5)mv1^2

    I rearranged and got.

    F=(0.5)mv2^2/cosθΔd

    F=(0.5)(12)(2.2^2)/cos38(3.4)

    F=10.8N


    How's that look?

    The question might seem simple but I've been out of high school for several year and am hoping to go back to school, so I am just trying to ensure I'm remembering all of the fundamentals.
    1. The problem statement, all variables and given/known data



    2. Relevant equations



    3. The attempt at a solution
     
  2. jcsd
  3. Mar 25, 2012 #2
    This basically what I did to rearrange the equation
     

    Attached Files:

  4. Mar 25, 2012 #3

    HallsofIvy

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    Staff Emeritus
    Science Advisor

    I don't understand why you are asking about energy when the problem asks about force.

    Force equals mass times acceleration so you can easily get the horizontal component of the applied component- the component that causes the acceleration. Then you need to use a little trig to get the applied force- cos(theta) is the horizontal force divided by the applied force- the hypotenuse in your force diagram.
     
  5. Mar 25, 2012 #4
    This question was in the kinetic energy section.

    But I think your method, and mine would equal the same thing?
     
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