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University Physics 1 Work Problem

  1. Oct 30, 2011 #1
    1. The problem statement, all variables and given/known data

    You apply a constant force F=(-68.0N)i + (36.0N)j to a 390kg car as the car travels 60.0m in a direction that is 240.0° counterclockwise from the x-axis.

    How much work does the force you apply do on the car?

    2. Relevant equations

    W=Fs
    W=K2-K1
    W=Fcos(θ)s
    F/m=a
    Vf^2=Vi^2 + 2Ad


    3. The attempt at a solution

    Im not sure what I am doing wrong here. I have calculated the magnitude of the Force as 76.9N.

    So I have tried:
    76.9(cos(240))*-60 (applying the force in the same direction as the car moves) = 2307J

    I also tried to use the Work energy theorem and found acceleration (76.9/390kg)=0.197m/s^2. I then used this in Vf^2=Vi^2 + 2Ad. So Vf = 4.86. I then use that in K2-K2=W=.5mVf^2-.5Vi^2. (I was assuming that Vi is 0 here, even though it didn't state from rest)

    I feel like this is something simple that I am missing. Any direction would be great!
     
  2. jcsd
  3. Oct 30, 2011 #2

    Doc Al

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    What's the angle between the force and the displacement?
     
  4. Oct 30, 2011 #3

    Simon Bridge

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    The dot-product of the displacement with the force is the Work.
    http://en.wikipedia.org/wiki/Work_(physics)#Mathematical_calculation

    Draw the diagram, 240 degrees anticlockwise from the +x axis is 60 degrees anticlockwise (pi/3 radiens) from the -x axis

    So what is the car moves -60sin(π/3)j-60cos(π/3)i (a 2:1:√3 triangle) gives the vector:

    d=(-30,-30√3)t
    F=(-68.0, 36.0)t

    You know how to evaluate a vector dot product?
    http://en.wikipedia.org/wiki/Dot_product
     
    Last edited: Oct 30, 2011
  5. Oct 30, 2011 #4
    Ahhhh ok. Thanks!

    I was using the angle of s and not the angle between s and F.

    So the angle of F is 152.1°
    240°-152.1°=87.9

    so W=(76.94N)cos(87.9)(60m) = 169N
     
  6. Oct 30, 2011 #5

    Doc Al

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    Much better!
     
  7. Oct 30, 2011 #6

    Simon Bridge

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    Well that's the other way to do it <sniff>.

    The dot product would have been:

    (-30)(-69)+(-30√3)(36)=169.39J

    <sulks>
     
  8. Oct 30, 2011 #7
    Haha, thanks for explaining that Simon! I guess I should have just looked at the W = F dot s to begin with and it would have made all of that a lot quicker. Im glad I have a new way of looking at that now.
    Thanks Again
     
  9. Oct 30, 2011 #8

    Simon Bridge

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    <perks up>
    It's the way forward with computers - wait till you start working with matlab, octave, and the like. After a while it is the first thing you think of.

    Basically, if you are given the angles - it's quicker to use the angles. If you are given coordinates, then the dot-product is faster and you don't have to worry about the deg/rad setting. In this case the angle of the displacement was simple, I didn't even need a calculator.
     
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