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Air Resistence

  1. Sep 21, 2008 #1
    A bicyclist rides 6.2 km due east, while the resistive force from the air has a magnitude of 2.7 N and points due west. The rider then turns around and rides 5.5 km due west. The resistive force from the air on the return trip has a magnitude of 2.2 N and point due east. Find the work done by the resistive force during the round trip.

    W = F*distance

    I figured I was given some vectors to work with:
    Vector A = 6.2 km E
    Vector B = 2.7 N W

    Vector C = 5.5 km W
    Vector D = 2.2 N E

    Vectors A and B directly affect each other, and Vectors C and D affect each other.

    Since W = F * distance:
    6.2 km * -2.7 N = -16.74 J (negative N because it's pointing due West)
    5.5 km * 2.2 N = 12.1 J (positive N because it's pointing East, positive km because distances can't be negative)

    -16.74 J + 12.1 J = -4.64 J (not the right answer)
    What am I doing wrong?
     
  2. jcsd
  3. Sep 21, 2008 #2
    You're correct in saying that distances can't be negative.

    However, it appears as though this problem is using displacement, not distance.

    What does that mean? ;)

     
  4. Sep 21, 2008 #3
    Okay, so the bicyclist only travels 0.7 km total. But where does the air resistance come into play then?
     
  5. Sep 21, 2008 #4
    Perhaps you misunderstood me.

    You wrote

    Since W = F * distance:
    6.2 km * -2.7 N = -16.74 J (negative N because it's pointing due West)
    5.5 km * 2.2 N = 12.1 J (positive N because it's pointing East, positive km because distances can't be negative)

    But this distance can be negative. In both cases, the work done by the air resistance should be negative because if the force is in the opposite direction as the displacement, then the work will be negative
     
  6. Sep 21, 2008 #5
    Oooooh. Now I get it. So -5.5 km * 2.2 N = -12.1 J. So I have -16.74 J and -12.1 J. Do I sum the vectors? I added the work done together and I don't think that's right. But they're not vectors, because they don't have direction. Work is a scalar, so how do I combine the two?
     
  7. Sep 21, 2008 #6
    You should be able to combine the two. Even though they're not vectors and do not have direction, work CAN be negative and so you should just be able to combine the two different values you got for work.
     
  8. Sep 21, 2008 #7
    work (a scalar) = the scalar product of distance and force (two vectors). Of course, the work for the second leg of the journey must be added to the work for the first leg of the journey.

    The scalar product of two vectors is the product of their magnitudes and the cosine of the angle between them. If the angle is 180 degrees, the cosine is -1, and the product is a negative scalar.

    Alternatively, the scalar product is the sum of the products of the separate components (three products in three dimesions). Distance in the x-direction times force in the x-direction, and so on. W = FxDx+FyDy + FzDz. Here, there is only one direction, but due east may be taken to be the positive x-direction, and due west the negative x-direction (or vice versa).
     
  9. Sep 21, 2008 #8
    So, if I'm understanding you correctly, Mattowander, I should be able to say the work done by air resistance is -28.84 J. And I think Almanzo is saying pretty much the same thing. There's only one problem. -28.84 J isn't right. (online homework and it gave me a big fat red NO when I entered that value)
     
  10. Sep 21, 2008 #9
    Now we must calculate.

    Work on the first leg of the journey is 6.2 km * -2.7 N.
    Work on the second leg is -5,5 km * 2.2 N.
    -16.74 kJoule + -12.10 kJoule = -28.84 kJoule. (A Joule is a Nm.)

    However, this is being too exact. The distances and forces were given in two decimals, so only the first two decimals of the answer are to be trusted. Round to -29 kJoule.
     
  11. Sep 21, 2008 #10
    kJoule? The units are J, not kJ. One of the things the website says is to enter at least 3 sig figs. "Unless otherwise noted, treat all numerical values given in CHIP problems as if they were exact. Use the physical and other constants as provided in the tables linked from CHIP or in your texts. Do not round the values at intermediate stages. Enter at least 3 significant figures in all numerical answers you put in CHIP answer boxes."
     
  12. Sep 21, 2008 #11
    But the distances you use are in kilometers, not meters. You need to convert. That's why his answers (and yours) are actually in kJ.
     
  13. Sep 21, 2008 #12
    ooooh. haha, it's always something like that that gets me. last time i forgot to change my calculator into degrees... thanks for the help. :)

    note to self: always read the problem a couple times!!
     
  14. Sep 21, 2008 #13
    No problem :) Glad I could help.
     
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