Work Done By Friction

  1. I have no clue how to do this problem :mad: :

    A flight attendant pulls her 75.2 N flight bag a distance of 268 m along a level airport floor at a constant velocity. The force she exerts is 33.0 N at an angle of 69.9 degrees above the horizontal. Find the work done by the force of friction on the flight bag. Answer in units of J.

    So far, I figured that the total the force parallel to the surface is 11.34 N...
     
  2. jcsd
  3. Ok I don't know what you mean by the parallel force (as I am a stupid Swede)
    Think about what friction is, the force of friction equals what?
    (I forgot the letters) F(friction)=mu(friction coeffecient)*Fn(the force from the ground repelling the bag) damn i need to work on my English... anywho
    So in order to calculate the friction force, which you will use to calculate the friction work since work = force*distance you will need to figure out Fn (assuming that you know the friction coeffecient).
    If you split up the force she exerts into x and y axises you can get the force she exerts upwards.
    What must the result of all forces working in the y axis be since the bag is only moving horisontally? From here you should be able to get Fn thus friction force thus friction work.
    If you don't understand something I'll gladly explain, elaborate
    I hope you understand that I do not want to give you a straight answer
     
  4. You're right, the X component of the pulling force is 11.34 N. And since the attendant is pulling her bag at a constant velocity, the sum of all forces in the X axis (and Y, but we don't care about that) must be 0. Therefore:

    [tex]\Sigma F_x = F_{pull}\cos \alpha - f_k = 0[/tex]

    So f(k) = 11.34 N. And what's the work of that force over a distance of 268 meters?

    [tex]W = \int _0^{268m}{\vec{f_k}\cdot d\vec{r}}[/tex]

    Just remebmer that f(k) is always in the opposite direction of dr, so the dot product will be -f(k)dr, and the work will be negative. Final answer should be 3039 J.
     
  5. hmmm..............
     
  6. what about the 75.2 N flight bag - since u can get the friction force parallel to the surface: (mg = 75.2N)

    [tex]F_{friction} = mg \sin \alpha[/tex]
     
  7. :approve: Thank you both so much.

    btw, the answer was in fact -3039
     
  8. so the 75.2 N put on the question does nothing...
     
  9. nevermind - i get it now :)
     
  10. I can't really understand why the included the bag's weight in the problem, unless one of the following questions would be "What's the coefficient of kinetic friction between the bag and the floor".
     
  11. That, by the way, was a problem found in the problem bank of hw.utexas.edu. I assume that the topic starter is a Stuy kid in SPXR, so a solution involving integration was probably not necessary :wink:.

    The weight of the bag tells you that the bag is, in fact, travelling parallel to the ground and is not being lifted off the ground :smile:. Of course, the weight of the bag is offset by the y-pull.

    Chen, you're right. The problem comes in three parts (my question had different numbers):

    A flight attendant pulls her 62.2 N flight bag
    a distance of 298 m along a level airport floor
    at a constant velocity. The force she exerts is
    38.1 N at an angle of 66.7° above the horizon-
    tal.
    a) Find the work she does on the flight bag.
    Answer in units of J.
    b) Find the work done by the force of friction
    on the flight bag. Answer in units of J.
    c) Find the coefficient of kinetic friction be-
    tween the flight bag and the floor.

    My correct answers:

    a) 4491 J
    b) -4491 J
    c) 0.5539
     
    Last edited: Nov 30, 2004
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