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Homework Help: Find the coefficient of kinetic friction problem

  1. Dec 9, 2004 #1
    Ok, here is my dilemma: A flight attendant pulls her 70.0 N flight bag a distance of 253 m along a level airport floor at a constant velocity. The force she exerts is 40.0 N at an angle of 52 degrees above the horizontal. Find the following:
    a) the work she does on the flight bag
    b) the work done on the flight bag
    c) the coefficient of kinetic friction between the flight bag and the floor

    What I did for a was, I multiplied the force of the bag (70)*the distance (253)*cos(theta). For the answere I got 10903.4 J.
    Then for b, I did the same thing, except used 40, and I got 6230.5
    Am I doing this right so far?
    For c, I have no clue how to solve for the coefficient of kinetic friction. Whenever I try an equation, it comes out greater than one, or, I have another variable to solve for.
    Is there something that I am forgettting in the equation?
  2. jcsd
  3. Dec 9, 2004 #2
    Did you get the right answer for a) and b) ??
  4. Dec 9, 2004 #3
    I think for a) all you have to do is:

    W = F_{she}\Delta x =( 40 \cos 52 )(253) = 6230.5

    for b)

    it should be -6230.5. just change the sign.

    for c) - u need to consider friction, remember that:

    [tex]F_{friction} = mg\mu = 700\mu[/tex]

    since, it is in constant velocity, this means that:

    [tex]F_{she} = F_{friction}[/tex]

    so the friction coefficient is... 0.0571428571
  5. Dec 10, 2004 #4


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    From the question it seems like the 70N is the bag's weight. ie mg=70N.

    For part a, I think you need to use 40N instead of 70N, since that is the force the flight attendant exerts on the flight bag.

    For part b, remember that total work done on a body is equal to its change in kinetic energy. How much does the kinetic energy change?

    For part 3, remember that acceleration in the horizontal direction is 0. So what is the net force in the horizontal direction? Use this to figure out the frictional force.

    To find the coefficient of friction you need the equation:
    [tex]f_{k}=u_{k}N [/tex] where N is normal force. To find out N, use the fact that the bag is not accelerating vertically. hint: there are 3 different forces in the vertical direction.

    Always use a free-body diagram for these problems. Makes things easier, and makes sure you don't miss anything.
    Last edited: Dec 10, 2004
  6. Dec 10, 2004 #5


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    The bag's weight is downward force. In calculating work, you use only the component of force in the direction of movement.
  7. Dec 10, 2004 #6


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    You need to remember that for the net work on the bag, you must include the other factors. The crate's weight, friction, normal force, and the force applied by the women. Because the normal force and weight are equal forces, and act in opposite directions they will cancel out.

    So you will have the equation:

    Work(net) = Work(gravity) + Work(normal) + Work(women) + Work(friction)

    Just calculate each and add.
    Last edited: Dec 10, 2004
  8. Dec 10, 2004 #7


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    This is incorrect. Normal force and weight are not equal in this case. There are 3 forces in the vertical direction: the normal force, the weight, AND the vertical component of the force exerted by the woman on the bag. ALL 3 add to 0.

    Always use a free body diagram. Otherwise you might miss something.
  9. Dec 10, 2004 #8
    To reinforce what Halls of Ivy told you, you always disregard any force that is 90 degrees away from the direction of displacement. If there were a force at 89 degrees, you would find the vector components and use that horizontal component, no matter how small. It's not because normal force and weight counteract each other. It is because work only occurs in there is displacement in a particular direction. Learning Physics is right, always use a vector diagram.
  10. Dec 10, 2004 #9
    For part c, I always remember to have Fun. Frictional Force=coeff of friction*normal force. Rearrange and solve for u. (I know, it's not a u, it's a mu, but who wants to have fmun?)
  11. Dec 10, 2004 #10


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    Thank you for pointing that out. I believe I calculated it neglecting the magnitude of the angle. So I just said that the forces perpendicular to the displacement cancel.

    But the equation is still right!
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