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A bug in a bowl, lol

  1. Feb 25, 2007 #1
    1. The problem statement, all variables and given/known data

    A bug slides back and forth in a bowl 11 cm ddep, starting from rest at the top. The bowl is frictionless except for a 1.5 cm wide sticky patch on its flat bottom, where the coefficient of frictino is .61. How many times does the bug cross the sticky region???

    Anybody have any ideas. I can't find an equation that uses the coefficient of friction and potential energy...
  2. jcsd
  3. Feb 25, 2007 #2
    Hi, Mark, welcome to the forums!

    What happens to the total energy of the bug when it crosses the sticky patch?
  4. Feb 25, 2007 #3
    Thanks for the welcome.

    It will decrease, the kenetic energy is at its peek when it reaches the patch, but the total energy will decrease after it passes the 1.5 in sticky area.
  5. Feb 25, 2007 #4
    Exactly! Some work is being done on the bug to decrease its energy. How would you calculate this work done?
  6. Feb 25, 2007 #5
    The integral of Fnetdx???
  7. Feb 25, 2007 #6
    Mmm...you wouldn't actually need an integral here, but the principle's the same.
    The force is constant (what is it?), and (presumably) it's going to travel in a straight line of 1.5 cm. So it boils to down to the simpler version, W = F(net)d.
  8. Feb 25, 2007 #7

    Well, we have a gravitational force, weight...and we also have the kenetic friction. It's weight is constant, but we don't know its weight because we don't have a mass.
  9. Feb 25, 2007 #8
    In the patch, the gravitational force is balanced by the normal force. The net force is due to friction.

    Actually, you don't need to know the mass. Once you look at the problem in its "entirety," the mass will vanish, so to speak.

    The bug starts with some energy at the top, then loses some in the sticky patch, and then it goes to the other side with less energy than it started with, and hence to a smaller height. All you need to know is that second height, and then repeat the process again (and again).
  10. Feb 25, 2007 #9

    I'm confused, so i take the work from the friction, and subtract that from the total energy..and then solve for height2?
    My Fnet is equal to (ma+Fsubk) right?
    And you said W=(Fnet)d. What is d?
  11. Feb 25, 2007 #10
    That's right.

    I assume Fsubk refers to the force due to kinetic friction.

    Fnet = Fk. What's the force due to friction?

    Dispalcement (or distance). Work = Force.Displacement. Remember, the force and displacement are in opposite directions.
  12. Feb 25, 2007 #11
    Fk=ukn where n is the normal force or the opposite of weight in this case.

    So the displacement is going to be 1.5cm?
  13. Feb 25, 2007 #12
    Yes. I think you now have enough info to solve for the height on the other side of the bowl. :smile:
  14. Feb 25, 2007 #13

    Ok, so i solved for work. W=-.915mg. Can i use U=mgh and then substitute U/h in for mg?? And i don't know my initial total energy. I'm still a little confused.
  15. Feb 25, 2007 #14
    Okay, that's how much it lost to friction.

    Let's say it started off at height h (which is eqaul to 11 cm, in this case). Therfore initial energy is mgh; it loses .915mg, and with the remaining energy goes up the other side to some height, say h'. Can you solve for h'?

    After that, it repeats the same thing again, but with "initial" energy decreased every time.
  16. Feb 25, 2007 #15

    Ok, so i start off with 11mg. Now i minus .915mg until i reach 0? I got it. Because it loses .915mg of energy every time it passes the sticky spot. So i minus that from the total each time it passes until the sticky spot completely stops the bug. Thanks for your help!
  17. Feb 25, 2007 #16

    Glad to help. :)
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