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Frictional force nudge

  1. Oct 25, 2006 #1
    Hey there,

    I need a nudge in the right direction on two problems. Can you anyone help me out.

    I will post the other question seperately in a few :)

    Here we go:

    1) A bead is sliding along a stiff wire track. The bead is released from rest at point A and comes to a stop (momentarily) at point B. The hieghts of these points above the horizontal are 50 cm and 30 cm, respectively. The length of the wire between then is 400 cm. If the bead is 3.0 g and the frictional force is approximatley constant throughotu the motion, find the value of the frictional force. Watch your units!

    WOW... brainblower!

    I take it they mean "cm" and "g" arent compatible so some conversions are needed. then... what...

    I was thinking this was momentum problem..but...not sure..

    can I have a nudge ??
  2. jcsd
  3. Oct 25, 2006 #2


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    Dearly Missed

    1. The most well-known value of "g" is given in m/s^2.

    2. Remember that frictional force acts along the track at all points.
    Big hint: Consider the work-energy theorem.
  4. Oct 25, 2006 #3
    ok... gotcha... let me go work it out... :)
  5. Oct 26, 2006 #4

    Isn't 3.0g mean the mass of the bead, thus, 0.003kg?
  6. Oct 26, 2006 #5


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    True enough, but you should have understood that what I meant by g was "little g", that is the acceleration due to gravity.
  7. Oct 26, 2006 #6
    Hey guys,

    I must be misunderstanding what the work engery therom is..

    I took it to be

    Wt= 1/2mvf^2 - 1/2mvi^2

    I dont have a velicity though?

    I know the 50 cm and 30 cm come in somewhere but how would any of that give me friction?
  8. Oct 26, 2006 #7


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    DOn't worry about the velocity.

    The bead has stopped by point B which means any kinetic energy it would otherwise have had has been lost to friction over a distance of 400cm.

    How does Work relate to Force?
  9. Oct 26, 2006 #8
    if you push on a wall and nothing happens then the Forces were equal, if your Work applied to the wall overcomes the Force you bust through it like superman.

    and the bead starts and stops...


    start & stop
    KE + PE = KE + PE
    0 + mgh1 = 1/2mv^2 + mgh2

    ??? I bet am I soo far off!
    Last edited: Oct 26, 2006
  10. Oct 26, 2006 #9


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    Energy is NOT conserved when friction forces do work. Your equation applies only when energy is conserved. This is not the work-energy theorem.
    You also noted earlier that
    Wt= 1/2mvf^2 - 1/2mvi^2. This is also a misapplication of the work energy theorem, since this one represents total net work done by all forces, including gravity and friction. You're interested in the work done by friction. So the question is, how must you apply the work energy theorem to get what you are looking for? Hint: Work done by friction is the change in total mechanical energy, i.e., change in kinetic energy plus change in potential energy, and since it has been established that there is no change in kinetic energy, then what does this tell you?
  11. Oct 26, 2006 #10
    ok.. The frictional force on the beads will actually change the KE making it slower or faster depending on how much friction is applied. Since the bead comes to a rest lower then the initail position.. the friction must be high.. so how much work is being done on the bead to slow it down.

    am I getting close?

    would that be the to TME = TMi + Work
    Last edited: Oct 26, 2006
  12. Oct 26, 2006 #11


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    Well, closer, I guess, but still off. Friction will slow its speed, but so might gravity. When you throw a ball straight up, it slows down to nothing at the top of its flight, even without friction or air resistance. This particular problem has a roller coaster type track that dips down and up, but you don't need to know that. You just have to apply the equation i hinted at to solve for the work done by friction, then apply the definition of work to calculate the friction force. Give it one more try.....
  13. Oct 27, 2006 #12
    okk.. I think I get it

    We are dealing with nothing but Potential engery and the work along the string,


    PE = PE - W
  14. Oct 27, 2006 #13


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    Ok, watch your subscripts
    [tex]W_f = PE_f - PE_i[/tex]
  15. Oct 27, 2006 #14

    I did take it a step farther in my notes.. didnt post it here by mistake.. but we had to move variables around solve for what the question was asking for.

    thanks for walking me through this :)
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