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Calculate the coefficient of kinetic friction for a sliding box

  1. Oct 21, 2015 #1
    • Mentor's note: Thread title changed to be decriptive of the problem
    1. The problem statement, all variables and given/known data
    An electric motor is used to pull a 125 kg box across a floor using a long cable. The tension in the cable is 350N and the box accelerates at 1.2m/s2 (forward) for 5.0s. The cable breaks and the box slows down and stops. Calculating the coefficient of kinetic force?

    Given:
    m=125 kg
    a= 1.2 m/s2
    t= 5.0s
    Ft=350 N

    2. Relevant equations
    uk=Fk / FN


    3. The attempt at a solution
    Fnet = ma
    Fnet = (125kg)(1.2m/s2)
    Fnet = 150

    Thought Fk would be 350N because that's the Ft.

    And then Uk would equal 350N divided by 150 but I got 2.3 when the answer is 0.16 not sure how to get answer?
     
  2. jcsd
  3. Oct 22, 2015 #2

    SteamKing

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    Ya mean, calculate the coefficient of kinetic friction?
     
  4. Oct 22, 2015 #3

    haruspex

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    You correctly got Fnet=150N, what forces is that the net of?
     
  5. Oct 22, 2015 #4
    I thought that since the object is 125 kg it would have a force of gravity of 1225N because 125kg•9.8m/s2
    so that'd mean that the normal force would also be 1225N and I know that the force of tension would be 350N. I think that Fg and FN would cancel out leaving me with only FT?
     
  6. Oct 22, 2015 #5

    haruspex

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    What about friction?
     
  7. Oct 22, 2015 #6
    friction would be 350N as well I'm assuming?
     
  8. Oct 22, 2015 #7

    haruspex

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    Why would you assume that?
     
  9. Oct 22, 2015 #8
    Because the force of tension is 350N so the opposite side is 350N as well?
     
  10. Oct 22, 2015 #9

    haruspex

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    That would make the net horizontal force zero, but you have already calculated that the net horizontal force is not zero.

    Do you understand what the equation ##\Sigma F_x = ma_x## means?
     
  11. Oct 22, 2015 #10
    Fnet = mass • acceleration that's all I know
     
  12. Oct 22, 2015 #11

    haruspex

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    OK, and you used that to calculate Fnet, which is to say the sum of the horizontal forces. The value was not zero. So why did you suggest in post #8 that the horizontal forces should balance, which would give a zero Fnet? Do you understand what net force means?
     
  13. Oct 22, 2015 #12
    Net force is the sum of all forces acting on an object
     
  14. Oct 22, 2015 #13

    haruspex

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    Right.
    As you said, the vertical forces balance, so it just leaves the horizontal forces: tension and friction.
    Write out the ##\Sigma F_x=F_{net}## equation using the known horizontal forces and net force.
     
  15. Oct 22, 2015 #14
    (350N) Fx = 150N
    Fx = 150N - 350N
    Fx = 200N [backwards]
     
  16. Oct 22, 2015 #15

    haruspex

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    I have no idea where you get that from.
    That is not quite what I meant, but it's close. "Fx" was meant to stand for a generic horizontal force. Let's use Ff for frictional force:
    ##150N=ma_x=F_{x net}=\Sigma F_x=##Tension+Friction##=350N+F_f##.
    So, yes, Ff has magnitude 200N.
    Now deduce the coefficient of friction.
     
  17. Oct 23, 2015 #16
    Do I use that entire formula and sub in all my information?
     
  18. Oct 23, 2015 #17

    haruspex

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    Do you mean the whole of this?
    No, that expresses the chain of logic. Do you understand each of the steps in there?
    It leads to result that ##150N=350N+F_f##, which I assume is what you meant by your equation in post #14.
    It remains to express ##F_f## in some way involving the coefficient of friction.
     
  19. Oct 23, 2015 #18
    150N = 350N + Ff is what I meant!
     
  20. Oct 23, 2015 #19
    Just not sure where to go after finding out that the magnitude of Ff equals 200N.
     
  21. Oct 23, 2015 #20
    How does your Ff relate to your Fk?
     
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