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Kinetic Friction Word Problem

  1. Oct 27, 2013 #1
    1. The problem statement, all variables and given/known data
    An electric motor is used to pull a 125kg box across a floor using a long cable. The tension in the cable is 350N and the box accelerates at 1.2 m/s/s [fwd] for 5.0s. The cable breaks and the box slows down and stops
    a )Calculate the coefficient of kinetic friction. [ans:0.16]
    b) How far does the box travel up to the moment the cable breaks? [ans: 15m]
    c) How far does the box travel from the moment the cable breaks until it stops? [ans: 11m]

    2. Relevant equations
    Kinematics equations

    3. The attempt at a solution
    Ft-Fk=ma
    350-Fk=(125)(1.2)
    Fk=200N

    μk=Fk/Fn
    =200/(125)(9.8)
    μk=0.16

    b) D=Vft-1/2at^2
    = 0-1/2(-1.2)(5)^2
    = 15m

    c)

    I know my solution for b) is correct. But I don't understand why the Vf is zero when the cable breaks because shouldn't the Vf be zero when the box actually comes to a stop and not when the cable breaks. Assuming I solved b) correct, and Vf = 0 then how would I solve c) when the box actually comes to a stop?
     
  2. jcsd
  3. Oct 27, 2013 #2

    Doc Al

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    Your formula is a bit off. That should be: D = vi + 1/2at2, where vi is the initial velocity. (But your answer is correct, just the same.)

    You'll need a different formula to calculate the distance the box slides after the cable breaks. (Or combine a few formulas.)
     
  4. Oct 27, 2013 #3
    I don't understand why the initial velocity is zero when the cable breaks. Also what formula would I use?
     
  5. Oct 27, 2013 #4

    arildno

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    In b), you are considering the time interval from t=0, to t=5.
    It is at t=5 the cable breaks, but t=0 determines the initial velocity to be 0
     
  6. Oct 27, 2013 #5

    arildno

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    You may, of course, use 0<=t<=5:
    [tex]v_{f}=v_{0}+at\to{v_{0}}=v_{f}-at[/tex]
    Thus, you can rewrite the standard formula for distance as:
    [tex]d=v_{f}t-\frac{1}{2}at^{2}[/tex]
    rather than:
    [tex]d=v_{0}t+\frac{1}{2}at^{2}[/tex]

    a>0
    --
    In your solution at b), you made two errors that cancelled each other's effects:
    You set v_f=0, and a=-1.2
    Thus, you got the correct result, but in an incorrect manner.
     
  7. Oct 27, 2013 #6
    Thanks, I understand that. But how would I solve c), would time still be 5s when it comes to a stop?
     
  8. Oct 27, 2013 #7

    arildno

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    No, it would not.
    You do not strictly need to find it, though, since for the new distance, you have the new initial&final velocities, plus the new acceleration.
     
  9. Oct 27, 2013 #8
    How would I find these new values?
     
  10. Oct 27, 2013 #9

    Doc Al

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    It's not. The box is moving when the cable breaks. (The final velocity will be zero.)

    Here's a list of standard kinematic formulas: Basic Equations of 1-D Kinematics
     
  11. Oct 27, 2013 #10

    arildno

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    You might ask yourself first:
    1. What forces act upon the car just after the cable breaks and beyond?
    2. What is the velocity of the car just after the cable breaks (or in that same moment if you like!), and how can you calculate it?
    3. What is the velocity of the car when it finally stops?
     
  12. Oct 27, 2013 #11
    Since the cable breaks, the net force would become Fk=200N since there is no tension.
    a= Fnet/m
    = 200/125
    a= 1.6 m/s/s

    To find the velocity when the cable breaks, I would have to determine the final velocity when t=5 and acceleration is 1.2m/s/s

    a = vf-vi/t
    1.2 = Vf-0/5
    Vf = 6 m/s

    This means this would now become the initial velocity after the cable breaks.

    So now I have enough information to determine the distance it travels after the cable breaks and comes to a stop.

    Vf2=Vi2+2ad
    d = Vf2-Vi2 / 2a
    = 0-(6)2/2(-1.6)
    = -36/-3.2
    = 11.25
    = 11m
    So is this correct? The answer that's given is rounded to 2 significant digits
     
  13. Oct 27, 2013 #12

    arildno

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