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Should be easy but I'm so confused! (Kinetic Friction)

  1. Sep 23, 2005 #1
    If anyone could help me with this, I would be so grateful! I know the answer to the problem is 4.1 m but I don't know how to get there and I have to show my work. Here's the problem:

    "A box is given a push so that it slides across the floor. How far will it go given that the coefficient of kinetic friction is .20 and the push imparts an initial speed of 4.0 m/s?"

    It does not seem like there is enough information. Friction Force(kinetic) = .20 X Force (Normal). The normal force is mass X force of gravity. But there is no mass given so I can't figure that out. I don't know the time so I don't know what to do with the velocity to get it to accelecration. I'm so lost. Please help me, thank you so much for ANY help!
     
  2. jcsd
  3. Sep 23, 2005 #2

    Päällikkö

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    Just keep the equations rolling, I predict m will have magically disappeared when you get to the final equation.

    Or you could use the conservation of energy (m's cancel out there too).
     
  4. Sep 23, 2005 #3
    Thanks for the reply, Päällikkö, but I still can't get it! I've been using all those velocitiy formulas, and solving for variables, and plugging them in and so forth, but I'm still stuck with time and acceleration being unknown. I'm so confused. I think there has to be something a lot simpler to this problem but I'm missing it.
     
  5. Sep 23, 2005 #4
    Well, the only force acting on the box is friction, which = .20 * mass * gravity. Also, F=ma, a=F/m.

    So acceleration = force / mass = .20 * gravity.

    You now know the accelleration, initial velocity and final velocity(0) so you can use your equations of motion to find the displacement.
     
  6. Sep 23, 2005 #5
    Ok phyicsisconfusing, this is a very simple property. I will first solve it for you using work-energy methods. This is one thing you need to realize. Work-energy is good for finding a.) final velocities, b.) distances moved. Equations of motion are good when you have to worry about a.) time.

    So lets use work-energy relations.

    You have 1/2mv^2 initial kinetic energy, and you LOOSE F*d of energy due to work loss of friction.

    *ENERGY MUST be conserved, so the net energy change is ZERO*

    So all the kinetic energy is lost into friciton, SOooooo......,

    1/2mv^2= F*d= (F=.2mg)*d= .2mgd

    so cancel out the m's,

    1/2v^2=.2gd

    and,

    d=1/2 (v^2) /(.2g)

    plug in numbers,

    d=.5 (16)/(.2*9.81)

    d= 4.07747 meters, or, 4.1 meters.

    NOW, lets see what using equations of motion tells us,

    F_x=ma = F_friction , because ONLY friction is the force here

    F=.2*(N)=.2(mg)

    so F=ma=.2mg or,

    a=.2g

    we know that this is constant DECELERATION so000000,

    v=v_0 + at, v =0 when it comes to a stop, sooooooo,

    o=4+at= 4-(.2g)t

    solving for t gives,

    t= 4/(.2*9.81)

    See how this told us how LONG it took.

    but we also know that a = dv/dt
    or that (.2*9.81)= dv/dt
    lets separate the differentials and find that:

    (.2*9.81)dt = dv

    lets integrate from t to t=0 and v to v=0, we get:

    (.2*9.81)t = v

    and v= ds/dt (s is distance)

    so (.2*9.81)t = ds/dt

    again move variables,

    so, (.2*9.81)t dt = ds

    integrate again and get

    (.2*9.81)(1/2t^2)=s

    plut in the time t= 4/(.2*9.81)

    and you get

    (.2*9.81)16/(.2*9.81)^2 (1/2)=s

    factor out (.2*9.81) and divide 16 by two and get,

    8/(.2*9.81)

    and AGAIN you STILL get 4.07747

    but look how much work we DID!

    my point, USE ENERGY METHODS!!!!!!!!!!!!!!!!!!!!!!!!!

    PLEASE PLEASE PLEASE recognize the importance of knowing if you should use ENERGY methods or if you should use Position equations!
     
  7. Sep 23, 2005 #6
    You can solve this without time, though I agree energy methods are the way to go in this case.

    [tex]v^2=u^2 + 2as[/tex]
    [tex]u^2=-2as[/tex]
    [tex]s=\frac{u^2}{-2a}[/tex]
    [tex]s=\frac{4^2}{2*0.2*9.8}[/tex]
    [tex]s=4.077[/tex]
     
  8. Sep 23, 2005 #7
    even more clever kazza, good job! I find the harmony in many methods beautiful! Realize, however, that [tex]v^2=u^2 + 2as[/tex]
    comes as a result of using my basic equations, so its one in the same in actuality.
     
  9. Sep 23, 2005 #8
    Ah, I see now! Thank you. My class hasn't gotten to the work-energy equation (it's 2 chapters after where we are now) so I assume he wants us to use the motion equation. Just as Päällikkö said, the mass cancelled out when you set those two eq. equal to eachother but I was too dopey to figure it out, lol. Anyway I completely understand it now...thank you again. This forum is great. :smile:
     
  10. Sep 23, 2005 #9
    Your welcome, thanks for the question, it was fun to solve.
     
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