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Friction problem of a box

  1. Dec 17, 2004 #1
    "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 4.0m/s?"

    I'm really stuck as to how to figure out the acceleration...
     
  2. jcsd
  3. Dec 17, 2004 #2

    dextercioby

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    The acceleration is easy to find.It's the acceleration due to kinetic friction.Use the definiton for the kinetic friction force and find the acceleration.
    UUse the 2 formulas:
    [tex] v(t)=v_{0} +at [/tex]
    [tex] x(t)=x_{0}+v_{0}t+\frac{at^{2}}{2} [/tex]
    Chose [itex] x_{0}=0 [/itex].

    Daniel.
     
  4. Dec 17, 2004 #3
    I'm still kind of stuck..how are you suppose to incorporate acceleration through those equations into f=ma?

    Alright, here's what I have so far.

    [tex]x = v_0 ((v-v_0)/a) + a*((v-v_0)/a)^2[/tex]

    How do I get the friction into that? And how do I make that equation look neater with the [tex] tag?
     
    Last edited: Dec 17, 2004
  5. Dec 17, 2004 #4

    dextercioby

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    It's actually the other way around.U use F=ma to find the acceleration and then plug the accelation in the first (kinetic) equation and find time.Plug time,acceleration and initial velocity in the second (kinematic) equation to find the distance on which the body moves and solve your problem.

    Daniel.

    PS.The F=ma is to be applied for the kinetic fricton force,bu first u have to know its definiton.I assume u do.
     
  6. Dec 17, 2004 #5
    But if you don't know the initial push, how can you set up F=ma?

    What I thought was:

    [tex]F_p - u_k mg = ma [/tex]
     
  7. Dec 17, 2004 #6

    dextercioby

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    No,no,no,the word 'push' just stands for an explanation to why your body has initial velocity.In your equation it shouldn't stand and therefore the force shoud be equaled to 0.
    Puttin' 0 in your equation above,it gives you the acceleration due to friction which is uded to determine the length.

    Daniel.

    PS.It should have been given the time in which that initial velocity is aquired.Only then u could have been able to compute the "push".
     
  8. Dec 17, 2004 #7
    Ah, I understand now! Thank you very much! The whole F_p thing was really throwing me off.
     
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