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Homework Help: Forces and friction

  1. Dec 8, 2009 #1
    1. The problem statement, all variables and given/known data

    A 1 kg box begins at rest and a hand applies a constant force of 2N along a 2 meter board. The first meter is frictionless, but the second meter has friction that makes the box come to a stop. Time is not give. Coefficient of friction is not given. My question is, is it possible to find acceleration at the 2nd meter and/or the coefficient of friction?


    2. Relevant equations

    $F=ma F(f)=uF(n) u= Coef. of friction

    3. The attempt at a solution

    $F= 2N-F(f) =(1kg)(a)

    2N-u(9.8N)=1kg(a) I get stuck with two variables.
     
  2. jcsd
  3. Dec 8, 2009 #2
    Ok so for the first meter you have a box, with a force of 2N being applied to it.

    F = ma, so the accel on the box is F/m = 2 m/s^2

    Now this box undergoes this accel from rest through 1 meter.

    v12=v02+2s(x1-x0)

    v12= 0 + 2(2 m / s2)(1m)
    v1 = 2m/s

    Now, for the box to decelerate from 2 m/s to zero through 1m, you need to have an acceleration of -2 m/s2. You can get this from just doing another kinematic equation or by reason (if it takes 2 m/s^2 to accel from rest to 2m/s through 1 meter, then it should take the same, but opposite to decel to go from 2m/s to rest through 1 meter)

    Ok, so now you know that the box must undergo a decel or 2m/s^2 to come to rest after 1 m.

    This means that when you sum the forces on the box [tex]\sum F[/tex]x = max = 1kg* (-2m/s^2)

    The net sum must therefore be -2N

    The forces acting on the box are the Fhand and the ffriction so sum Fx = Fhand - ffriction = -2N

    so ffriction = 4N = [tex]\mu[/tex]k*N = [tex]\mu[/tex]k*mg

    so [tex]\mu[/tex]k = about .4
     
  4. Dec 8, 2009 #3
    of course.. Thanks
     
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