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Going through a rough spot

  1. Jul 18, 2006 #1
    I am wondering if someone could guide me on this. I am getting a vf of 4.6m/s, which can't be right, considering vi is only 3.8m/s.

    A box slides across a frictionless floor with an initial speed v = 3.8
    m/s. It encounters a rough region where the coefficient of friction is
    µk = 0.7. If the strip is 0.52 m long, with what speed does the box
    leave the strip?

    My thoughts:
    vi=3.8 m/s

    Using work-energy theorem:
    F*d = .5*m*vf^2 - .5*m*vi^2
    6.867*d*m = m*(.5*vf^2 - .5*vi^2)
    m's cancel
    6.867*.52 + .5*vi^2 =.5*vf^2
    3.4335 + .5*(3.8)^2 =.5*vf^2
    (3.4335 + 7.22)*2=vf^2
    sqrt (21.307) = vf
    4.62 = vf

    Ridiculous. What am I doing wrong?
  2. jcsd
  3. Jul 18, 2006 #2


    User Avatar
    Staff Emeritus
    Science Advisor
    Gold Member

    Because the force of friction is doing work against the motion, it should actually be:

    -F*d= .5mvf^2 - .5*m*vi^2

    When calculating work, it's force*distance*cos(theta), where theta is the angle between the force and the movement. In this case, theta is 180 degrees, so cos(theta) = -1
  4. Jul 18, 2006 #3
    Thank you so much.
  5. Jul 18, 2006 #4
    You must realise that, by definition, F.d is negative. The force is not in the direction of motion.

    EDIT: Guess I was a bit late. ;)
  6. Jul 18, 2006 #5
    I'm going to venture a guess and say that since you know that logically the object will slow down, vf will be less than vi, so in the work-energy equation you have, the right hand side would have to be negative to satisfy this condition. Hence the left hand side would have to be negative to make satisfy the equality. Also, since work = F*d*cos(theta) and theta here is 180 degrees since it is acting in the direction opposite of the motion. I'm not too sure on any of this, but when i put in a negative 3.43 i get a final velocity around 2.something m/s, which makes more sense

    EDIT: Guess I was even later...
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