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Homework Help: Work and Energy with 2 variables

  1. Jun 26, 2011 #1
    1. The problem statement, all variables and given/known data
    I feel as though I am missing information but I have been assured I'm not can anyone help me out?

    At an accident scene on a level road, investigators measure a car's skid mark to be d=88m long. The accident occurred on a rainy day, and the coefficient of friction was estimated to be Mu=0.34. Use this data to determine the speed of the car, v when the driver slammed on (and locked) the breaks. Take g=10m/s^2.

    2. Relevant equations
    What is the speed of the car?
    F=ma Ft=m(delta)v

    3. The attempt at a solution
    1. The problem statement, all variables and given/known data

    2. Relevant equations

    3. The attempt at a solution
    Last edited: Jun 26, 2011
  2. jcsd
  3. Jun 26, 2011 #2


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    Homework Helper

    Initially the car has KE mv2/2

    when the car has gone a distance of 88 m the velocity is zero.

    So applying your second formula you will get

    mv2/2 = Work done by friction.

    You should know how to get the frictional force (so you will see why you don't need the mass of the car).
  4. Jun 26, 2011 #3
    ok so if mv^2/v = WFfr and Ffr is the same as Fn*mu = mg*mu in the opposite direction then mass cancels out and leaves me with v^2/2=g*mu and when I solve for V I'm getting v=2.89ms. yes?
  5. Jun 26, 2011 #4
    You're on the right track, but I see one error in your attempt to solve the problem.

    Namely the statement in bold here:
    E1 = E1'+W*Ffr

    As a quick review, we know that

    W = Fd


    W = KE


    KE = Fd

    Well, heat caused by friction is a form of kinetic energy, so we can surmise the following:

    Eh = Ffrd

    Can you see how this can apply to your equation?

    Please note, however, that this only accounts for the transfer of your original kinetic energy into heat energy. That isn't a problem in introductory physics, but sound and other forms of energy may be important in later years.
  6. Jun 26, 2011 #5
    2.89m/s is equivalent to 6.46 mi/h (10.4 km/h), it seems unlikely that a vehicle would skid 88 meters with that kind of velocity.

    Logic would conclude that your answer is missing a factor.

    Where did you get the term " W*Ffr " and what does it define?
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