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Work and Friction Problem

Discussion in 'MCAT Study Question Q&A' started by NYKnick, Aug 8, 2011.

  1. NYKnick

    5+ Year Member

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    A car moving at 35m/s on dry pavement, skids to a stop in 7 seconds. What is the coefficient of friction between the car's tires and the pavement?

    a. .2
    b. .5
    c. 1
    d. 2
     
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  3. andrewsmack05

    andrewsmack05 SDN Lifetime Donor
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    Hey,

    Here is what I got:

    Force = - mu(coefficient of friction) x Normal Force

    simplifying:

    mass x acceleration = -mu x (mass x gravity)

    thus mass will cancel:

    acceleration = -mu x gravity

    since it is decelerating, acceleration is negative

    plugging in the numbers we get:

    -(35 m/s) / (7s) = -mu x (10 m/s^2)

    (-5 m/s^2) = -mu x (10 m/s^2)
    -0.5 = -mu

    mu=0.5
     
  4. mikexima

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    yup that works. think this was a prolem in EK.

    if you had the mass of the car you could solve it as a work problem. W = change in energy = change in momentum
     
  5. zhemnis

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    . A car moving at 35 m/s on dry pavement, skids to a stop over 175 m. What is the coefficient of friction between the car’s tires and the pavement?
     
  6. DrknoSDN

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    I would start by finding time by using average velocity = Vfinal + Vinitial /2 = 35+0/2 = 17.5 m/s. And if D=175m then Time=10 seconds

    If you have 10 seconds you know the car is slowing down at 3.5 m/s.
    Since Force(kinetic) = u(k)*Fnormal
    Mass is constant, and if we are assuming the normal force on the car is (m*g) you can just say u(k) = 3.5/10 = 0.35 is the coefficient of kinetic friction.
     
  7. DrknoSDN

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    Lol, Xishen... Gotta read the question man.
    I was responding to zhemnis, and when you change variables you can't assume that the acceleration is going to be constant between 2 problems.
    Zhemnis asked a new question 3 years later, with entirely new variables.
    You are not given acceleration in his problem so assuming -5m/s accel is going to give you a wrong answer. Acceleration is going to be different for these problems.

    And,,, your kidding right? The equation is correct so I am not sure where the error occurred... but I'll help you with the arithmetic.

    Vf^2 = Vo^2 + 2ad
    0 = 35^2 + 2(a)(175m)
    0 = (1225) + 2(a)(175m)
    -1225 = 2(a)(175)
    -1225/(2*175) = a
    -1225/350=a
    ... get this... A = -3.5m/s.

    I just didn't use the equation cause it's faster to do simple math in your head. Just find average velocity (moving velocity/2), and then you can find how long it takes to travel D... Once you have time, then A=((deltaV)/T)

    I'm not guaranteeing that I'm correct.. Just that your misunderstanding was probably from an incorrect assumption about acceleration.. followed by incorrect algebra. Sorry to be harsh sounding but that's just a lot of mistakes in one post.
     
  8. ipmed

    ipmed ipmed
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    You dont need mass to solve this problem. 1/2mv^2=MuGD and the M cancel
     

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