1. Dismiss Notice
  2. Download free Tapatalk for iPhone or Tapatalk for Android for your phone and follow the SDN forums with push notifications.
    Dismiss Notice
Dismiss Notice
Visit Interview Feedback to view and submit interview information.

Work and Friction Problem

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

  1. NYKnick

    5+ Year Member

    Joined:
    Jun 20, 2011
    Messages:
    65
    Likes Received:
    2
    Status:
    Medical Student
    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
     
  2. Note: SDN Members do not see this ad.

  3. andrewsmack05

    andrewsmack05 SDN Lifetime Donor
    Lifetime Donor Classifieds Approved 7+ Year Member

    Joined:
    Oct 21, 2009
    Messages:
    312
    Likes Received:
    6
    Status:
    Medical Student
    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

    7+ Year Member

    Joined:
    Jul 2, 2007
    Messages:
    227
    Likes Received:
    1
    Status:
    Pre-Dental
    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

    Joined:
    Mar 9, 2014
    Messages:
    2
    Likes Received:
    0
    Status:
    Fellow [Any Field]
    . 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

    Joined:
    Feb 21, 2014
    Messages:
    450
    Likes Received:
    104
    Status:
    Pre-Medical
    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

    Joined:
    Feb 21, 2014
    Messages:
    450
    Likes Received:
    104
    Status:
    Pre-Medical
    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
    5+ Year Member

    Joined:
    Mar 3, 2012
    Messages:
    70
    Likes Received:
    19
    Status:
    Pre-Medical
    You dont need mass to solve this problem. 1/2mv^2=MuGD and the M cancel
     

Share This Page