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Friction Force and Motion

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  1. Oct 27, 2013 #1
    1. The problem statement, all variables and given/known data
    You testify as an "expert witness" in a case involving an accident in which car A slid into the rear of car B, which was stopped at a red light along a road headed down a hill. You find that the slope of the hill is θ = 12.0°, that the cars were separated by distance d = 24.0 m when the driver of car A put the car into a slide (it lacked any automatic anti-brake-lock system), and that the speed of car A at the onset of braking was v0 = 18.0 m/s.

    (a) With what speed did car A hit car B if the coefficient of kinetic friction was 0.60 (dry road surface)?

    v = ? m/s

    16fef727-0b8d-42a2-bf97-94d2c25b2a24.jpe

    2. Relevant equations

    F = ma

    Kinetic Friction Ff = μFn

    3. The attempt at a solution

    I've actually figured out where I went wrong, I just don't know why my way was wrong.

    I set up my FBD with the positive x direction going up the hill so that my acceleration and frictional force (Ff) is positive and going down the hill is negative x direction so that mgsinθ is negative.

    This would give me Fnet x components as Ff - mgsinθ = m(+a)

    But apparently this is wrong as I only get the correct answer if I switch the signs on the x axis. Why should going down the hill be positive and going up be negative?
     
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  3. Oct 27, 2013 #2

    Simon Bridge

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    Shouldn't make any difference - it just changes the signs around.
    I think you made your mistake elsewhere... maybe you forgot to change the sign in the initial velocity?
     
  4. Oct 27, 2013 #3
    I have to use v(final)^2 = v(initial)^2 + 2ad

    The velocity gets squared so it wouldn't matter. The way I used gave me a positive acceleration ≈ 3.7ms^-2

    The other method gives negative acceleration ≈ -3.7ms^-2
     
  5. Oct 27, 2013 #4

    Simon Bridge

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    That is the correct equation and you are right, the sign of the velocity does not matter.
    What is the sign of the displacement? ("d" is also a vector.)

    It's usually best to pick your axis so that the expected movement, rather than the acceleration, is positive. Less chance of missing a minus sign.
     
  6. Oct 27, 2013 #5
    Oh wow I think it might be the displacement. The car is moving down the hill so my value of d should be negative, right? If so, thanks for your help.
     
  7. Oct 27, 2013 #6

    Simon Bridge

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    Well done :)
    Since you defined +x to be "up the slope", the car starts out at ##x_i=0##, and ends at ##x_f=-s:s>0##, then the displacement is ##\vec{d}=x_f-x_i = -s##

    You'll also see that if you construct the velocity-time graph.
     
  8. Oct 27, 2013 #7
    Great, thanks for your help! I knew it would be some small difference but couldn't figure it out.
     
  9. Oct 29, 2013 #8
    Actually I just realized another mistake I was making. I was assuming that acceleration was pointing up the hill because the car was slowing down so my acceleration sign was wrong.

    I thought if an object was slowing down, the acceleration was in the opposite direction of velocity (which in this case is pointing down the hill)....
     
  10. Oct 29, 2013 #9

    Simon Bridge

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    That's correct - there is no "deceleration" in physics, only a negative acceleration... which is a positive acceleration pointing in the negative direction.

    Since the vehicle is slowing down in the -x direction, the acceleration is positive in the +x direction.
    Double-negatives are trouble like that.

    It is easy to get confused, which is why it is usually best practice to align the axis so the expected displacement is positive and let the acceleration fall where it will. Like, for objects falling it is common to pick "down" as positive.
     
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