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Homework Help: Dynamics Question: Finding mass using Vi, Vf, D and acceleration

  1. Jul 21, 2012 #1
    Hi everyone, I'm new to forum and am a 3rd year university student who has never taken physics until now.. I'm currently studying for a dynamics unit test and came upon this question....

    1. The problem statement, all variables and given/known data

    A car whose brakes are locked skids to a stop in 70m from an initial velocity of 22.22m/s. Find the coefficient of kinetic friction.

    Vi: 22.22m/s
    Vf: 0m/s
    d: 70m
    a(deceleration): -3.526m/s^2
    g= 9.81m/s^2
    μ=?

    2. Relevant equations

    - Found deceleration using kinematics formula Vf^2=Vi^2 =2ad
    - Ff= μFn
    - Fg=mg
    - Fnet(y)=ma ; a(up/down)=0 ; Fnet(y)=0 ; Fn-Fg=0 ; Fn=Fg

    3. The attempt at a solution

    In order to find Fn, in order to be able to solve for μ, I need to know the mass of the car... I have no idea where to even begin solving for the mass of the car using the currently known values :(

    If anyone has time, please point me in the right direction.
     

    Attached Files:

    Last edited: Jul 21, 2012
  2. jcsd
  3. Jul 21, 2012 #2
    I suppose it should be Fnet(x)=ma.
    What is Fnet(x) here? (Here only one force is acting in the direction of motion of car i.e force of friction)

    EDIT: Oh, btw, welcome to PF!
     
  4. Jul 21, 2012 #3
    Hi there :)
    I added an attachment.
    Fnet(x)=ma

    We have the acceleration but still not the mass. That's where I lost it :uhh:

    EDIT: Thanks very much! I've been circling around this forum for weeks now and finally decided to join since magic tutor couldn't save me and the test is on Monday :P
     
  5. Jul 21, 2012 #4
    I haven't checked your calculations but can you write [itex]μmg=ma[/itex]? See the m cancels on both the sides, you don't require the mass of car now.
     
  6. Jul 21, 2012 #5
    I'm not sure why you think you need the mass. If you already have the acceleration, then you know that the force of kinetic friction [itex]F_k = ma[/itex] is also equal to [itex]\mu F_g = \mu m g[/itex]. The mass cancels out.
     
  7. Jul 21, 2012 #6
    OMG! I can't believe I missed that! I see what you did there :wink:
    Grazie mille!:!!)
     
  8. Jul 21, 2012 #7
    Because logic and the obvious escape me :wink:
    This is what happens when a mathematically challenged biology student tries to do physics :redface: Thanks for your reply!
     
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