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Just a quick check for my solutions - ice patch friction

  1. Oct 9, 2013 #1
    SOLVED - Just a quick check for my solutions - ice patch friction

    1. The problem statement, all variables and given/known data
    Say you (65kg) take a brisk ice skate across a frictionless lake when suddenly you hit a rough patch of ice that's 10 m long with a coefficient of friction (0.02). If you were traveling at 75 m/s before you hit the patch, how fast would you be traveling ater the patch? How long would the patch have to be in order for you to stop?


    2. Relevant equations

    Work equations

    3. The attempt at a solution

    I got 74.97 m/s after the 10m ice-patch, that the ice-patch would need to be 14335.09762m long for you to stop completely, and as a bonus, I calculated the acceleration that you would experience to be -0.1962 m/s²

    I just wanted to check if my answers are correct for this - thank you in advance :)
     
    Last edited: Oct 9, 2013
  2. jcsd
  3. Oct 9, 2013 #2

    arildno

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    It seems roughly correct to me, but:
    What you ALWAYS should include at PF are the EQUATIONS you use (many students get wrong answers because they miss out on that).

    What you do NOT need to include is every step of the way from the equations to the final answers!
    ---
    I'll make a check on your first (you can see if this agrees with how you did the problem!):

    We need to determine WHEN the skater is past the rough patch:

    We have then:
    10 m=75m/s*t+1/2*(-0,02*g)*t^2

    Then, let's say our solution time is T, we can find the velocity afterwards, by:
    v=75-0.02*g*T
    ------------
    If this is how you proceeded, I'm sure your answers are correct. :smile:
     
  4. Oct 9, 2013 #3
    I used F = μ*g*m, multiplied that by 10 to get the work
    Then calculated the acceleration work (mv²/2), subtracted the friction work from that, and calculated backwards to get the new v
    For the length of the patch, I simply set friction work = acceleration work and for the acceleration itself I set acceleration work = m*a*s
     
  5. Oct 9, 2013 #4

    gneill

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    Hi Elpinetos, Welcome to Physics Forums.

    It looks like you've made the right calculations. How many significant figures to report for each value is a bit of a puzzler, but five decimal places on the long patch of rough ice is definitely over the top :smile:
     
  6. Oct 9, 2013 #5

    arildno

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    I was just about to say you didn't need that damn time, anyway; seems you made it all right on your own. :smile:
     
  7. Oct 9, 2013 #6
    Thanks guys for your quick responses :)
     
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