1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Friction problem 3

  1. Feb 17, 2012 #1
    1. The problem statement, all variables and given/known data
    An object takes 12.3 meters to stop because of friction. Assume no skidding. If the coefficient of static friction is 0.5, what is the initial velocity of the object?




    2. The attempt at a solution
    My teacher said that in this task the mass doesn't matter, so we would have
    ##F_{net}=ma##
    ##F-ug=a##
    The object is on motion and we need to find v0 so we would have
    ##v_0^2=v_f^2-2ax##.
    Now for this we need the acceleration, so
    ##a=F-ug##, I have the coefficient and the gravity, but no applying force, so how can I find the acceleration, and how will I find the v0, If I have no acceleration?
     
  2. jcsd
  3. Feb 17, 2012 #2
    Do you know about the work/energy relation?
    If you do this problem becomes a lot simpler.

    Regarding your attempt;
    What is this F you have in your force equation
    [itex]F-ug=a[/itex]
    From the conditions given I don't see what other external force is being applied other than that friction.
     
  4. Feb 17, 2012 #3
    Sorry for double posting I think I got it:
    In this case the force of friction is the net force so (thanks to genericusrnme)
    ##ug=a##
    ##=0.5(9.81\frac{m}{s^2})=a##
    ##4.905\frac{m}{s^2}=a##
    ##v_0^2=v^2-2ax##
    ##v_0^2=0\frac{m^2}{s^2} - 2(4.905\frac{m}{s^2})(12.3m)##
    ##v_0^2=-120.663\frac{m}{s}## (to the left)
    ##v_0=\sqrt{120.663}##
    ##v_0=10.98467113754436\frac{m}{s}##
    Is that right?
     
    Last edited: Feb 17, 2012
  5. Feb 17, 2012 #4
    Yes, that is correct
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook




Similar Discussions: Friction problem 3
  1. 3 blocks with friction (Replies: 1)

Loading...