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Force of Friction

  1. Jun 4, 2015 #1
    A driver of a 1,000kg car is traveling at a speed of 30m/sec when he sees an obstruction in the road. It takes driver .75sec to apply foot to brakes. The car begins to slow down at max rate possible for a coefficient of friction of .80 between the road and tires.

    My question is: what is the force of friction acting to stop the car after the brakes are applied?

    I would think the equation would be Ff=.8(1000kg) but I thought the force of friction needed to be in newtons and that doesn't give me a final answer in newtons.
     
  2. jcsd
  3. Jun 4, 2015 #2
    Nevermind, I forgot I left out 9.81m/sec2
     
  4. Jul 6, 2015 #3
    Here the frictional force=retarding force. frictional force=f=(coefficient of friction)×(mg)=μmg; if stopping distance=s then v2=u2-2as where a=f/m=μg. as the car finally stops so v=0; put all the values....................

    Hope you will find this link useful : https://problemsofphysics.wordpress.com/2015/07/06/more-than-50-thought-provoking-problems-on-friction/ [Broken]
     
    Last edited by a moderator: May 7, 2017
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