# Force of Friction

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1. Jun 4, 2015

### PhysxMakesMeCry

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. Jun 4, 2015

### PhysxMakesMeCry

Nevermind, I forgot I left out 9.81m/sec2

3. Jul 6, 2015

### arnab.ghosh

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