Force of Friction: Stopping 1,000kg Car at .8 Coeff.

[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.

 

[PhysxMakesMeCry]

Nevermind, I forgot I left out 9.81m/sec2

 

[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/