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.