- #1
PhysxMakesMeCry
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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.
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.