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

AI Thread Summary
The discussion centers on calculating the force of friction required to stop a 1,000kg car traveling at 30m/sec, given a coefficient of friction of 0.80. The force of friction can be determined using the formula Ff = μmg, where μ is the coefficient of friction, m is the mass of the car, and g is the acceleration due to gravity (9.81 m/sec²). The user initially miscalculated by omitting the gravitational acceleration but later clarified the correct approach. The relevant equations for determining stopping distance and deceleration are also mentioned. This analysis highlights the importance of including all variables in physics calculations.
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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.
 
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Nevermind, I forgot I left out 9.81m/sec2
 
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/
 
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