I need to find the coeffecient of kinetic friction for the following problem... all the stuff after the dashes is what I've done... The answer is approximately half the answer I keep getting, so I think I'm missing something...(adsbygoogle = window.adsbygoogle || []).push({});

The driver of a 1200kg truck travelling 45km/h [E] on a slippery road applies the brakes, skidding to a stop in 35m. Determine the coeffecient between the road and the car tires.

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45km/h = 12.5m/s

find the time it took to stop to find acceleration:

v=d/t 12.5 = 35/t t=2.56s

find acceleration:

a=v/t a=12.5/2.56

a=4.88 m/s^2

find force:

a=f/m 4.88= f / 1200 f=5856N[E]

find normal force:

mg = 1200 (9.81) = 11772N

find friction coeff.

Force of friction = coeff * Normal Force

5856N = x * 11772

x = .49

The answer in the book is .23... I don't understand how to find friction force. I substituted the force of the truck for the friction because I remember seeing that somewhere with a similar question. I'd appreciate any help.

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# Homework Help: Friction coeffecient

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