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An old car is traveling down a long, straight, dry road at 25.0 m/s when the driver slams on the brakes, locking the wheels. The car comes to a complete stop after sliding 275 m in a straight line. If the car has a mass of 755 kg, what is the coefficient of kinetic friction between the tires and the road?

1 Answer

3 votes

Answer:


\mu=0.11

Step-by-step explanation:

It is given that,

Initial speed of the car, u = 25 m/s

Final speed of the car, v = 0 (it stops)

Distance travelled by the car when it slides, d = 275 m

Mass of the car, m = 755 kg

Let a is the acceleration of the car. Using the third equation of motion to find it as :


a=(v^2-u^2)/(2d)


a=(0-(25)^2)/(2* 275)


a=-1.13\ m/s^2

The car is decelerating.

Let
\mu is the coefficient of kinetic friction between the tires and the road. So,


\mu mg=ma


\mu=(a)/(g)


\mu=(1.13)/(9.8)


\mu=0.11

So, the coefficient of kinetic friction between the tires and the road is 0.11. Hence, this is the required solution.

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