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Physics Problem?
An old car is traveling down a long, straight, dry road hill (angle of incline 7.00°) at 25.0 m/s when the driver slams on the brakes, locking the wheels. The car comes to a complete stop after sliding 44.0 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
- az_lenderLv 72 years ago
The along-slope component of car's weight is
(755 kg)(9.8 m/s^2)*sin(7 deg) = 901.7 N.
The normal force is the cosine component,
(755 kg)(9.8 m/s^2)*cos(7 deg) = 7343.8 N.
The observed acceleration is found from:
2ad = (vf)^2 - (v0)^2 =>
a = -(25.0 m/s)^2 / (88.0 m) = -7.10 m/s^2.
This acceleration must have been produced by a force of (7.10 m/s^2)(755 kg) = 5360.5 N.
So the retarding force of friction was
5360.5 N + 901.7 N = 6262.2 N,
and the coefficient of friction is
6262.2/7343.8 = 0.853.
