Brake or turn? Figure 6-45 depicts an overhead view of a car\'s path as the car

Question

Brake or turn? Figure 6-45 depicts an overhead view of a car's path as the car travels toward a wall. Assume that the driver begins to brake the car when the distance to the wall is d = 109 m, and take the car's mass as m = 1430 kg, its initial speed as v0 = 37.0 m/s, and the coefficient of static friction as s = 0.530. Assume that the car's weight is distributed evenly on the four wheels, even during braking. (a) What magnitude of frictional force is needed (between tires and road) to stop the car just as it reaches the wall? (b) What is the maximum possible static friction fs, max? (c) If the coefficient of kinetic friction between the (sliding) tires and the road is k = 0.420, at what speed will the car hit the wall? To avoid the crash, a driver could elect to turn the car so that it just barely misses the wall, as shown in the figure. (d) What magnitude of frictional force would be required to keep the car in a circular path of radius d and at the given speed v0?

Explanation / Answer

The car's kinetic energy is (1/2)(1430 kg)(37 m/s)2 = 978835 J. To stop before the wall requires that friction do that many joules of work. W = Fd so

F = W/d = (978835 J) / (109 m) = 8980 N. That's (a).

Maximum possible static friction is mg = (0.53)(1430 kg)(9.81 m/s2) = 7435 N. That's (b).

Kinetic friction is (0.42)(1430 kg)(9.81 m/s2) = 5892 N which will perform (5892 N)(109 m) = 642216 J of work before impact, leaving 336619 J of kinetic energy at impact, which equates to a velocity of:

v = (2K/m)1/2 = 21.7 m/s.   That's (c).

Centripetal force = mv2/r = (1430 kg)(37 m/s)2/(109 m) = 17960 N. That is (d), the amount of frictional force to keep it in a circular path.

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