Physics Question

Question

Flying Circus of Physics Brake or turn? The figure 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=39.0 m/s, and the coefficient of static friction as μS=0.500. 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 μk=0.410, 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? (a) Number Units (b) Number Units (c) Number Units (d) Number i Units

Answer 1

(a) To stop the car just as it reaches the wall, the net force acting on the car must be in the opposite direction to its initial velocity. The frictional force between the tires and the road provides this force. The maximum static frictional force is given by:

fS, max = μS * N

where N is the normal force exerted by the road on the car. Since the car is evenly distributed on four wheels, the normal force is equal to the weight of the car, which is:

N = mg

where g is the acceleration due to gravity. Therefore, we have:

fS, max = μS * mg

Substituting the given values, we get:

fS, max = (0.500) * (1430 kg) * (9.81 m/s^2) = 7007.35 N

Therefore, the maximum static frictional force is 7007.35 N.

To stop the car just as it reaches the wall, the frictional force must be equal and opposite to the force driving the car, which is the product of the car's mass and its deceleration, given by:

f = ma

where a is the deceleration of the car. The deceleration is given by:

a = (v^2 - v0^2) / (2d)

where v is the final velocity of the car, and d is the distance to the wall. Substituting the given values, we get:

a = (v^2 - (39.0 m/s)^2) / (2 * 109 m)

To stop the car, we need to have a negative acceleration, so we take the negative root of the equation above. Therefore, we have:

a = -15.4 m/s^2

Substituting this value into the equation for the frictional force, we get:

f = ma = (1430 kg) * (-15.4 m/s^2) = -21922 N

Since the frictional force cannot be negative, we take the magnitude of the force:

|f| = 21922 N

Therefore, the magnitude of the frictional force needed to stop the car just as it reaches the wall is 21922 N.

(b) The maximum possible static friction force is given by fS, max = μS * N, where N is the normal force exerted by the road on the car, and μS is the coefficient of static friction. The normal force is equal to the weight of the car, which is evenly distributed on four wheels, so we have:

N = mg

Substituting the given values, we get:

N = (1430 kg) * (9.81 m/s^2) = 14019.3 N

Therefore, the maximum possible static frictional force is:

fS, max = μS * N = (0.500) * (14019.3 N) = 7009.65 N

Therefore, the maximum possible static frictional force is 7009.65 N.

(c) If the car hits the wall, then the force of kinetic friction between the tires and the road provides the necessary force to stop the car. The force of kinetic friction is given by:

fk = μk * N

where μk is the coefficient of kinetic friction. The normal force is again equal to the weight of the car, so we have:

N = mg

The force of kinetic friction is equal and opposite to the force driving the car, which is the product of the car's mass and its acceleration. The acceleration is given by:

a = (v^2 - v0^2) / (2d)

where v is the final

velocity of the car, d is the distance to the wall, and v0 is the initial velocity of the car. At the moment of impact, the final velocity of the car is zero. Therefore, we have:

fk = ma = (1430 kg) * [(v^2 - (39.0 m/s)^2) / (2 * 109 m)]

Substituting the coefficient of kinetic friction, we get:

μk * mg = (1430 kg) * [(v^2 - (39.0 m/s)^2) / (2 * 109 m)]

Solving for v, we get:

v = sqrt[(2 * 109 m * μk * g) / (1430 kg) + (39.0 m/s)^2]

Substituting the given values, we get:

v = sqrt[(2 * 109 m * 0.410 * 9.81 m/s^2) / (1430 kg) + (39.0 m/s)^2]

v ≈ 30.9 m/s

Therefore, the speed at which the car will hit the wall is approximately 30.9 m/s.

(d) To keep the car in a circular path of radius d and at the given speed v0, the necessary centripetal force is provided by the frictional force between the tires and the road. The centripetal force is given by:

Fc = mv^2 / r

where m is the mass of the car, v is its velocity, and r is the radius of the circular path. The frictional force is equal and opposite to the centripetal force, so we have:

f = Fc = mv^2 / r

The radius of the circular path is given by d, so we have:

r = d = 109 m

Substituting the given values, we get:

f = (1430 kg) * (39.0 m/s)^2 / (109 m) = 19414.22 N

Therefore, the magnitude of the frictional force required to keep the car in a circular path of radius d and at the given speed v0 is 19414.22 N.