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The driver of a car with a total of 1800 kg mass is traveling at 23 m/s when he slams on the brakes, locking the wheels on the dry pavement. The coefficient of kinetic friction between rubber and dry concrete is typically 0.7. Use the work–energy principle to calculate how far the car will travel before stopping.

123.4 m
378 m
1.68 m
38.6

User Jupaol
by
4.5k points

2 Answers

5 votes

Final answer:

Using the work-energy principle, a car with a mass of 1800 kg and traveling at 23 m/s will travel approximately 38.6 meters before stopping when the coefficient of kinetic friction on dry pavement is 0.7.

Step-by-step explanation:

The subject of the question is a calculation using the work-energy principle in Physics to determine the stopping distance of a car. The coefficient of kinetic friction (μ) is 0.7 and the initial speed (v) is 23 m/s. To find the stopping distance (d), we will use the work-energy principle which states that the work done by friction is equal to the change in kinetic energy of the car.

We know that the work done by friction (W) is equal to the frictional force (F) times the distance (d), and F is μ times the normal force (N), which in this case is equal to the weight of the car (mg). Thus, W = Fd = μmgd.

The initial kinetic energy (KE) of the car is ½mv². The car comes to a stop so its final kinetic energy is zero. Thus, the work done by friction is equal to the initial kinetic energy: μmgd = ½mv².

Let's solve for d:

  • Calculate initial kinetic energy: KE = ½ * 1800 kg * (23 m/s)²
  • Calculate work done by friction: W = KE
  • Use the relation μmgd = W to find d
  • d = W / (μmg)

Substituting the values into the above equation, we get:

  • d = (½ * 1800 kg * (23 m/s)²) / (0.7 * 9.81 m/s² * 1800 kg)
  • d = (0.5 * 1800 * 529) / (0.7 * 9.81 * 1800)
  • d = 477180 / 12348.6
  • d ≈ 38.6 meters

Therefore, the car will travel approximately 38.6 meters before coming to a stop.

1

User DaddyRatel
by
5.0k points
3 votes

Answer: 38.6 m

Step-by-step explanation:

1/2(1800)23^2 - (.7)(1800)(9.8 m/s^2)

476,100/121,100 = 3.93 which closest answer was 38.6 and I got it right

User Claude Houle
by
5.1k points