Question

A car (m=1500-kg) driving on level ground at 20.0 m/s slams on its brakes and skids to a halt. If the coefficient of kinetic friction between the car's tires and the road is 0.750, how far does the car skid before stopping?

Answer 1

Using the work-energy principle and the coefficient of kinetic friction, the car will skid approximately 40.9 meters before stopping. The friction force was calculated using the car's mass, gravity, and friction coefficient, and the work done by friction was equated to the kinetic energy lost.

Step-by-step explanation:

To calculate how far the car skids before stopping, we can use the work-energy principle. The work done by the friction force is equal to the change in kinetic energy of the car. We have the following:

• Mass of car m = 1500 kg
• Initial velocity v = 20.0 m/s
• Coefficient of kinetic friction μ_k = 0.750

First, let's find the friction force (F_f):

F_f = μ_k × m × g

Where g is the acceleration due to gravity and is approximately 9.8 m/s². The friction force can be calculated as:

F_f = 0.750 × 1500 kg × 9.8 m/s² = 11025 N

Next, we use the work-energy principle where the work done by friction is equal to the kinetic energy lost:

Work done by friction (W) = Change in kinetic energy

W = ½ × m × v²

Since the work done by friction is also F_f × distance (d), we have:

11025 N × d = ½ × 1500 kg × (20.0 m/s)²

Solving for d gives us:

d = ½ × 1500 kg × (20.0 m/s)² / 11025 N = 40.9 meters (approximately)

Therefore, the car skids approximately 40.9 meters before stopping.