Problem 19 - College Physics

Question

\(\cdot\) \(\cdot\) Stopping distance of a car. The driver of an 1800 \(\mathrm{kg}\) car (including passengers) traveling at 23.0 \(\mathrm{m} / \mathrm{s}\) slams on the brakes, locking the wheels on the dry pavement. The coeffi- cient of kinetic friction between rubber and dry concrete is typically 0.700 . (a) Use the work-energy principle to calculate how far the car will travel before stopping. (b) How far would the car travel if it were going twice as fast? (c) What happened to the car's original kinetic energy?

Step-by-Step Solution

Answer

(a) Car travels 37.8 m before stopping. (b) At double speed, it travels 151.2 m. (c) Kinetic energy was converted to heat and sound.

Step 1: Calculate Initial Kinetic Energy

First, calculate the car's initial kinetic energy \( KE_i \) using the formula: \( KE = \frac{1}{2}mv^2 \), where \( m = 1800 \) kg and \( v = 23.0 \) m/s. \( KE_i = \frac{1}{2} \times 1800 \times (23.0)^2 \).

Step 2: Calculate the Work Done by Friction

The work done by friction \( W_f \) is equal to the kinetic friction force times the stopping distance \( d \). First, find the friction force \( f_k \) using \( f_k = \mu_k mg \), where \( \mu_k = 0.700 \) and \( g = 9.8 \) m/s\(^2\). Then express \( W_f = f_k \times d \), and set \( W_f = KE_i \) to find \( d \).

Step 3: Solve for Stopping Distance \(d\)

Set the work done by friction equal to the initial kinetic energy: \( \mu_k mgd = \frac{1}{2}mv^2 \). Simplify and solve for \( d \): \( d = \frac{v^2}{2\mu_kg} \), substitute \( v = 23.0 \), \( \mu_k = 0.700 \), and \( g = 9.8 \).

Step 4: Calculate Stopping Distance at Double the Speed

When the car's speed is doubled (i.e., \( 2v = 46.0 \) m/s), substitute into the stopping distance formula: \( d' = \frac{(2v)^2}{2\mu_kg} \). Simplify to show that \( d' = 4d \).

Step 5: Explain the Fate of the Kinetic Energy

The original kinetic energy of the car is converted into work done against friction, which is dissipated as heat and sound, bringing the car to a stop.

Key Concepts

Kinetic Friction

Kinetic friction is an essential concept when dealing with objects in motion, like a car in this exercise. When a driver slams on the brakes, kinetic friction comes into play. It is the resistive force that acts against the motion of the car. This type of friction occurs between two surfaces sliding past each other. In our scenario, it's between the car's tires and the road.
The force of kinetic friction can be calculated using the coefficient of kinetic friction ( \( \mu_k \)), which is a measure of how easily one surface slides over another. This value varies depending on materials involved. For rubber on dry concrete, it is typically 0.700. Kinetic frictional force ( \( f_k \)) is then given by:

• \( f_k = \mu_k mg \)
• Here, \( m \) is the mass of the car and \( g \) the acceleration due to gravity.

This force does the work required to stop the car, as energy is converted into heat through friction.

Kinetic Energy

Before a car comes to a stop, it possesses kinetic energy. This energy is due to its motion. When moving at a speed of 23.0 m/s, the car's kinetic energy can be calculated using the formula: \( KE = \frac{1}{2}mv^2 \).
Kinetic energy increases with both the car's mass and the square of its velocity:

• The mass (\( m \)) for the car is 1800 kg.
• Velocity (\( v \)) is 23.0 m/s at the given instant.

Calculating the initial kinetic energy helps us understand how much energy needs to be dissipated for the car to stop. If the car were to travel at double the speed, this energy would multiply by fourfold, significantly affecting stopping distance and required force to bring it to a halt. Thus, speed plays a critical role in safety.

Frictional Force

Frictional force is crucial in decelerating a moving car. It provides the necessary resistance to oppose the car's kinetic energy, eventually bringing it to a standstill. Unlike static friction that acts on stationary objects, kinetic friction is at work once the wheels are locked during a braking event.
The force exerted by kinetic friction uses the coefficient of kinetic friction (\( \mu_k \)) determined by the interacting materials, as well as the object's weight (mass × gravity). This force is described by:

• \( f_k = \mu_k mg \)
• Rubbery tires on dry concrete typically yield a \( \mu_k \) of 0.700.

Ultimately, the frictional force dictates how rapidly a vehicle can halt, emphasizing the importance of tire quality and road conditions in braking efficiency.

Stopping Distance

Stopping distance is the space a vehicle travels during the process of coming to a complete stop. It's crucial for road safety, and it relies heavily on factors such as speed, mass, braking force, and friction.
For our car traveling at 23.0 m/s, the stopping distance is derived from the work-energy principle, relating the initial kinetic energy to the work done by friction:

• \( d = \frac{v^2}{2\mu_k g} \)
• With \( v = 23.0 \) m/s, \( \mu_k = 0.700 \), and \( g = 9.8 \) m/s\(^2\).

Should the car travel at twice the speed (46.0 m/s), the stopping distance doesn't just double; it quadruples! This illustrates how higher speeds dramatically increase the distance required to stop, endangering both the driver and others on the road. Hence, understanding kinetic energy's transformation into other forms through frictional forces helps predict and enhance safety measures.