Problem 12 - Physics for Scientists and Engineers with Modern Physics

Question

(II) \((a)\) Show that the minimum stopping distance for an automobile traveling at speed \(v\) is equal to \(v^{2} / 2 \mu_{\mathrm{s}} g,\) where \(\mu_{\mathrm{s}}\) is the coefficient of static friction between the tires and the road, and \(g\) is the acceleration of gravity. (b) What is this distance for a \(1200-\mathrm{kg}\) car traveling \(95 \mathrm{~km} / \mathrm{h}\) if \(\mu_{\mathrm{s}}=0.65 ?\) (c) What would it be if the car were on the Moon (the acceleration of gravity on the Moon is about \(g / 6\) ) but all else stayed the same?

Step-by-Step Solution

Answer

(a) The stopping distance is \( \frac{v^2}{2\mu_{\mathrm{s}}g} \). (b) On Earth it's 84.0 m. (c) On the Moon, it's 503.0 m.

Step 1: Set up the problem

To find the minimum stopping distance, we first consider the forces acting on the car. The braking force due to friction is the key force involved here. The maximum static frictional force that can act is given by \( F = \mu_{\mathrm{s}} m g \), where \( \mu_{\mathrm{s}} \) is the static friction coefficient, \( m \) is the mass of the car, and \( g \) is the acceleration due to gravity. Using Newton's second law, this force will cause deceleration.

Step 2: Calculate deceleration due to friction

Using Newton's second law, \( F = ma \), where \( a \) is the acceleration (or deceleration in this case). Therefore, \( \mu_{\mathrm{s}} m g = ma \). This simplifies to the deceleration \( a = \mu_{\mathrm{s}} g \).

Step 3: Use kinematic equation for stopping distance

We apply the kinematic equation \( v^2 = u^2 + 2as \) (where \( v \) is the final velocity, \( u \) is the initial velocity, \( a \) is acceleration, and \( s \) is the stopping distance). Here final velocity \( v = 0 \), initial velocity \( u = v \), and \( a = -\mu_{\mathrm{s}} g \) because it's a deceleration. This reduces to \( 0 = v^2 - 2\mu_{\mathrm{s}} gs \). Solving for \( s \), we get \( s = \frac{v^2}{2\mu_{\mathrm{s}} g} \).

Step 4: Compute the stopping distance on Earth

Given: \( v = 95 \mathrm{~km/h} = \frac{95 \times 1000}{3600} = 26.39 \mathrm{~m/s} \), \( \mu_{\mathrm{s}} = 0.65 \), and \( g = 9.81 \mathrm{~m/s^2} \). Substituting into \( s = \frac{v^2}{2\mu_{\mathrm{s}} g} \), we find \( s = \frac{(26.39)^2}{2 \times 0.65 \times 9.81} \approx 84.0 \mathrm{~m} \).

Step 5: Calculate stopping distance on the Moon

On the Moon, \( g = \frac{9.81}{6} \approx 1.635 \mathrm{~m/s^2} \). Using the same values for \( v \) and \( \mu_{\mathrm{s}} \), we use the formula \( s = \frac{v^2}{2\mu_{\mathrm{s}} g} \). Substituting, \( s = \frac{(26.39)^2}{2 \times 0.65 \times 1.635} \approx 503.0 \mathrm{~m} \).

Key Concepts

Static Friction

Static friction is a force that keeps surfaces from sliding past each other. This frictional force acts when two objects are in contact but do not move relative to each other. In the context of a car coming to a stop, static friction plays an essential role in maintaining traction between the car tires and the road. When a car attempts to stop, the tires begin to exert a force on the road. This force needs to be met equally and oppositely by the static friction for the car to halt effectively.

In this problem, static friction is defined by the equation \[ F_{friction} = \ \ \mu_{s} \times m \times g \]where:

• \( \mu_{s} \) is the coefficient of static friction,
• \( m \) is the mass of the car, and
• \( g \) is the acceleration due to gravity.

Breaking force acts as a decelerating force, crucial for stopping the car within a minimum distance.

Kinematics

Kinematics deals with the motion of objects without considering the forces that cause this motion. When determining the stopping distance, kinematics helps us understand how distance, speed, and acceleration relate to each other. For a car to stop, we utilize a kinematic equation that connects these elements:\[ v^2 = u^2 + 2as \]In this equation:

• \( v \) is the final velocity, which is 0 when the car stops,
• \( u \) is the initial velocity of the car,
• \( a \) is the acceleration (negative in the case of deceleration), and
• \( s \) is the stopping distance we want to find.

By using kinematic relations, we see how initial velocity and frictional deceleration impact the stopping distance. The ultimate formula derived from rearranging the kinematic equation:\[ s = \frac{u^2}{2 \mu_{s} g} \]shows that the square of initial velocity directly affects the stopping distance as it increases quadratically.

Newton's Second Law

Newton's Second Law is fundamental in understanding motion changes when forces are applied. The law is encapsulated in the famous formula:\[ F = m \times a \]where:

• \( F \) is the net force acting on an object,
• \( m \) is the object's mass, and
• \( a \) is the acceleration produced by the net force.

When applied to the problem of a stopping car, the force of static friction is responsible for the car's deceleration. According to Newton's Second Law, the deceleration the car experiences can be illustrated as:\[ \mu_{s} \times m \times g = m \times a \]By simplifying, we obtain:\[ a = \mu_{s} \times g \]This demonstrates that the acceleration (or deceleration) is independent of mass, showing how static friction and gravity exclusively determine the car's ability to stop.

Deceleration

Deceleration refers to the reduction in velocity or speed of an object. When a car brakes, it undergoes deceleration to reach a stop, primarily due to friction between the tires and the road. In this exercise, the deceleration caused by friction is critical for defining how quickly a car can stop from a given speed.

Deceleration can be regarded as a negative acceleration since it involves a decrease in speed. As calculated using Newton’s Second Law in the context of friction:\[ a = -\mu_{s} \times g \]This outcome emphasizes that

• deceleration is influenced by the coefficient of static friction \( \mu_{s} \), and
• the gravitational acceleration \( g \),

though it operates in the opposite direction of the car's motion, allowing it to stop smoothly and safely as expected.

Acceleration Due to Gravity

Acceleration due to gravity \( g \) is the rate at which objects accelerate towards Earth. On Earth, this is approximately \( 9.81 \, \text{m/s}^2 \). It plays a pivotal role in calculating the deceleration of a car when it stops because it contributes to the force of static friction.

On the Earth's surface, objects experience this gravitational pull, influencing how static friction acts during a stop. However, if the scenario shifts to a place like the Moon, where gravity is weaker (\( g/6 \)), this affects the stopping distance significantly.

In the Moon's reduced gravitational field, the same car would take much longer to stop due to a decrease in the static friction force, thereby increasing the stopping distance. This reveals the crucial element gravity plays in mechanical systems, especially when differential scenarios like lunar conditions are involved.