Problem 33 - University Physics with Modern Physics

Question

(a) If the coefficient of kinetic friction between tires and dry pavement is \(0.80,\) what is the shortest distance in which you can stop an automobile by locking the brakes when traveling at 28.7 \(\mathrm{m} / \mathrm{s}\) (about 65 \(\mathrm{mi} / \mathrm{h} ) ?\) (b) On wet pavement the coefficient of kinetic friction may be only \(0.25 .\) How fast should you drive on wet pavement in order to be able to stop in the same distance as in part (a)? (Note: Locking the brakes is not the safest way to stop.)

Step-by-Step Solution

Answer

(a) 52.5 m; (b) approximately 16.07 m/s.

Step 1: Identify Given Variables for Part (a)

We need to find the stopping distance using the given coefficient of kinetic friction (\( \mu_k = 0.80 \)) and initial velocity (\( v_i = 28.7 \text{ m/s} \)).

Step 2: Apply the Kinetic Friction Formula

Kinetic friction provides the maximum deceleration when brakes are locked. The deceleration \(a\) can be computed using the formula: \( a = \mu_k \cdot g \), where \(g\) is the acceleration due to gravity (\(9.8 \text{ m/s}^2\)).

Step 3: Calculate Deceleration (Part a)

Substitute \( \mu_k = 0.80 \) and \( g = 9.8 \text{ m/s}^2 \) into the formula: \[ a = 0.80 \times 9.8 = 7.84 \text{ m/s}^2 \] (Note: This deceleration will be negative because it's a stopping force.)

Step 4: Find the Stopping Distance Using Deceleration (Part a)

The stopping distance \(d\) can be found using the equation \( v^2 = u^2 + 2ad \), where \(v\) is the final velocity (\(0 \text{ m/s}\)) and \(u\) is the initial velocity: \[ 0 = (28.7)^2 + 2(-7.84)d \] Solve for \(d\): \[ 28.7^2 = 2 \times 7.84 \times d \] \[ d = \frac{28.7^2}{2 \times 7.84} \approx 52.5 \text{ meters} \]

Step 5: Identify Variable for Part (b)

We need to find the initial velocity on wet pavement so that the stopping distance remains the same (\(d = 52.5 \text{ meters} \)), with \( \mu_k = 0.25 \).

Step 6: Calculate Deceleration (Part b)

Using the new coefficient of kinetic friction: \( a = 0.25 \times 9.8 = 2.45 \text{ m/s}^2 \).

Step 7: Determine Initial Velocity Using Stopping Distance (Part b)

Using the same distance formula, solve for \(u\): \[ 0 = u^2 + 2(-2.45)(52.5) \] \[ "u^2 = 2 \times 2.45 \times 52.5 \] \[ u = \sqrt{2 \times 2.45 \times 52.5} \approx 16.07 \text{ m/s} \]

Key Concepts

Coefficient of Friction

The coefficient of friction (\( \mu \)) is a value that represents the frictional force between two surfaces. It is a dimensionless scalar, meaning it has no units. This value is crucial in determining how much frictional force exists between the tires of a vehicle and the road surface.

When discussing kinetic friction, specifically, we're talking about the frictional force when there is relative motion between the surfaces in contact. In the given exercise, the coefficients of kinetic friction are \( \mu_k = 0.80 \) for dry pavement and \( \mu_k = 0.25 \) for wet pavement. These values highlight how different surfaces affect the force needed to slow down a moving object.

The higher the coefficient, the greater the frictional force; this explains why vehicles can stop more quickly on dry surfaces compared to wet ones. It is imperative for students to understand this concept to evaluate real-world scenarios, such as how road conditions affect stopping distances.

Deceleration

Deceleration is essentially negative acceleration. While acceleration refers to an increase in velocity, deceleration refers to a decrease in velocity. When a vehicle is slowing down, it is decelerating. This is crucial when calculating stopping distances since it directly affects how quickly a vehicle can come to a complete stop.

In the exercise, deceleration is calculated using the formula:

• \( a = \mu_k \cdot g \)

where:

• \( a \) is the deceleration,
• \( \mu_k \) is the coefficient of kinetic friction, and
• \( g \) is the acceleration due to gravity (approximately \( 9.8 \text{ m/s}^2 \)).

In the problem's context, differing road conditions affect \( \mu_k \) and hence the deceleration; on dry surfaces, \( a = 0.80 \times 9.8 = 7.84 \text{ m/s}^2 \), and on wet surfaces, \( a = 0.25 \times 9.8 = 2.45 \text{ m/s}^2 \).

To properly handle questions related to vehicle stopping, one must take into account how deceleration varies with different surface conditions.

Stopping Distance

Stopping distance is the total distance a vehicle travels from the moment the brakes are applied until it comes to a complete stop. It is a critical aspect of driving safety, and understanding how various factors affect it can save lives. In physics, stopping distance can be determined using the equation:

• \( v^2 = u^2 + 2ad \)

where:

• \( v \)is the final velocity (usually \( 0 \text{ m/s} \) when a vehicle stops completely),
• \( u \) is the initial velocity, and
• \( d \) is the stopping distance.

In the exercise, for a vehicle traveling at an initial velocity of \( 28.7 \text{ m/s} \) with a coefficient of friction of \( 0.80 \), the stopping distance is approximately \( 52.5 \text{ meters} \) on dry pavement.

For the same stopping distance on wet pavement with a coefficient of friction of \( 0.25 \), the initial speed must be reduced to approximately \( 16.07 \text{ m/s} \). This calculation highlights the importance of adjusting driving speed depending on road conditions to maintain safety.