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Chapter 5: Problem 33

(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.)

Short Answer

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

Step by step solution

01

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 (\( v_i = 28.7 \text{ m/s} \)).
02

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 (\(9.8 \text{ m/s}^2\)).
03

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.)
04

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 \(d\): \[ 28.7^2 = 2 \times 7.84 \times d \] \[ d = \frac{28.7^2}{2 \times 7.84} \approx 52.5 \text{ meters} \]
05

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 \( \mu_k = 0.25 \).
06

Calculate Deceleration (Part b)

Using the new coefficient of kinetic friction: \( a = 0.25 \times 9.8 = 2.45 \text{ m/s}^2 \).
07

Determine Initial Velocity Using Stopping Distance (Part b)

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

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

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.

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Most popular questions from this chapter

People (especially the elderly) who are prone to falling can wear hip pads to cushion the impact on their hip from a fall. Experiments have shown that if the speed at impact can be reduced to 1.3 \(\mathrm{m} / \mathrm{s}\) or less, the hip will usually not fracture. Let us investigate the worst-case scenario in which a 55 -kg person completely loses her footing (such as on icy pavement) and falls a distance of \(1.0 \mathrm{m},\) the distance from her hip to the ground. We shall assume that the person's entire body has the same acceleration, which, in reality, would not quite be true. (a) With what speed does her hip reach the ground? (b) A typical hip pad can reduce the person's speed to 1.3 \(\mathrm{m} / \mathrm{s}\) over a distance of 2.0 \(\mathrm{cm} .\) Find the acceleration (assumed to be constant) of this person's hip while she is slowing down and the force the pad exerts on it. (c) The force in part (b) is very large. To see whether it is likely to cause injury, calculate how long it lasts.

A 45.0 -kg crate of tools rests on a horizontal floor. You exert a gradually increasing horizontal push on it and observe that the crate just begins to move when your force exceeds 313 \(\mathrm{N}\) . After that you must reduce your push to 208 \(\mathrm{N}\) to keep it moving at a steady 25.0 \(\mathrm{cm} / \mathrm{s}\) (a) What are the coefficients of static and kinetic friction between the crate and the floor? (b) What push must you exert to give it an acceleration of 1.10 \(\mathrm{m} / \mathrm{s}^{2} ?\) (c) Suppose you were performing the same experiment on this crate but were doing it on the moon instead, where the acceleration due to gravity is 1.62 \(\mathrm{m} / \mathrm{s}^{2}\) . (i) What magnitude push would cause it to move? (ii) What would its acceleration be if you maintained the push in part (b)?

A box with mass \(m\) is dragged across a level floor having a coefficient of kinetic friction \(\mu_{k}\) by a rope that is pulled upward at an angle \(\theta\) above the horizontal with a force of magnitude \(F .\) (a) In terms of \(m, \mu_{k}, \theta,\) and \(g,\) obtain an expression for the magnitude of the force required to move the box with constant speed. (b)Knowing that you are studying physics, a CPR instructor asks you how much force it would take to slide a 90 -kg patient across a floor at constant speed by pulling on him at an angle of \(25^{\circ}\) above the horizontal. By dragging some weights wrapped in an old pair of pants down the hall with a spring balance, you find that \(\mu_{\mathrm{k}}=0.35 .\) Use the result of part (a) to answer the instructor's question.

You are called as an expert witness in the trial of a traffic violation. The facts are these: A driver slammed on his brakes and came to a stop with constant acceleration. Measurements of his tires and the skid marks on the pavement indicate that he locked his car's wheels, the car traveled 192 ft before stopping, and the coefficient of kinetic friction between the road and his tires was \(0.750 .\) The charge is that he was speeding in a \(45-\mathrm{mi} / \mathrm{h}\) zone. He pleads innocent. What is your conclusion, guilty or innocent? How fast was he going when he hit his brakes?

You are riding your motorcycle one day down a wet street that slopes downward at an angle of \(20^{\circ}\) below the horizontal. As you start to ride down the hill, you notice a construction crew has dug a deep hole in the street at the bottom of the hill. A siberian tiger, escaped from the City Zoo, has taken up residence in the hole. You apply the brakes and lock your wheels at the top of the hill, where you are moving with a speed of 20 \(\mathrm{m} / \mathrm{s} .\) The inclined street in front of you is 40 \(\mathrm{m}\) long. (a) Will you plunge into the hole and become the tiger's lunch, or do you skid to a stop before you reach the hole? (The coefficients of friction between your motorcycle tires and the wet pavement are \(\mu_{\mathrm{s}}=0.90\) and \(\mu_{\mathrm{k}}=0.70 .\) ) ( b ) What must your initial speed be if you are to stop just before reaching the hole?
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