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Chapter 5: Problem 36

\(\bullet$$\bullet\) Stopping distance of a car. (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 lock- ing the brakes when traveling at 29.1 \(\mathrm{m} / \mathrm{s}\) (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 ordel 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) 54 meters; (b) Drive at 16.3 m/s on wet pavement.

Step by step solution

01

Understand Given Information

We need to find the stopping distance for an automobile under two different friction conditions. In part (a), the coefficient of kinetic friction on dry pavement is given as \(\mu = 0.80\) and the initial speed is \(v_0 = 29.1\, \text{m/s}\). In part (b), the coefficient of kinetic friction on wet pavement is \(\mu = 0.25\). We have to find the speed such that the stopping distance is the same as in part (a).
02

Determine Deceleration Due to Friction

The deceleration due to friction \(a\) can be calculated using the formula \(a = \mu g\), where \(g = 9.8 \, \text{m/s}^2\) is the acceleration due to gravity. For part (a):\[ a = 0.80 \times 9.8 = 7.84 \, \text{m/s}^2. \] In part (b), we only need this value to keep calculations similar.
03

Use Kinematic Equation to Calculate Stopping Distance

The stopping distance \(d\) can be determined using the kinematic equation \(v^2 = v_0^2 + 2ad\). Since the final velocity \(v = 0\) when the car stops: \[ 0 = (29.1)^2 + 2(-7.84)d. \] Solving this for \(d\), we get: \[ d = \frac{(29.1)^2}{2(7.84)} = 54 \text{ m}. \]
04

Calculate Initial Speed on Wet Pavement

For part (b), we use the same kinematic equation but adjust it for the new \(\mu = 0.25\). Since the stopping distance \(d\) is 54 m, we have \(a = 0.25 \times 9.8 = 2.45 \, \text{m/s}^2\): \[ 0 = v_{wet}^2 - 2(2.45)(54). \] Solving for \(v_{wet}\), we find: \[ v_{wet} = \sqrt{2 \times 2.45 \times 54} \approx 16.3 \, \text{m/s}. \]
05

Finalize the Results

In summary, the shortest stopping distance on dry pavement with a friction coefficient of 0.80 is 54 meters. To achieve this stopping distance on wet pavement (friction coefficient 0.25), the driving speed should be reduced to approximately 16.3 m/s.

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Key Concepts

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

Kinetic Friction
Kinetic friction is the force that resists the movement of two objects sliding against each other. In the context of stopping a car, it is a crucial factor because it determines how effectively the car can come to a halt.
Unlike static friction, which acts on objects that are not sliding, kinetic friction acts when there is relative motion between the surfaces. The kinetic friction force can be calculated using the formula:
  • \( f_k = \mu_k N \)
where \( f_k \) is the kinetic friction force, \( \mu_k \) is the coefficient of kinetic friction, and \( N \) is the normal force, typically the weight of the car due to gravity.
When the brakes are locked, the car skids, and kinetic friction becomes the dominant force opposing the car's motion. This friction force is directly proportional to the coefficient of kinetic friction, meaning the higher the value, the greater the resistance, shortening the stopping distance.
Deceleration
Deceleration is essentially negative acceleration, which means a decrease in speed. When a car brakes, it experiences deceleration due to the force of friction acting against its motion.
This force leads to a rapid drop in speed, eventually bringing the vehicle to a stop. The rate of deceleration due to friction can be calculated using the formula:
  • \( a = \mu g \)
In this formula, \( a \) is the deceleration, \( \mu \) is the coefficient of friction, and \( g \) is the acceleration due to gravity (about 9.8 m/s虏 on Earth).
A higher coefficient of friction leads to a greater deceleration, as seen on dry pavement with a friction coefficient of 0.80, resulting in an effective deceleration that allows for shorter stopping distances, while a lower coefficient seen on wet roads leads to decreased deceleration.
Kinematic Equations
Kinematic equations are fundamental in physics for describing the motion of objects. They connect concepts like displacement, initial velocity, final velocity, acceleration, and time.
In the case of stopping a car, the most relevant kinematic equation is:
  • \( v^2 = v_0^2 + 2ad \)
where \( v \) is the final velocity (zero when the car stops), \( v_0 \) is the initial velocity, \( a \) is acceleration (or deceleration in this context), and \( d \) is the stopping distance.
This equation is used to calculate how far the car will travel until it completely stops given a certain deceleration. By substituting known values, such as initial speed and deceleration, one can solve for the stopping distance or the necessary initial speed to achieve a particular stopping distance under different conditions.
Coefficient of Friction
The coefficient of friction is a dimensionless number that represents the frictional properties between two surfaces.
There are two main types: static and kinetic, but here, kinetic friction is specially considered, as it applies to objects in motion.
A higher coefficient of kinetic friction indicates that more force is required to keep the motion going, which is helpful in stopping a car quickly on dry pavement. The coefficient is affected by:
  • Type of materials in contact (e.g., rubber tires on asphalt)
  • Surface conditions (e.g., dry vs. wet pavement)
  • Presence of lubricants or contaminants
In this exercise, dry pavement has a coefficient of 0.80, indicating stronger frictional force compared to wet pavement with a coefficient of 0.25. As a result, more force is needed to stop a car on dry pavement, which allows for shorter stopping distances.

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Most popular questions from this chapter

\(\bullet$$\bullet\) Atwood's Machine. A 15.0 -kg load of bricks hangs from one end of a rope that passes over a small, frictionless pulley. A 28.0 -kg counterweight is suspended from the other end of the rope, as shown in Fig. 5.60 . The system is released from rest. (a) Draw two free-body diagrams, one for the load of bricks and one for the counterweight. (b) What is the magnitude of the upward acceleration of the load of bricks? (c) What is the ten- sion in the rope while the load is moving? How does the ten- sion compare to the weight of the load of bricks? To the weight of the counterweight?

\(\bullet$$\bullet\) You've attached a bungee cord to a wagon and are using it to pull your little sister while you take her for a jaunt. The bungee's unstretched length is \(1.3 \mathrm{m},\) and you happen to know that your little sister weighs 220 \(\mathrm{N}\) and the wagon weighs 75 \(\mathrm{N}\) . Crossing a street, you accelerate from rest to your normal walk- ing speed of 1.5 \(\mathrm{m} / \mathrm{s}\) in 2.0 \(\mathrm{s}\) , and you notice that while you're accelerating, the bungee's length increases to about 2.0 \(\mathrm{m}\) . What's the force constant of the bungee cord, assuming it obeys Hooke's law?

\(\bullet$$\bullet \mathrm{A} 75,600 \mathrm{N}\) spaceship comes in for a vertical landing. From an initial speed of \(1.00 \mathrm{km} / \mathrm{s},\) it comes to rest in 2.00 \(\mathrm{min}\) . with uniform acceleration. (a) Make a free-body diagram of this ship as it is coming in. (b) What braking force must its rockets provide? Ignore air resistance.

\(\bullet$$\bullet\) The monkey and her bananas. \(\Lambda 20 \mathrm{kg}\) monkey has a firm hold on a light rope that passes over a frictionless pulley and is attached to a 20 \(\mathrm{kg}\) bunch of bananas (Figure 5.72 ). The monkey looks up, sees the bananas, and starts to climb the rope to get them. (a) As the monkey climbs, do the bananas move up, move down, or remain at rest? (b) As the monkey climbs, does the distance between the monkey and the ba- nanas decrease, increase, or remain con- stant? (c) The monkey releases her hold on - the rope. What happens to the distance be- tween the monkey and the bananas while she is falling? (d) Before reaching the ground, the monkey grabs the rope to stop her fall. What do the bananas do?

\(\bullet$$\bullet\) Friction in an elevator. You are riding in an elevator on the way to the 18 th floor of your dormitory. The elevator is accelerating upward with \(a=1.90 \mathrm{m} / \mathrm{s}^{2} .\) Beside you is the box containing your new computer; the box and its contents have a total mass of 28.0 \(\mathrm{kg} .\) While the elevator is accelerating upward, you push horizontally on the box to slide it at constant speed toward the elevator door. If the coefficient of kinetic friction between the box and the elevator floor is \(\mu_{\mathrm{k}}=0.32\) what magnitude of force must you apply?
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