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Chapter 5: Problem 35

Stopping Distance. (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 a car by locking the brakes when the car is 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 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
The shortest distance the car can stop in when moving at \(28.7 m/s\) on dry pavement is about \(41.7 m\). If the pavement is wet, the car can be moving up to about \(16.4 m/s\) and still stop in the same distance.

Step by step solution

01

Determine the stopping distance

Start by applying Newton's second law, which states that the force equals mass times acceleration (\(F=ma\)). Since in this case, the only horizontal force acting on the car is kinetic friction, the formula changes to \(ma=渭kmg\), where \(渭k\) is the coefficient of kinetic friction, \(m\) is the car's mass, and \(g\) is the acceleration due to gravity. The mass of the car is cancelled out on both sides, leaving \(a=渭kg\). Now, use the equation \(vf^2 = vi^2 - 2ad\), where \(vf\) is the final velocity (in this case, 0), \(vi\) is the initial velocity, \(a\) is the acceleration (negative because it's slowing down), and \(d\) is the distance. Solving for \(d\), we get \(d = (vi^2 - vf^2) / 2a\). Substitute \(a\) with \(渭kg\), so \(d = (vi^2 - vf^2) / 2渭kg\). For part (a), substitute \(渭k = 0.80\), \(vi = 28.7 m/s\), \(vf = 0 m/s\), and \(g = 9.8 m/s^2\) into the equation to solve for the stopping distance.
02

Find the maximum speed for stopping within the same distance on wet pavement

In part (b), we need to determine the maximum speed at which the car can be moving and still stop in the same distance as in part (a), but now the coefficient of kinetic friction is only \(0.25\). We still use the equation from step 1, but solve for \(vi\) instead of \(d\). Rearranging the equation, we get \(vi = sqrt(vf^2 + 2渭kgd)\). Since we are using the same stopping distance as in part (a), we know the value for \(d\), and we can substitute \(渭k = 0.25\), \(vf = 0 m/s\), \(g = 9.8 m/s^2\), and the stopping distance from part (a) to find the maximum permissible initial speed.

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Key Concepts

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

Kinetic Friction
When an object slides across a surface, kinetic friction acts against its motion. In the context of a moving car, kinetic friction between the tires and the road surface is what allows the car to come to a stop when the brakes are locked. It's a resistive force described by the equation \( F_k = \mu_k \times N \), where \( F_k \) is the kinetic frictional force, \( \mu_k \) is the coefficient of kinetic friction, and \( N \) is the normal force, often equal to the object's weight. The higher the coefficient, the greater the frictional force.
In our exercise, two scenarios were given, one with dry pavement (\( \mu_k = 0.80 \)) and one with wet pavement (\( \mu_k = 0.25 \)), showing how the variance in \( \mu_k \) produces a significant change in the car's stopping distance. Understanding this relationship is vital for not just physics problems but also for practical driving decisions, like adjusting speed in different weather conditions to maintain control of the vehicle.
Newton's Second Law
Newton's second law fundamentally states that \( F = ma \), which means that force is equal to mass multiplied by acceleration. It connects force directly to the changes in velocity that an object experiences. In our stopping distance problem, this law is at play when we substitute the force of kinetic friction for \( F \) in the equation, leading to \( ma = \mu_kmg \).
This simplifies to \( a = \mu_kg \) after canceling out mass, illustrating how the car's acceleration (or deceleration in this case) is directly dependent on the coefficient of kinetic friction and gravity, not the mass of the car. Hence, for any vehicle mass, as long as the tires and surface conditions remain the same, the deceleration value remains constant. It's critical for students to grasp this concept as it links the abstract principles of forces to the very tangible experience of stopping a moving car.
Deceleration Calculation
Deceleration, the rate at which an object slows down, is a key element in solving stopping distance problems. Calculating deceleration involves determining the negative acceleration necessary to bring a moving object to a stop. The equation \( vf^2 = vi^2 - 2ad \) allows us to do just that. The initial velocity (\/\br(vi\/)) is known, the final velocity (\/\br(vf\/)) is zero since the car comes to a stop, and \( a \) can be expressed in terms of kinetic friction and gravity (\( a = \mu_kg \) from Newton's second law).
The equation can then be rearranged to solve for \( d \) (stopping distance) or \( vi \) (initial velocity), depending on what the problem asks for. Recognizing and performing this deceleration calculation is an essential skill in physics, helping students not only answer homework questions but also understand real-world motion scenarios, such as estimating stopping distances when driving under different conditions.

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Most popular questions from this chapter

A bowling ball weighing \(71.2 \mathrm{~N}(16.0 \mathrm{lb})\) is attached to the ceiling by a \(3.80 \mathrm{~m}\) rope. The ball is pulled to one side and released; it then swings back and forth as a pendulum. As the rope swings through the vertical, the speed of the bowling ball is \(4.20 \mathrm{~m} / \mathrm{s}\). At this instant, what are (a) the acceleration of the bowling ball, in magnitude and direction, and (b) the tension in the rope?

\(\mathrm{A} 45.0 \mathrm{~kg}\) crate of tools rests on a horizontal floor. You exert a gradually increasing horizontal push on it, and the crate just begins to move when your force exceeds \(313 \mathrm{~N}\). Then 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 the moon, 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 large crate with mass \(m\) rests on a horizontal floor. The coefficients of friction between the crate and the floor are \(\mu_{\mathrm{s}}\) and \(\mu_{\mathrm{k}} .\) A woman pushes downward with a force \(\overrightarrow{\boldsymbol{F}}\) on the crate at an angle \(\theta\) below the horizontal. (a) What magnitude of force \(\vec{F}\) is required to keep the crate moving at constant velocity? (b) If \(\mu_{\mathrm{s}}\) is greater than some critical value, the woman cannot start the crate moving no matter how hard she pushes. Calculate this critical value of \(\mu_{\mathrm{s}}\)

A racetrack curve has radius \(120.0 \mathrm{~m}\) and is banked at an angle of \(18.0^{\circ} .\) The coefficient of static friction between the tires and the roadway is \(0.300 .\) A race car with mass \(900 \mathrm{~kg}\) rounds the curve with the minimum speed needed to not slide down the banking. (a) As the car rounds the curve, what is the normal force exerted on it by the road? (b) What is the car's speed?

Double Atwood's Machine. In Fig. \(\mathbf{P 5 . 1 1 4}\) masses \(m_{1}\) and \(m_{2}\) are connected by a light string \(A\) over a light, frictionless pulley \(B .\) The axle of pulley \(B\) is connected by a light string \(C\) over a light, frictionless pulley \(D\) to a mass \(m_{3} .\) Pulley \(D\) is suspended from the ceiling by an attachment to its axle. The system is released from rest. In terms of \(m_{1}, m_{2}, m_{3},\) and \(g,\) what are (a) the acceleration of block \(m_{3}\) (b) the acceleration of pulley \(B ;\) (c) the acceleration of block \(m_{1}\) (d) the acceleration of block \(m_{2} ;\) (e) the tension in string \(A ;\) (f) the tension in string \(C ?\) (g) What do your expressions give for the special case of \(m_{1}=m_{2}\) and \(m_{3}=m_{1}+m_{2} ?\) Is this reasonable?
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