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Chapter 4: Problem 71

A city planner is working on the redesign of a hilly portion of a city. An important consideration is how steep the roads can be so that even low-powered cars can get up the hills without slowing down. A particular small car, with a mass of 920 kg, can accelerate on a level road from rest to 21 m/s (75 km/h) in 12.5 s. Using these data, calculate the maximum steepness of a hill.

Short Answer

Expert verified
The maximum steepness of the hill is \( 9.85^\circ \).

Step by step solution

01

Calculate the Force Required on a Level Road

First, find the force required for the car to accelerate on a level road. Use the formula for force: \[ F = ma \]Where:- \( m = 920 \text{ kg} \) (mass of the car)- \( a \) is the accelerationTo find \( a \), use the formula for acceleration: \[ a = \frac{v_f - v_i}{t} \]Here, \( v_i = 0 \text{ m/s} \), \( v_f = 21 \text{ m/s} \), and \( t = 12.5 \text{ s} \).Thus, \[ a = \frac{21 - 0}{12.5} = 1.68 \text{ m/s}^2 \]Substitute the values into the force equation:\[ F = 920 \times 1.68 = 1545.6 \text{ N} \]
02

Calculate the Gravitational Component Along the Hill

For a hilly road, the maximum force available to overcome gravitational pull is the same as the force used on the level road. The gravitational component along the hill is given by:\[ F_{gravity} = mg\sin(\theta) \]Here, \( g \) is the gravitational acceleration (\(9.81 \text{ m/s}^2\)) and \( \theta \) is the inclination angle of the hill. In the most challenging situation, \[ F_{gravity} = 1545.6 \text{ N} \]
03

Solve for the Maximum Hill Steepness

Equate the gravitational force equation to the force calculated for the level surface:\[ mg\sin(\theta) = 1545.6 \]Now, substitute the known values of \( m \) and \( g \):\[ 920 \times 9.81 \times \sin(\theta) = 1545.6 \]\[ 9025.2 \times \sin(\theta) = 1545.6 \]Solve for \( \sin(\theta) \):\[ \sin(\theta) = \frac{1545.6}{9025.2} \approx 0.171 \]Determine \( \theta \) using the inverse sine function:\[ \theta = \sin^{-1}(0.171) \approx 9.85^\circ \]
04

Conclusion

The maximum steepness of the hill that the car can climb without slowing down is approximately \( 9.85^\circ \).

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Key Concepts

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

Newton's laws of motion
Newton's Laws of Motion are fundamental principles that describe how objects move and interact with forces. These laws were formulated by Sir Isaac Newton in the 17th century and remain essential in physics today. Here鈥檚 a brief overview of each:
  • First Law: An object at rest stays at rest, and an object in motion stays in motion at a constant velocity unless acted upon by a net external force.
  • Second Law: The acceleration of an object is directly proportional to the net force acting on it and inversely proportional to its mass, articulated as \( F = ma \).
  • Third Law: For every action, there is an equal and opposite reaction.
These laws explain why a car needs a force to overcome road resistance to start moving or to continue moving uphill. In our exercise, the car utilizes the second law of motion, where force is calculated based on its mass and the acceleration required to climb a hill. Understanding these laws helps in determining how forces need to be managed to maintain motion on inclined planes.
acceleration
Acceleration is the rate at which an object changes its velocity. It is an essential aspect of understanding motion dynamics, especially for vehicles on roads. In physics, acceleration is calculated using the formula:\[a = \frac{v_f - v_i}{t}\]where:
  • \( a \) = acceleration
  • \( v_f \) = final velocity
  • \( v_i \) = initial velocity
  • \( t \) = time taken to change velocity
In the given problem, the car's acceleration is 1.68 m/s虏, which is critical to determining how it can perform on both level and inclined roads. Mastery of acceleration concepts helps predict how quickly a vehicle can reach desired speeds and how much force is necessary for that process.
gravitational force
Gravitational force is the force by which a planet or other massive object attracts another object towards its center. For Earth, this force is quantified by the acceleration due to gravity \( g \), which is approximately 9.81 m/s虏 at the Earth's surface. When dealing with inclined planes, like hills, the gravitational force acting on an object has components:
  • Parallel Component : This is the force pulling directly down the slope, calculated as \( mg\sin(\theta) \).
  • Perpendicular Component : Acts perpendicular to the slope, calculated as \( mg\cos(\theta) \), and does not influence movement directly.
In the exercise, the parallel component \( mg\sin(\theta) \) is used to find the maximum angle \( \theta \). This component becomes crucial for understanding how much of the car's force is needed to overcome gravity on the hill.
inclined planes
Inclined planes are surfaces tilted at an angle, other than the horizontal, used to lift or lower objects. They're a staple concept in physics because they demonstrate the practical application of forces. On an inclined plane, different forces come into play:
  • Downward Force: Often driven by gravity, this force compels the object to slide down.
  • Normal Force: Acts perpendicular to the surface of the plane and is crucial in balancing forces.
Analyzing inclined planes involves breaking down forces into components parallel and perpendicular to the surface. In our exercise, understanding the physics of inclined planes is key to calculating the maximum steepness a car can manage. The slope angle \( \theta \), determined by \( \sin^{-1}\left(\frac{1545.6}{9025.2}\right) \), ensures the car can ascend without hindrance.

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Most popular questions from this chapter

(II) A 2.0-kg silverware drawer does not slide readily. The owner gradually pulls with more and more force, and when the applied force reaches 9.0 N, the drawer suddenly opens, throwing all the utensils to the floor. What is the coefficient of static friction between the drawer and the cabinet?

A motorcyclist is coasting with the engine off at a steady speed of 20.0 m/s but enters a sandy stretch where the coefficient of kinetic friction is 0.70.Will the cyclist emerge from the sandy stretch without having to start the engine if the sand lasts for 15 m? If so, what will be the speed upon emerging?

(II) On an icy day, you worry about parking your car in your driveway, which has an incline of 12\(^\circ\). Your neighbor's driveway has an incline of 9.0\(^\circ\), and the driveway across the street is at 6.0\(^\circ\). The coefficient of static friction between tire rubber and ice is 0.15. Which driveway(s) will be safe to park in?

Piles of snow on slippery roofs can become dangerous projectiles as they melt. Consider a chunk of snow at the ridge of a roof with a slope of 34\(^\circ\). (\(a\)) What is the minimum value of the coefficient of static friction that will keep the snow from sliding down? (\(b\)) As the snow begins to melt, the coefficient of static friction decreases and the snow finally slips. Assuming that the distance from the chunk to the edge of the roof is 4.0 m and the coefficient of kinetic friction is 0.10, calculate the speed of the snow chunk when it slides off the roof. (\(c\)) If the roof edge is 10.0 m above ground, estimate the speed of the snow when it hits the ground.

(II) Can cars "stop on a dime"? Calculate the acceleration of a 1400-kg car if it can stop from 35 km/h on a dime (diameter \(=\) 1.7 cm). How many \(g\)'s is this? What is the force felt by the 68-kg occupant of the car?
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