/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 50 Multiple-Concept Example 17 revi... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 4: Problem 50

Multiple-Concept Example 17 reviews the basic concepts involved in this problem. Air rushing over the wings of high-performance race cars generates unwanted horizontal air resistance but also causes vertical downforce, which helps the cars hug the track more securely. The coefficient of static friction between the track and the tires of a \(690-\mathrm{kg}\) race car is \(0.87 .\) What is the magnitude of the maximum acceleration at which the car can speed up without its tires slipping when a \(4060-\mathrm{N}\) downforce and an \(1190-\mathrm{N}\) horizontal-air-resistance force act on it?

Short Answer

Expert verified
The maximum acceleration is approximately \(11.80\,\text{m/s}^2\).

Step by step solution

01

Identify Forces

List all the forces acting on the race car. The weight of the car is a force downwards, calculated as \( W = mg \), where \( m = 690 \, \text{kg} \) and \( g = 9.8 \, \text{m/s}^2 \). There is also the downforce of \( 4060 \, \text{N} \) acting downwards and the horizontal air resistance force of \( 1190 \, \text{N} \) acting horizontally.
02

Calculate Normal Force

The normal force \( N \) is the sum of the car's weight and the downward downforce: \[ N = mg + \text{downforce} = (690 \, \text{kg})(9.8 \, \text{m/s}^2) + 4060 \, \text{N} = 10722 \, \text{N}. \]
03

Calculate Maximum Static Friction Force

The maximum static friction force \( f_s \) can be calculated using the coefficient of static friction \( \mu_s = 0.87 \) and the normal force \( N \): \[ f_s = \mu_s N = 0.87 \times 10722 \, \text{N} = 9328.14 \, \text{N}. \]
04

Calculate Net Horizontal Force

The net horizontal force \( F_{\text{net}} \) is the difference between the maximum static friction and the horizontal air resistance: \[ F_{\text{net}} = f_s - F_{\text{resistance}} = 9328.14 \, \text{N} - 1190 \, \text{N} = 8138.14 \, \text{N}. \]
05

Calculate Maximum Acceleration

Use Newton's second law \( F = ma \) to find the maximum acceleration \( a \). Solve for \( a \): \[ a = \frac{F_{\text{net}}}{m} = \frac{8138.14 \, \text{N}}{690 \, \text{kg}} \approx 11.80 \, \text{m/s}^2. \]

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91影视!

Key Concepts

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

Normal Force
The normal force is a crucial concept in understanding how objects interact with surfaces. It is the force exerted by a surface to support the weight of an object resting on it. This force acts perpendicular to the surface.
In our case, the normal force is not just the weight of the car. The downforce from the race car's wings also contributes. This additional downforce is crucial in motorsports as it helps the car maintain traction by increasing the force pressing the tires onto the track.
  • The car's weight is calculated using the formula: \( W = mg \), where \( m \) is the mass, and \( g \) is the gravitational acceleration \( (9.8 \, \text{m/s}^2) \).
  • The total normal force is the sum of the car's weight and the downforce: \( N = mg + \text{downforce} \).
This increased normal force allows for greater friction between the tires and the track, which is vital for preventing slipping.
Downforce
Downforce is a downward-directed force created mainly by aerodynamic features like wings on a vehicle. In high-speed race cars, downforce enhances the car's grip on the track.
This force helps in "pushing" the car onto the track. With 4060 N of downforce in our case, it plays an essential role in how much frictional force the tires can generate.
  • Downforce effectively increases the normal force which in turn increases the maximum possible static friction.
  • More static friction means the car can accelerate harder without the tires slipping.
The strategic use of downforce in racing cars is a careful balance 鈥� maximizing grip without causing excessive drag.
Newton's Second Law
Newton's Second Law is foundational in understanding motion and forces. It states that the force acting on an object is equal to the mass of that object times its acceleration: \( F = ma \).
In this exercise, we apply Newton's Second Law to find the maximum acceleration the race car can achieve without its tires slipping. The net force used here is the force of static friction minus any opposing forces like air resistance.
  • We calculate the net force using the formula: \( F_{\text{net}} = f_s - F_{\text{resistance}} \).
  • By rearranging Newton's Second Law, we solve for acceleration: \( a = \frac{F_{\text{net}}}{m} \).
This allows us to determine how quickly the car can speed up, effectively translating the force into motion.
Maximum Acceleration
Maximum acceleration is the greatest rate of change of velocity that an object can achieve under given conditions without losing traction. For our race car, it is determined by the forces acting on it and the available friction.
Static friction plays a key role here, as it provides the grip necessary for acceleration without slipping.
  • The maximum static friction force \( f_s \) is based on the normal force and the coefficient of static friction \( \mu_s \).
  • Once we calculate the net force, Newton's second law helps us find the maximum acceleration: \( a = \frac{F_{\text{net}}}{m} \).
This value is crucial for the car's performance on the track, ensuring it can achieve optimal speeds safely.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

When a 58 -g tennis ball is served, it accelerates from rest to a speed of 45 \(\mathrm{m} / \mathrm{s}\) . The impact with the racket gives the ball a constant acceleration over a distance of 44 \(\mathrm{cm} .\) What is the magnitude of the net force acting on the ball?

A stuntman is being pulled along a rough road at a constant velocity by a cable attached to a moving truck. The cable is parallel to the ground. The mass of the stuntman is 109 kg, and the coefficient of kinetic friction between the road and him is 0.870. Find the tension in the cable.

ssm A 6.00-kg box is sliding across the horizontal floor of an elevator. The coefficient of kinetic friction between the box and the floor is 0.360. Determine the kinetic frictional force that acts on the box when the elevator is (a) stationary, (b) accelerating upward with an acceleration whose magnitude is 1.20 \(\mathrm{m} / \mathrm{s}^{2}\) , and \((\mathrm{c})\) accelerating downward with an acceleration whose magnitude is 1.20 \(\mathrm{m} / \mathrm{s}^{2}\) .

A skater with an initial speed of 7.60 \(\mathrm{m}/\mathrm{s}\) stops propelling himself and begins to coast across the ice, eventually coming to rest. Air resistance is negligible. (a) The coefficient of kinetic friction between the ice and the skate blades is 0.100. Find the deceleration caused by kinetic friction. (b) How far will the skater travel before coming to rest?

An airplane has a mass of \(3.1 \times 10^{4} \mathrm{kg}\) and takes off under the influence of a constant net force of \(3.7 \times 10^{4} \mathrm{N}\) . What is the net force that acts on the plane's \(78-\mathrm{kg}\) pilot?
See all solutions

Recommended explanations on Physics Textbooks

Radiation

Read Explanation

Fields in Physics

Read Explanation

Physics of Motion

Read Explanation

Energy Physics

Read Explanation

Oscillations

Read Explanation

Nuclear Physics

Read Explanation
View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.