/*! 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 4 In a performance test, each of t... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

In a performance test, each of two cars takes 9.0 s to accelerate from rest to 27 m/s. Car A has a mass of 1400 kg, and car B has a mass of 1900 kg. Find the net average force that acts on each car during the test.

Short Answer

Expert verified
Car A: 4200 N, Car B: 5700 N.

Step by step solution

01

Understand the Problem

We are tasked with finding the net average force that acts on each car during acceleration. We know the initial velocity is 0 m/s (since they start from rest), final velocity is 27 m/s, time taken is 9.0 s, and we have the masses of both cars.
02

Use the Formula for Force

We can calculate the net average force on each car by using Newton's second law, which states \( F = ma \). First, we need to find the acceleration \( a \) for the cars as \( a = \frac{v_f - v_i}{t} \) where \( v_f = 27 \text{ m/s} \), \( v_i = 0 \text{ m/s} \), and \( t = 9.0 \text{ s}\).
03

Calculate the Acceleration

Substitute the known values into the acceleration formula: \( a = \frac{27 \text{ m/s} - 0 \text{ m/s}}{9.0 \text{ s}} = 3 \text{ m/s}^2 \). The cars have the same acceleration \( 3 \text{ m/s}^2 \).
04

Compute the Force for Car A

For Car A, the mass \( m = 1400 \text{ kg} \). Use the force formula \( F = ma \) to find the force: \( F = 1400 \text{ kg} \times 3 \text{ m/s}^2 = 4200 \text{ N} \).
05

Compute the Force for Car B

For Car B, the mass \( m = 1900 \text{ kg} \). Use the same force formula \( F = ma \) to calculate: \( F = 1900 \text{ kg} \times 3 \text{ m/s}^2 = 5700 \text{ N} \).

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 91Ó°ÊÓ!

Key Concepts

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

Force Calculation
Calculating force is a fundamental concept in physics, particularly when dealing with motion and Newton's second law. The formula used is \( F = ma \), where \( F \) is the force, \( m \) is the mass, and \( a \) is the acceleration. In this exercise, both car A and car B are accelerating from rest, which means they are initially stationary.
To find the net average force that acts on each car, first determine the acceleration (which remains constant in this scenario) and then use this acceleration value to compute the force.
  • For car A, with a mass of 1400 kg, the force was calculated to be 4200 N.
  • For car B, with a mass of 1900 kg, the force was found to be 5700 N.
The differing forces arise from the difference in mass between the two cars, demonstrating that more massive objects require greater force to achieve the same acceleration.
Acceleration Determination
Acceleration is a key parameter in motion problems, especially when using Newton's second law. When objects accelerate from a standstill, determining their acceleration helps us calculate the acting force.
In this case, both cars accelerated from an initial velocity of 0 m/s to a final velocity of 27 m/s over a 9.0-second interval.
The formula to calculate acceleration is:\[ a = \frac{v_f - v_i}{t} \] where:
  • \( v_f \) is the final velocity = 27 m/s
  • \( v_i \) is the initial velocity = 0 m/s
  • \( t \) is the time taken = 9.0 s
Substituting those values, we get:\[ a = \frac{27 \text{ m/s} - 0 \text{ m/s}}{9.0 \text{ s}} = 3 \text{ m/s}^2 \]This shows each car experiences an identical acceleration of 3 m/s², which is crucial for calculating the net average forces acting on them.
Mass and Force Relation
The relationship between mass and force is a core concept of Newton's second law. This relationship demonstrates how the same acceleration affects objects of different masses.
Given the formula \( F = ma \), it is evident that force is directly proportional to both mass and acceleration. When two objects experience the same acceleration, the one with more mass will require a larger force to achieve that acceleration.
  • Car A, with a mass of 1400 kg, required a force of 4200 N to accelerate.
  • Car B, being heavier with a mass of 1900 kg, needed 5700 N of force to reach the same acceleration.
This shows that as mass increases, the required force for the same amount of acceleration also increases. Understanding this mass-force relationship helps reinforce the principles behind calculating the force for any accelerating object.

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

A wagon is rolling forward on level ground. Friction is negligible. The person sitting in the wagon is holding a rock. The total mass of the wagon, rider, and rock is 95.0 \(\mathrm{kg}\) . The mass of the rock is 0.300 \(\mathrm{kg}\) . Initially the wagon is rolling forward at a speed of 0.500 \(\mathrm{m} / \mathrm{s}\) . Then the person throws the rock with a speed of 16.0 \(\mathrm{m} / \mathrm{s}\) . Both speeds are relative to the ground. Find the speed of the wagon after the rock is thrown directly forward in one case and directly backward in another.

mmh For tests using a ballistocardiograph, a patient lies on a horizontal platform that is supported on jets of air. Because of the air jets, the friction impeding the horizontal motion of the plafform is negligible. Each time the heart beats, blood is pushed out of the heart in a direction that is nearly parallel to the platform. Since momentum must be conserved, the body and the platform recoil, and this recoil can be detected to provide information about the heart. For each beat, suppose that 0.050 \(\mathrm{kg}\) of blood is pushed out of the heart with a velocity of \(+0.25 \mathrm{m} / \mathrm{s}\) and that the mass of the pattient and platform is 85 \(\mathrm{kg}\) . Assuming that the patient does not slip with respect to the platform, and that the patient and platform start from rest, determine the recoil velocity.

ssm \(\mathrm{A} 50.0\) -kg skater is traveling due east at a speed of 3.00 \(\mathrm{m} / \mathrm{s} . \mathrm{A}\) 70.0-kg skater is moving due south at a speed of 7.00 \(\mathrm{m} / \mathrm{s}\) . They collide and hold on to each other after the collision, managing to move off at an angle \(\theta\) south of east, with a speed of \(v_{\mathrm{f}}\) . Find \(\quad\) (a) the angle \(\theta\) and (b) the speed \(v_{\mathrm{f}},\) assuming that friction can be ignored.

@ One object is at rest, and another is moving. The two collide in a one- dimensional, completely inelastic collision. In other words, they stick together after the collision and move off with a common velocity. Momentum is conserved. The speed of the object that is moving initially is 25 \(\mathrm{m} / \mathrm{s}\) . The masses of the two objects are 3.0 and 8.0 \(\mathrm{kg}\) . Determine the final speed of the two-object system after the collision for the case when the large-mass object is the one moving initially and the case when the small- mass object is the one moving initially.

mmh A stream of water strikes a stationary turbine blade horizon- tally, as the drawing illustrates. The incident water stream has a velocity of 16.0 m/s, while the exiting water stream has a velocity of 16.0 m/s. The mass of water per second that strikes the blade is 30.0 kg/s. Find the magnitude of the average force exerted on the water by the blade.

See all solutions

Recommended explanations on Physics Textbooks

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.