/*! 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 83 A certain force gives an object ... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

A certain force gives an object of mass \(m_{1}\) an acceleration of \(12.0 \mathrm{~m} / \mathrm{s}^{2}\) and an object of mass \(m_{2}\) an acceleration of \(3.30\) \(\mathrm{m} / \mathrm{s}^{2} .\) What acceleration would the force give to an object of mass (a) \(m_{2}-m_{1}\) and (b) \(m_{2}+m_{1}\) ?

Short Answer

Expert verified
(a) 20.4 m/s², (b) 3.00 m/s².

Step by step solution

01

Understand the problem

We are given a force that accelerates objects of different masses at different rates. Specifically, the force causes an object of mass \( m_1 \) to accelerate at \( 12.0 \, \text{m/s}^2 \), and an object of mass \( m_2 \) to accelerate at \( 3.30 \, \text{m/s}^2 \). We need to find the acceleration of the force on objects of mass \( m_2 - m_1 \) and \( m_2 + m_1 \).
02

Calculate the force using mass \( m_1 \)

Using Newton's second law, \( F = m_1 \cdot a_1 \), where \( a_1 = 12.0 \, \text{m/s}^2 \). Therefore, the force \( F \) is given by:\[ F = m_1 \times 12.0 \, \text{m/s}^2 \]
03

Calculate the force using mass \( m_2 \)

Similarly, using Newton's second law for mass \( m_2 \):\[ F = m_2 \times 3.30 \, \text{m/s}^2 \]
04

Relate mass \( m_1 \) to mass \( m_2 \) using forces

Since the force \( F \) is the same, equate the two expressions:\[ m_1 \times 12.0 = m_2 \times 3.30 \]Solve this equation to express \( m_2 \) in terms of \( m_1 \), or vice versa, depending on which quantity is convenient for the problem.
05

Find acceleration for mass \( m_2 - m_1 \)

The force \( F \) is still the same by the equation \( F = (m_2 - m_1) \times a \). Substitute:\[ (m_2 - m_1) \times a = m_1 \times 12 \]Use the expression for \( m_2 \) from the previous step to solve for the new acceleration \( a \).
06

Find acceleration for mass \( m_2 + m_1 \)

Using the relation:\[ (m_2 + m_1) \times a' = m_1 \times 12 \]Again, use the expression for \( m_2 \) from previous steps to solve for \( a' \).
07

Solution to part (a)

Solve \((m_2 - m_1) \times a = m_1 \times 12 \)Using \( m_2 = \frac{12}{3.30} m_1 \),Find \( a = 20.4 \, \text{m/s}^2 \).
08

Solution to part (b)

Using the expression for \( m_2 + m_1 \), solve the equation:\((m_2 + m_1) \times a' = m_1 \times 12 \)To find \( a' = 3.00 \, \text{m/s}^2 \).

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 91Ó°ÊÓ!

Key Concepts

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

Mass and Acceleration
Understanding the relationship between mass and acceleration is crucial to grasping Newton's second law. This law states that the force acting on an object is equal to the mass of the object multiplied by its acceleration. This can be represented mathematically as: \( F = m \cdot a \).
In simple terms:
  • Mass \( (m) \) is a measure of the amount of matter in an object. It helps determine how the object will respond to an applied force.
  • Acceleration \( (a) \) is the rate at which an object's velocity changes with time.

In any given scenario under consistent force, a larger mass will accelerate more slowly than a smaller mass. This is because greater mass requires more effort (or force) to bring about the same change in speed.
For instance, in the exercise, the force impacts two different masses: \( m_1 \) accelerates at \( 12.0 \ \text{m/s}^2 \) and \( m_2 \) at \( 3.30 \ \text{m/s}^2 \).
Understanding this inverse relationship is key to predicting the behavior of objects when forces are applied.
Force Calculations
Calculating forces using Newton's second law involves understanding how force, mass, and acceleration are intertwined. The formula \( F = m \cdot a \) guides this computation.
Consider the following steps in force calculation:
  • Identify the mass \( (m) \) and acceleration \( (a) \) of the object.
  • Substitute these values into the formula to find the force \( (F) \).
In the exercise, the same force \( F \) is described using two scenarios: one involving mass \( m_1 \) and the other with mass \( m_2 \). The calculations determine the force by equating the effects of that force on the two masses using:
  • \( F = m_1 \times 12.0 \ \text{m/s}^2 \)
  • \( F = m_2 \times 3.30 \ \text{m/s}^2 \)

This approach helps compare how different masses respond to the same force and allows solving for unknown parameters.
Object Motion Under Force
The motion of an object under a force is a fundamental aspect of dynamics. Newton's second law lays the foundation for determining how objects move when forces are applied. In the given exercise, we explore how objects of different masses are affected by identical forces.
Key concepts to consider include:
  • Forces applied to an object can change its velocity, hence its motion.
  • The direction of acceleration matches the direction of the force applied.
Using the data provided, we determine the new accelerations for different hypothetical masses \( m_2 - m_1 \) and \( m_2 + m_1 \). The solutions show that adjusting mass results in different accelerations:
  • A reduced mass \((m_2 - m_1)\) led to an increased acceleration of \(20.4 \ \text{m/s}^2\), showing that less mass moves faster under the same force.
  • An increased mass \((m_2 + m_1)\) resulted in a decreased acceleration of \(3.00 \ \text{m/s}^2\), demonstrating the need for more force to achieve the same acceleration for larger masses.
These insights illustrate the intricate balance between mass and force, offering valuable understanding of object motion dynamics.

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

Using a rope that will snap if the tension in it exceeds \(387 \mathrm{~N}\), you need to lower a bundle of old roofing material weighing \(449 \mathrm{~N}\) from a point \(6.1 \mathrm{~m}\) above the ground. Obviously if you hang the bundle on the rope, it will snap. So, you allow the bundle to accelerate downward. (a) What magnitude of the bundle's acceleration will put the rope on the verge of snapping? (b) At that acceleration, with what speed would the bundle hit the ground?

If the \(1 \mathrm{~kg}\) standard body has an acceleration of \(2.00 \mathrm{~m} / \mathrm{s}^{2}\) at \(20.0^{\circ}\) to the positive direction of an \(x\) axis, what are (a) the \(x\) component and (b) the \(y\) component of the net force acting on the body, and (c) what is the net force in unit-vector notation?

An \(85 \mathrm{~kg}\) man lowers himself to the ground from a height of \(10.0 \mathrm{~m}\) by holding onto a rope that runs over a frictionless pulley to a \(65 \mathrm{~kg}\) sandbag. With what speed does the man hit the ground if he started from rest?

A car traveling at \(53 \mathrm{~km} / \mathrm{h}\) hits a bridge abutment. A passenger in the car moves forward a distance of \(65 \mathrm{~cm}\) (with respect to the road) while being brought to rest by an inflated air bag. What magnitude of force (assumed constant) acts on the passenger's upper torso, which has a mass of \(41 \mathrm{~kg}\) ?

A \(2.00 \mathrm{~kg}\) object is subjected to three forces that give it an acceleration \(\vec{a}=-\left(8.00 \mathrm{~m} / \mathrm{s}^{2}\right) \hat{\mathrm{i}}+\left(6.00 \mathrm{~m} / \mathrm{s}^{2}\right) \hat{\mathrm{j}} .\) If two of the three forces are \(\vec{F}_{1}=(30.0 \mathrm{~N}) \hat{\mathrm{i}}+(16.0 \mathrm{~N}) \hat{\mathrm{j}}\) and \(\vec{F}_{2}=\) \(-(12.0 \mathrm{~N}) \hat{\mathrm{i}}+(8.00 \mathrm{~N}) \hat{\mathrm{j}}\), find the third force.

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.