/*! 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 9 A \(20.0 \mathrm{~kg}\) object t... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 3: Problem 9

A \(20.0 \mathrm{~kg}\) object that can move freely is subjected to a resultant force of \(45.0 \mathrm{~N}\) in the \(-x\) -direction. Find the acceleration of the object. We make use of the second law in component form, \(\sum F_{x}=m a_{x}\), with \(\sum F_{x}=-45.0 \mathrm{~N}\) and \(m=20.0 \mathrm{~kg}\). Then $$ a_{x}=\frac{\sum F_{x}}{m}=\frac{-45.0 \mathrm{~N}}{20.0 \mathrm{~kg}}=-2.25 \mathrm{~N} / \mathrm{kg}=-2.25 \mathrm{~m} / \mathrm{s}^{2} $$ where we have used the fact that \(1 \mathrm{~N}=1 \mathrm{~kg} \cdot \mathrm{m} / \mathrm{s}^{2}\). Because the resultant force on the object is in the \(-x\) -direction, its acceleration is also in that direction.

Short Answer

Expert verified
The acceleration of the object is \(-2.25 \mathrm{~m/s^2}\) in the \(-x\)-direction.

Step by step solution

01

Understand the Problem

We need to determine the acceleration of a moving object given the force applied to it and its mass. The force is given as \(-45.0 \mathrm{~N}\) and is directed in the \(-x\)-direction. The mass of the object is \(20.0 \mathrm{~kg}\).
02

Recall Newton's Second Law

Newton's Second Law states that the force acting on an object is equal to the mass of the object multiplied by its acceleration, which can be written as \( F = ma \).
03

Identify the Known Values

From the problem, we are given \( \sum F_{x} = -45.0 \mathrm{~N}\) and \( m = 20.0 \mathrm{~kg}\).
04

Setup the Equation for Acceleration

Using the formula \( a_{x} = \frac{\sum F_{x}}{m} \), we substitute in the known values: \( a_{x} = \frac{-45.0 \mathrm{~N}}{20.0 \mathrm{~kg}} \).
05

Calculate the Acceleration

Perform the calculation: \( a_{x} = \frac{-45.0}{20.0} = -2.25 \mathrm{~m/s^2} \). The negative sign indicates the direction of acceleration is in the \(-x\)-direction.

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.

Force and Acceleration
Newton's Second Law is a fundamental concept that links force, mass, and acceleration. It tells us that the acceleration of an object is directly proportional to the net force acting on it and inversely proportional to its mass. This is expressed by the equation \( F = ma \).
In the context of the exercise, the object experiences a force of \(-45.0 \, \mathrm{N}\), which is applied in the \(-x\)-direction. Since we know the force and the mass (\(20.0 \, \mathrm{kg}\)), we can find the acceleration using the formula:
  • \( a = \frac{F}{m} \)
  • Substitute the values: \( a = \frac{-45.0 \, \mathrm{N}}{20.0 \, \mathrm{kg}} \)
  • The calculated acceleration is \(-2.25 \, \mathrm{m/s^2}\)
The negative sign in our answer indicates that the acceleration is in the same direction as the force, in this case, the \(-x\)-direction.
This relationship highlights how forces can alter motion by changing the speed or direction of an object.
Mass and Force Relationship
The relationship between mass and force is crucial in understanding how objects move. Mass is a measure of the amount of matter in an object. It's a scalar quantity and does not change regardless of the object's location.
In contrast, force is a vector quantity, meaning it has both magnitude and direction. This distinction is important because it means that when a force acts on a given mass, it doesn't just cause acceleration; it also affects the direction in which the object moves.
Newton's Second Law shows that for a constant force, a larger mass results in a smaller acceleration. This is because the force must "push" harder to move more mass. The equation that describes this is \( F = ma \). Therefore, we can rearrange this to find the acceleration \( a = \frac{F}{m} \).
By increasing the mass while keeping the force constant, the acceleration decreases. This intuitive understanding is essential in physics, especially when analyzing motion.
Vector Components in Physics
In physics, vectors are used to represent quantities that have both magnitude and direction. Understanding vectors and their components is essential for problems involving motion and forces like the one in the exercise.
When a vector such as force or velocity acts at an angle, it can be broken down into its components along the coordinate system axes. In this exercise, the force is applied solely in the \(-x\)-direction, so we only consider the \(x\)-component.
  • The total force vector \(\vec{F}\) can be described by its components: \(\vec{F} = F_x \hat{i} + F_y \hat{j} + F_z \hat{k}\)
  • Here, only \(F_x\) is non-zero: \(-45.0 \, \mathrm{N}\)
Decomposing forces into vector components allows for a clearer understanding of their effects on motion. This technique is not restricted to forces; it applies similarly to velocity and acceleration. Recognizing and using vector components is a fundamental skill in physics, vital for solving problems involving direction and magnitude.

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 \(12-\mathrm{kg}\) box is released from the top of an incline that is \(5.0 \mathrm{~m}\) long and makes an angle of \(40^{\circ}\) to the horizontal. A 60-N friction force impedes the motion of the box. ( \(a\) ) What will be the acceleration of the box, and \((b)\) how long will it take to reach the bottom of the incline?

A 200-N wagon is to be pulled up a \(30^{\circ}\) incline at constant speed. How large a force parallel to the incline is needed if friction effects are negligible? The situation is shown in Fig. \(3-16(a)\). Because the wagon moves at a constant speed along a straight line, its velocity vector is constant. Therefore, the wagon is in translational equilibrium, and the first condition for equilibrium applies to it. We isolate the wagon as the object. Three non-negligible forces act on it: (1) the pull of gravity \(F_{W}\) (its weight), directed straight down; (2) the applied force \(F\) exerted on the wagon parallel to the incline to pull it up the incline; (3) the push \(F_{N}\) of the incline that supports the wagon. These three forces are shown in the free-body diagram in Fig. \(3-16\). For situations involving inclines, it is convenient to take the \(x\) -axis parallel to the incline and the \(y\) -axis perpendicular to it. After taking components along these axes, we can write the first condition for equilibrium: $$ \begin{array}{lll} +\sum F_{x}=0 & \text { becomes } & F-0.50 F_{W}=0 \\ +\sum F_{y}=0 & \text { becomes } & F_{N}-0.866 F_{W}=0 \end{array} $$ Solving the first equation and recalling that \(F_{W}=200 \mathrm{~N}\), we find that \(F=0.50 F_{W}\). The required pulling force to two significant figures is \(0.10 \mathrm{kN}\).

A car whose weight is \(F_{W}\) is on a ramp which makes an angle \(\theta\) to the horizontal. How large a perpendicular force must the ramp withstand if it is not to break under the car's weight? As rendered in Fig. \(3-6\), the car's weight is a force \(\overrightarrow{\mathbf{F}}_{W}\) that pulls straight down on the car. We take components of \(\overrightarrow{\mathbf{F}}\) along the incline and perpendicular to it. The ramp must balance the force component \(F_{W} \cos \theta\) if the car is not to crash through the ramp. In other words, the force exerted on the car by the ramp, upwardly perpendicular to the ramp, is \(F_{N}\) and \(F_{N}=F_{W} \cos \theta\).

Two blocks, of masses \(m_{1}\) and \(m_{2}\), moving in the \(x\) -direction are pushed by a force \(F\) as shown in Fig. 3-19. The coefficient of friction between each block and the table is \(0.40 .(a)\) What must be the value of \(F\) if the blocks are to have an acceleration of \(200 \mathrm{~cm} / \mathrm{s}^{2} ?\) How large a force does \(m_{1}\) then exert on \(m_{2}\) ? Use \(m_{1}=300 \mathrm{~g}\) and \(m_{2}=500 \mathrm{~g}\). Remember to work in SI units. The friction forces on the blocks are \(F_{f 1}=0.40 m_{1} g\) and \(F_{f 2}=0.40 m_{2} g\). Take the two blocks in combination as the object for discussion; the horizontal forces on the object from outside (i.e., the external forces on it) are \(F\), \(F_{f 1}\), and \(F_{f 2}\). Although the two blocks do push on each other, those pushes are internal forces; they are not part of the unbalanced external force on the two-mass object. For that object, $$ \pm \sum F_{x}=m a_{x} \text { becomes } F-F_{f 1}-F_{\mathrm{f} 2}=\left(m_{1}+m_{2}\right) a_{x} $$ (a) Solving for \(F\) and substituting known values $$ F=0.40 g\left(m_{1}+m_{2}\right)+\left(m_{1}+m_{2}\right) a_{x}=3.14 \mathrm{~N}+1.60 \mathrm{~N}=4.7 \mathrm{~N} $$ (b) Now consider block \(m_{2}\) alone. The forces acting on it in the \(x\) -direction are the push of block \(m_{1}\) on it (which we represent by \(F_{b}\) ) and the retarding friction force \(F_{12}=0.40 m_{2} g\). Then, for it, $$ \pm \sum F_{x}=m a_{x} \quad \text { becomes } \quad F_{b}-F_{f 2}=m_{2} a_{x} $$ We know that \(a_{x}=2.0 \mathrm{~m} / \mathrm{s}^{2}\) and so $$ F_{b}=F_{\mathrm{f} 2}+m_{2} a_{x}=1.96 \mathrm{~N}+1.00 \mathrm{~N}=2.96 \mathrm{~N}=3.0 \mathrm{~N} $$

The Earth's radius is about \(6370 \mathrm{~km}\). An object that has a mass of \(20 \mathrm{~kg}\) is taken to a height of \(160 \mathrm{~km}\) above the Earth's surface. ( \(a\) ) What is the object's mass at this height? ( \(b\) ) How much does the object weigh (i.e. how large a gravitational force does it experience) at this height?
See all solutions

Recommended explanations on Physics Textbooks

Space Physics

Read Explanation

Magnetism

Read Explanation

Translational Dynamics

Read Explanation

Electrostatics

Read Explanation

Electric Charge Field and Potential

Read Explanation

Torque and Rotational Motion

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.