/*! 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 3 A \(3.00-\mathrm{kg}\) object un... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

A \(3.00-\mathrm{kg}\) object undergoes an acceleration given by \(\mathbf{a}=(2.00 \mathbf{i}+5.00 \hat{\mathbf{j}}) \mathrm{m} / \mathrm{s}^{2} .\) Find the resultant force acting on it and the magnitude of the resultant force.

Short Answer

Expert verified
The resultant force acting on the object is \(\vec{F}=(6.00 \mathbf{i}+15.00 \hat{\mathbf{j}}) \mathrm{N}\) and the magnitude of the force is \(\sqrt{6.00^2+15.00^2} \approx 16.16 \mathrm{N}\).

Step by step solution

01

Identify the given parameters

Note the mass of the object, which is 3.00 kg, and the acceleration as a vector \(\vec{a} = 2.00 \mathbf{i} + 5.00 \hat{\mathbf{j}}\) in units of \(\mathrm{m}/\mathrm{s}^{2}\).
02

Apply Newton's Second Law

Use Newton's Second Law of Motion, \(\vec{F} = m \vec{a}\), to find the force vector. Multiply the given mass by the acceleration components.
03

Calculate the force vector

Calculate the force vector components: \(F_x = m \cdot a_x = 3.00 \, \mathrm{kg} \cdot 2.00 \, \mathrm{m}/\mathrm{s}^{2}\) for the i direction and \(F_y = m \cdot a_y = 3.00 \, \mathrm{kg} \cdot 5.00 \, \mathrm{m}/\mathrm{s}^{2}\) for the j direction.
04

Write the force vector

Combine the calculated force components to express the force vector as \(\vec{F} = F_x \mathbf{i} + F_y \hat{\mathbf{j}}\).
05

Determine the magnitude of the resultant force

Magnitude of the force vector is given by \(|\vec{F}| = \sqrt{F_x^2 + F_y^2}\). Substitute the calculated force components to find the magnitude.

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.

Newton's Second Law
Understanding Newton's Second Law is crucial for solving problems related to force and motion. According to this principle, the force acting on an object is equal to the mass of that object multiplied by its acceleration. In simple terms, more massive objects require more force to accelerate. This law can be mathematically expressed as
\( \vec{F} = m \vec{a} \),
where \( \vec{F} \) represents the force vector applied on the object, \( m \) denotes the object's mass, and \( \vec{a} \) symbolizes the acceleration vector.
In our exercise, by knowing the mass (3.00 kg) and acceleration (\(2.00 \mathbf{i} + 5.00 \hat{\mathbf{j}}\) meters per second squared), we can use this law to calculate the force vector. Remember, each variable and symbol has a precise meaning, and using the correct units is paramount for the accuracy of the results.
Vector Mathematics
To solve problems involving forces, a solid grasp of vector mathematics is essential. Vectors are mathematical objects with both magnitude and direction, which is perfect for representing quantities like force and acceleration. In two dimensions, a vector \( \vec{v} \) can be broken down into its horizontal component \( v_x \) along the \( \mathbf{i} \), or x-axis, and its vertical component \( v_y \) along the \( \hat{\mathbf{j}} \), or y-axis. These components are treated as separate values that can be manipulated algebraically.
In the given solution, force components \( F_x \) and \( F_y \) were calculated independently before being recombined to form the resultant force vector \( \vec{F} \). Vector addition is like putting two displacement steps together: one in the x direction and one in the y direction, to navigate to a new position. When visualized, the vector components span from the origin to their respective points along the axes, and the resultant vector stretches diagonally from the origin to a point that encapsulates both displacements.
Force Magnitude
The magnitude of a force — often simply referred to as 'force' — is a measurement of its strength without considering its direction. It's a scalar quantity, which means it has only magnitude and no direction. Calculating the force's magnitude involves using the Pythagorean Theorem to combine the two perpendicular force vector components into a single resultant force value. This is why, as done in Step 5 of the solution, we use the equation
\(|\vec{F}| = \sqrt{F_x^2 + F_y^2}\).
This equation tells us how to find the length of the diagonal in a right-angled triangle when we know the lengths of the two other sides. The 'sides' in this context are the horizontal and vertical components of the force vector, which, when squared and summed, give us the square of the magnitude of the resultant force. Taking the square root of this sum provides the actual strength of the force without any associated direction, making it a foundational calculation in physics problems involving vectors.

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

Draw a free-body diagram of a block which slides down a frictionless plane having an inclination of \(\theta=15.0^{\circ}\) (Fig. P5.22). The block starts from rest at the top and the length of the incline is 2.00 \(\mathrm{m}\) . Find (a) the acceleration of the block and \((b)\) its speed when it reaches the bottom of the incline.

An electron of mass \(9.11 \times 10^{-31} \mathrm{kg}\) has an initial speed of \(3.00 \times 10^{5} \mathrm{m} / \mathrm{s}\) . It travels in a straight line, and its speed increases to \(7.00 \times 10^{5} \mathrm{m} / \mathrm{s}\) in a distance of \(5.00 \mathrm{cm} .\) Assuming its acceleration is constant, (a) determine the force exerted on the electron and (b) compare this force with the weight of the electron, which we neglected.

Three forces acting on an object are given by \(\mathbf{F}_{1}=(-2.00 \hat{\mathbf{i}}+2.00 \hat{\mathbf{j}}) \mathrm{N}, \mathbf{F}_{2}=(5.00 \hat{\mathbf{i}}-3.00 \hat{\mathbf{j}}) \mathrm{N},\) and \(\mathbf{F}_{3}=(-45.0 \hat{\mathbf{i}}) \mathrm{N} .\) The object experiences an acceleration of magnitude \(3.75 \mathrm{m} / \mathrm{s}^{2} .\) (a) What is the direction of the acceleration? (b) What is the mass of the object? (c) If the object is initially at rest, what is its speed after 10.0 \(\mathrm{s}\) ? (d) What are the velocity components of the object after 10.0 \(\mathrm{s}\) ?

Three blocks are in contact with each other on a frictionless, horizontal surface, as in Figure \(\mathrm{P} 5.54 .\) A horizontal force \(\mathbf{F}\) is applied to \(m_{1} .\) Take \(m_{1}=2.00 \mathrm{kg}, m_{2}=\) \(3.00 \mathrm{kg}, m_{3}=4.00 \mathrm{kg},\) and \(F=18.0 \mathrm{N} .\) Draw a separate free-body diagram for each block and find (a) the acceleration of the blocks, (b) the resultant force on each block, and (c) the magnitudes of the contact forces between the blocks. (d) You are working on a construction project. A coworker is nailing up plasterboard on one side of a light partition, and you are on the opposite side, providing "backing" by leaning against the wall with your back pushing on it. Every blow makes your back sting. The supervisor helps you to put a heavy block of wood between the wall and your back. Using the situation analyzed in parts (a), \((b),\) and \((c)\) as a model, explain how this works to make your job more comfortable.

If a man weighs 900 \(\mathrm{N}\) on the Earth, what would he weigh on Jupiter, where the acceleration due to gravity is 25.9 \(\mathrm{m} / \mathrm{s}^{2}\) ?

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.