/*! 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 44 A force, \(\mathbf{F}=2 \hat{\ma... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 3: Problem 44

A force, \(\mathbf{F}=2 \hat{\mathbf{i}}+2 \hat{\mathbf{j}} \mathbf{N}\) displaces a particle through \(\mathbf{S}=2 \hat{\mathbf{i}}+2 \hat{\mathbf{k}} \mathbf{m}\) in \(16 \mathrm{~s}\). The power developed by \(\mathbf{F}\) is (a) \(0.25 \mathrm{~J} \mathrm{~s}^{-1}\) (b) \(25 \mathrm{~J} \mathrm{~s}^{-1}\) (c) \(225 \mathrm{~J} \mathrm{~s}^{-1}\) (d) \(450 \mathrm{~J} \mathrm{~s}^{-1}\)

Short Answer

Expert verified
The power developed is \(0.25 \text{ J/s}\)

Step by step solution

01

Calculate Work Done

The work done by the force \( \mathbf{F} \) on the particle is given by the dot product of the force and displacement vectors: \( W = \mathbf{F} \cdot \mathbf{S} \). First, substituting the given vectors, we have: \( \mathbf{F} = 2 \hat{\mathbf{i}} + 2 \hat{\mathbf{j}} \) and \( \mathbf{S} = 2 \hat{\mathbf{i}} + 2 \hat{\mathbf{k}} \). Calculate the dot product: \( W = (2 \hat{\mathbf{i}} + 2 \hat{\mathbf{j}}) \cdot (2 \hat{\mathbf{i}} + 2 \hat{\mathbf{k}}) \).
02

Solve for Dot Product

Perform the dot product operation: \( W = 2 \cdot 2 + 2 \cdot 0 + 0 \cdot 2 + 0 \cdot 2 = 4 \). Since the terms involving \( \hat{\mathbf{j}} \cdot \hat{\mathbf{k}} \) and \( \hat{\mathbf{k}} \cdot \hat{\mathbf{i}} \) are zero, they do not contribute to the dot product.
03

Determine Power Developed

Power is defined as the work done per unit time. Use the formula \( P = \frac{W}{t} \), where \( W \) is the work and \( t \) is the time. Given that \( W = 4 \text{ J} \) and \( t = 16 \text{ s} \), calculate \( P = \frac{4}{16} = 0.25 \text{ J/s} \).

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.

Dot Product in Physics
The dot product is a fundamental operation in vector analysis and is essential for calculating work done by a force. It involves taking two vectors and returning a scalar quantity. In our context, the force \( \mathbf{F} = 2 \hat{\mathbf{i}} + 2 \hat{\mathbf{j}} \) and displacement \( \mathbf{S} = 2 \hat{\mathbf{i}} + 2 \hat{\mathbf{k}} \) are the two vectors that need to be considered.
  • Definition: The dot product \( \mathbf{A} \cdot \mathbf{B} \) is calculated as \( A_xB_x + A_yB_y + A_zB_z \) for vectors \( \mathbf{A} \) and \( \mathbf{B} \).
  • Component Analysis: In our example, each vector has components along different unit vectors \( \hat{\mathbf{i}}, \hat{\mathbf{j}}, \hat{\mathbf{k}} \).
  • Relevant Calculation: Calculate the dot product as\[W = (2 \cdot 2) + (2 \cdot 0) + (0 \cdot 2) = 4\]This results in a scalar quantity indicating the work done.
The dot product plays a crucial role in finding how much of one vector lies in the direction of another, which is key in determining the actual contribution of a force to doing work.
Work Done by Force
Work done by a force is a key concept in physics. It refers to the energy transferred when a force acts upon an object to cause displacement. In simpler terms, it measures how much "effort" a force uses to move an object over a distance. For the calculation of work done, the dot product of force and displacement vectors is employed.
  • Formula: \( W = \mathbf{F} \cdot \mathbf{S} \), where \( \mathbf{F} \) is the force vector, and \( \mathbf{S} \) is the displacement vector.
  • Units: The work done is measured in joules (\( \text{J} \)).
  • Context in Problem: For the vectors \( \mathbf{F} = 2 \hat{\mathbf{i}} + 2 \hat{\mathbf{j}} \) and \( \mathbf{S} = 2 \hat{\mathbf{i}} + 2 \hat{\mathbf{k}} \), the work done \( W \) is calculated using their dot product, giving us 4 J.
Understanding the concept of work done is crucial for solving problems related to energy transfer, as it indicates how effectively a force contributes to an object's movement.
Vector Analysis in Physics
Vector analysis is an essential tool in physics, allowing us to work with quantities that have both magnitude and direction, such as displacement and force. Knowing how to manipulate these vectors is crucial to solving many physics problems, like the one presented where we calculated how force creates work.
  • Key Concepts:
    • Vectors: Entities with both magnitude and direction, typically represented in terms of unit vectors \( \hat{\mathbf{i}}, \hat{\mathbf{j}}, \hat{\mathbf{k}} \).
    • Operations: In vector analysis, addition, subtraction, dot product, and cross product are crucial operations.
  • Application in Given Problem: The problem involves force and displacement vectors, each plotted in a three-dimensional coordinate system with components along \( \hat{\mathbf{i}}, \hat{\mathbf{j}}, \hat{\mathbf{k}} \). Each contributes differently based on its direction and magnitude.
  • Result Interpretation: With vector analysis, we can determine how effective a force is in moving an object in a particular direction. This is exemplified by the work calculation through the dot product.
Vector analysis not only aids in calculations but also helps in visualizing the physical significance behind forces and movements in physics, making it a critical part of studying mechanics.

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

The \(X\) and \(Y\) components of vector A have numerical values 6 and 6 respectively and that of \((\mathbf{A}+\mathbf{B})\) have numerical values 10 and \(9 .\) What is the numerical value of \(\mathbf{B} ?\) (a) 2 (b) 3 (c) 4 (d) 5

The \(x\) and \(y\) components of a force are \(2 \mathrm{~N}\) and \(-3 \mathrm{~N}\). The force is (a) \(2 \hat{\mathrm{i}}-3 \hat{\mathrm{j}}\) (b) \(2 \hat{\mathrm{i}}+3 \hat{\mathrm{j}}\) (c) \(-2 \hat{i}-3 \hat{j}\) (d) \(3 \hat{\mathrm{i}}+2 \hat{\mathrm{j}}\)

If vectors \(\mathbf{A}\) and \(\mathbf{B}\) are given by \(\mathbf{A}=5 \hat{\mathbf{i}}+6 \hat{\mathbf{j}}+3 \mathbf{k}\) and \(\mathbf{B}=6 \hat{\mathbf{i}}-2 \hat{\mathbf{j}}-6 \hat{\mathbf{k}}\). Which is/are of the following correct? (a) A and B are mutually perpendicular (b) Product of \(A \times B\) is the same \(B \times A\) (c) The magnitude of \(\mathrm{A}\) and \(\mathrm{B}\) are equal (d) The magnitude of \(A \cdot B\) is zero

Which one of the following statements is true? (a) A scalar quantity is the one that is conserved in a process (b) A scalar quantity is the one that can never take negative values (c) A scalar quantity is the one that does not vary from one point to another in space (d) A scalar quantity has the same value for observers with different orientations of the axes

A body constrained to move in \(y\)-direction, is subjected to a force given by \(\mathbf{F}=(-2 \hat{\mathbf{i}}+15 \hat{\mathbf{j}}+6 \hat{\mathbf{k}})\) done by this force in moving the body through a distance of \(10 \mathrm{~m}\) along \(y\)-axis? (a) \(190 \mathrm{~J}\) (b) \(160 \mathrm{~J}\) (c) \(150 \mathrm{~J}\) (d) \(20 \mathrm{~J}\)
See all solutions

Recommended explanations on Physics Textbooks

Wave Optics

Read Explanation

Oscillations

Read Explanation

Medical Physics

Read Explanation

Fields in Physics

Read Explanation

Linear Momentum

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.