/*! 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 56 An initially stationary \(2.0 \m... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 7: Problem 56

An initially stationary \(2.0 \mathrm{~kg}\) object accelerates horizontally and uniformly to a speed of \(10 \mathrm{~m} / \mathrm{s}\) in \(3.0 \mathrm{~s}\). (a) In that \(3.0 \mathrm{~s}\) interval, how much work is done on the object by the force accelerating it? What is the instantaneous power due to that force (b) at the end of the interval and (c) at the end of the first half of the interval?

Short Answer

Expert verified
The work done is 99.9 J. Instantaneous power is 66.6 W at 3 s, and 33.3 W at 1.5 s.

Step by step solution

01

Calculate the Acceleration

First, we need to determine the acceleration of the object. The formula to calculate acceleration when an object accelerates uniformly is given by:\[ a = \frac{v_f - v_i}{t} \]where:\( v_f \) is the final velocity, \( 10 \mathrm{~m/s} \),\( v_i \) is the initial velocity, which is \( 0 \mathrm{~m/s} \), and\( t \) is the time, \( 3.0 \mathrm{~s} \).Substituting the values, we get:\[ a = \frac{10 \mathrm{~m/s} - 0 \mathrm{~m/s}}{3.0 \mathrm{~s}} = \frac{10}{3} \mathrm{~m/s^2} \approx 3.33 \mathrm{~m/s^2} \]
02

Calculate the Net Force

Using Newton's second law of motion, we can calculate the net force \( F \) acting on the object:\[ F = m \cdot a \]where:\( m = 2.0 \mathrm{~kg} \) is the mass of the object,and \( a \approx 3.33 \mathrm{~m/s^2} \).Substituting the values, we find:\[ F = 2.0 \mathrm{~kg} \times 3.33 \mathrm{~m/s^2} \approx 6.66 \mathrm{~N} \]
03

Calculate the Work Done

The work done \( W \) on the object can be calculated using the formula:\[ W = F \cdot d \]where \( d \) is the distance covered in the interval. Since we don't have \( d \) directly, we use:\[ d = \frac{1}{2} \cdot a \cdot t^2 \]Substituting the known values:\[ d = \frac{1}{2} \times 3.33 \mathrm{~m/s^2} \times (3.0 \mathrm{~s})^2 = 15 \mathrm{~m} \]So the work done is:\[ W = 6.66 \mathrm{~N} \times 15 \mathrm{~m} = 99.9 \mathrm{~J} \]
04

Calculate Instantaneous Power at the End of the Interval

Instantaneous power \( P \) is given by:\[ P = F \cdot v \]At the end of the interval, the speed \( v = 10 \mathrm{~m/s} \):\[ P = 6.66 \mathrm{~N} \times 10 \mathrm{~m/s} = 66.6 \mathrm{~W} \]
05

Calculate Instantaneous Power at the End of the First Half of the Interval

At the end of \(1.5\) s (half the interval), we find the velocity using:\[ v = v_i + a \cdot t = 0 + 3.33 \mathrm{~m/s^2} \times 1.5 \mathrm{~s} \approx 5 \mathrm{~m/s} \]So the power at the midpoint is:\[ P = F \cdot v = 6.66 \mathrm{~N} \times 5 \mathrm{~m/s} = 33.3 \mathrm{~W} \]

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.

Uniform Acceleration
When an object undergoes uniform acceleration, it means that the velocity of the object changes at a constant rate over time. In simpler terms, the speed of the object increases (or decreases) steadily. This is described mathematically by the formula for acceleration:
  • \[ a = \frac{v_f - v_i}{t} \]
Here, \( a \) represents acceleration, \( v_f \) is the final velocity, \( v_i \) is the initial velocity, and \( t \) is the time period over which the change occurs. If an object starts from rest, \( v_i \) is zero, simplifying our calculations.

For example, if a 2.0 kg object starts from rest and reaches a speed of 10 m/s in 3.0 seconds, like in our exercise, the acceleration can be computed as approximately 3.33 m/s虏. This steady change in speed is what characterizes uniform acceleration.
Newton's Second Law
Newton's Second Law of motion is a fundamental concept that describes how forces cause changes in motion. It is expressed by the formula:
  • \[ F = m \cdot a \]
In this equation, \( F \) is the force applied to an object, \( m \) is the mass of the object, and \( a \) is the acceleration produced by the force. This formula tells us that the force exerted on an object is equal to the mass of the object multiplied by the acceleration.

In the given exercise, the object has a mass of 2.0 kg and an acceleration of approximately 3.33 m/s虏. Using Newton's Second Law, we find that the force needed to produce this acceleration is about 6.66 N. This provides a way to predict how an object will move when subjected to certain forces, highlighting the relationship between mass, acceleration, and force.
Instantaneous Power
Instantaneous power is a measure of how quickly work is done at a specific moment in time. It is defined by the formula:
  • \[ P = F \cdot v \]
In this equation, \( P \) is the power, \( F \) is the force, and \( v \) is the velocity of the object. This tells us how much power is being used to move the object at that exact moment.

For this exercise, we're asked to calculate the instantaneous power at the end of a 3.0 second interval and at the halfway point. At the end of the interval, the object's velocity is 10 m/s, thus power is 66.6 W. At the halfway point, the velocity is approximately 5 m/s, leading to a power of 33.3 W. By understanding how power varies at different points in time, we gain insight into how energy is applied and converted in dynamic systems.

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 force of \(5.0 \mathrm{~N}\) acts on a \(15 \mathrm{~kg}\) body initially at rest. Compute the work done by the force in (a) the first, (b) the second, and (c) the third seconds and (d) the instantaneous power due to the force at the end of the third second.

At \(t=0,\) force \(\vec{F}=(-5.00 \hat{\mathrm{i}}+5.00 \hat{\mathrm{j}}+4.00 \hat{\mathrm{k}}) \mathrm{N}\) begins to act on a \(2.00 \mathrm{~kg}\) particle with an initial speed of \(4.00 \mathrm{~m} / \mathrm{s} .\) What is the particle's speed when its displacement from the initial point is \(\vec{d}=(2.00 \hat{\mathrm{i}}+2.00 \hat{\mathrm{j}}+7.00 \hat{\mathrm{k}}) \mathrm{m} ?\)

(a) At a certain instant, a particle-like object is acted on by a force \(\vec{F}=(4.0 \mathrm{~N}) \hat{\mathrm{i}}-(2.0 \mathrm{~N}) \hat{\mathrm{j}}+(9.0 \mathrm{~N}) \hat{\mathrm{k}}\) while the object's velocity is \(\vec{v}=-(2.0 \mathrm{~m} / \mathrm{s}) \hat{\mathrm{i}}+(4.0 \mathrm{~m} / \mathrm{s}) \hat{\mathrm{k}} .\) What is the instantaneous rate at which the force does work on the object? (b) At some other time, the velocity consists of only a \(y\) component. If the force is unchanged and the instantaneous power is \(-12 \mathrm{~W},\) what is the velocity of the object?

A proton (mass \(m=1.67 \times 10^{-27} \mathrm{~kg}\) ) is being accelerated along a straight line at \(3.6 \times 10^{15} \mathrm{~m} / \mathrm{s}^{2}\) in a machine. If the proton has an initial speed of \(2.4 \times 10^{7} \mathrm{~m} / \mathrm{s}\) and travels \(3.5 \mathrm{~cm},\) what then is (a) its speed and (b) the increase in its kinetic energy?

A spring with a spring constant of \(18.0 \mathrm{~N} / \mathrm{cm}\) has a cage attached to its free end. (a) How much work does the spring force do on the cage when the spring is stretched from its relaxed length by 7.60 \(\mathrm{mm} ?\) (b) How much additional work is done by the spring force when the spring is stretched by an additional \(7.60 \mathrm{~mm} ?\)
See all solutions

Recommended explanations on Physics Textbooks

Fundamentals of Physics

Read Explanation

Oscillations

Read Explanation

Linear Momentum

Read Explanation

Force

Read Explanation

Electromagnetism

Read Explanation

Dynamics

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.