/*! 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 21 A body of mass \(6 \mathrm{~kg}\... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 5: Problem 21

A body of mass \(6 \mathrm{~kg}\) is acted upon by a force which causes a displacement in it given by \(x=\frac{t^{2}}{4} \mathrm{~m}\) where \(t\) is the time in second. The work done by the force in 2 second is (a) \(12 \mathrm{~J}\) (b) \(9 \mathrm{~J}\) (c) \(6 \mathrm{~J}\) (d) \(3 \mathrm{~J}\)

Short Answer

Expert verified
The work done by the force in 2 seconds is 3 J.

Step by step solution

01

Determine Displacement

The given displacement equation is \( x = \frac{t^2}{4} \). We will calculate the displacement after 2 seconds by substituting \( t = 2 \) into the equation.\[\begin{align*} x &= \frac{2^2}{4} \ &= \frac{4}{4} \ &= 1 \text{ m} \end{align*}\]So, the displacement after 2 seconds is 1 m.
02

Calculate Velocity

To find the velocity, we need to differentiate the displacement equation with respect to time. The displacement equation is \( x = \frac{t^2}{4} \).The derivative of \( x \) with respect to \( t \) is \( v = \frac{dx}{dt} = \frac{d}{dt}\left(\frac{t^2}{4}\right) = \frac{t}{2} \). Now, substitute \( t = 2 \) to find the velocity at 2 seconds:\[\begin{align*} v &= \frac{2}{2} \ &= 1 \, \text{m/s} \end{align*}\]So, the velocity at 2 seconds is 1 m/s.
03

Determine Acceleration

The acceleration can be found by differentiating the velocity equation with respect to time. From the previous step, \( v = \frac{t}{2} \).Differentiate \( v \) with respect to \( t \) to find acceleration \( a \):\[a = \frac{dv}{dt} = \frac{d}{dt}\left(\frac{t}{2}\right) = \frac{1}{2}\]The acceleration is constant at \( \frac{1}{2} \text{ m/s}^2 \).
04

Calculate Force

We apply Newton's second law, \( F = ma \), to find the force on the body. Given the mass \( m = 6 \text{ kg} \) and acceleration \( a = \frac{1}{2} \text{ m/s}^2 \), we can calculate:\[ F = 6 \times \frac{1}{2} = 3 \text{ N} \]Thus, the force applied on the body is 3 N.
05

Calculate Work Done

The work done by a force is calculated using the formula \( W = Fd \), where \( F \) is the force and \( d \) is the displacement. From previous steps, \( F = 3 \text{ N} \) and \( d = 1 \text{ m} \):\[ W = 3 \times 1 = 3 \text{ J} \]Therefore, the work done by the force in 2 seconds is 3 J.

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.

Newton's Second Law
Newton's Second Law is fundamental to understanding motion and forces in physics. It states that the force acting on an object is equal to the mass of that object multiplied by its acceleration, represented as \( F = ma \).This equation unveils the direct relationship between force, mass, and acceleration, which are key elements in dynamics.
  • Mass \( (m) \) is a measure of the amount of matter in an object and does not change regardless of its motion.
  • Acceleration \( (a) \) is the rate of change of velocity over time, dictating how quickly an object speeds up or slows down.
  • Force \( (F) \) is what causes an object to accelerate. It is a vector quantity, meaning it has both magnitude and direction.
In the context of the exercise, the mass is provided as 6 kg, and the constant acceleration derived is \( \frac{1}{2} \text{ m/s}^2 \). By applying Newton's Second Law, we computed the force exerted on the body to be 3 N.
Kinematics Equations
Kinematics is the branch of mechanics that focuses on the motion of objects without considering the forces causing the motion. The basic kinematic equations describe how position, velocity, and acceleration relate to one another over time.
  • Displacement: Kinematics deals with how an object's position changes over time, often represented by the variable \( x \).
  • Velocity: Describes how fast an object's position is changing, which can be determined from displacement using differentiation.
  • Acceleration: Describes how fast an object's velocity is changing, which can also be assessed via differentiation.
In this exercise, the given displacement \( x = \frac{t^2}{4} \) allows us to determine both velocity and acceleration through differentiation. Initially, we assessed the displacement to find it as 1 m after 2 seconds, which served as a key component in computing work done.
Differentiation in Physics
Differentiation is a powerful mathematical tool used in physics to analyze changes.In kinematics, differentiation helps us determine the rate of change of physical quantities like displacement and velocity.
  • From Displacement to Velocity: Differentiation of the displacement function \( x = \frac{t^2}{4} \) with respect to time \( t \) gives us the velocity \( v = \frac{dx}{dt} = \frac{t}{2} \).
  • From Velocity to Acceleration: Further differentiation of velocity \( v = \frac{t}{2} \) results in a constant acceleration of \( a = \frac{dv}{dt} = \frac{1}{2} \text{ m/s}^2 \).
Differentiation allows us to explore how motion parameters change, enabling calculations of instantaneous velocity and constant acceleration. This progression from differentiation is central to solving kinematics problems, such as determining the force and work done as demonstrated in the exercise above.

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

If the potential energy of a gas molecule is $$ U=\frac{M}{r^{6}}-\frac{N}{r^{12}} $$ \(M\) and \(N\) being positive constants, then the potential energy at equilibrium must be (a) zero (b) \(M^{2} / 4 N\) (c) \(N^{2} / 4 M\) (d) \(M N^{2} / 4\)

If \(\vec{F}=\hat{i} F_{x}+\hat{j} F_{y}+\hat{k} F_{t}\) is conservative, then (a) \(\frac{\partial F_{x}}{\partial y}=\frac{\partial F_{y}}{\partial x} ; \frac{\partial F_{y}}{\partial z}=\frac{\partial F_{z}}{\partial y} ; \frac{\partial F_{z}}{\partial x}=\frac{\partial F_{x}}{\partial z}\) (b) \(\frac{\partial F_{x}}{\partial y} \neq \frac{\partial F_{y}}{\partial x} ; \frac{\partial F_{y}}{\partial z} \neq \frac{\partial F_{z}}{\partial y} ; \frac{\partial F_{z}}{\partial x} \neq \frac{\partial F_{x}}{\partial z}\) (c) \(\frac{\partial F_{x}}{\partial x}=\frac{\partial F_{y}}{\partial y}=\frac{\partial F_{z}}{\partial z} \neq 0\) (d) \(\frac{\partial F_{y}}{\partial y}=\frac{\partial F_{x}}{\partial x}+\frac{\partial F_{z}}{\partial z}\)

A body of mass \(m\) accelerates uniformly from rest to \(v_{1}\) in time \(t_{1}\). The instantaneous power delivered to the body as a function of time \(t\) is (a) \(\frac{m v_{1} t}{t_{1}}\) (b) \(\frac{m v_{1}^{2} l}{t_{1}^{2}}\) (c) \(\frac{m v_{1} t^{2}}{t_{1}}\) (d) \(\frac{m v_{1}^{2} t}{t_{1}}\)

An ideal spring with spring constant \(k\) is hung from the ceiling and a block of mass \(m\) is attached to its lower end. The mass is released with the spring initially unstreched. The maximum extension in the spring is (a) \(\frac{4 m g}{k}\) (b) \(\frac{2 m g}{k}\) (c) \(\frac{m g}{k}\) (d) \(\frac{m g}{2 k}\)

A body is moving along a straight line by a machine delivering a constant power. The distance moved by the body in time \(t\) is proportional to (a) \(t^{1 / 2}\) (b) \(t\) (c) \(t^{1 / 2}\) (d) \(t^{2}\)
See all solutions

Recommended explanations on Physics Textbooks

Fluids

Read Explanation

Magnetism and Electromagnetic Induction

Read Explanation

Nuclear Physics

Read Explanation

Electric Charge Field and Potential

Read Explanation

Physical Quantities And Units

Read Explanation

Kinematics 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.