/*! 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 22 When a man increases his speed b... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 6: Problem 22

When a man increases his speed by \(2 \cdot \mathrm{m} / \mathrm{s}\), he finds that his kinetic energy is doubled, the original speed of the man is: (a) \(2(\sqrt{2}-1) \mathrm{m} / \mathrm{s}\) (b) \(2(\sqrt{2}+1) \mathrm{m} / \mathrm{s}\) (c) \(4.5 \mathrm{~m} / \mathrm{s}\) (d) none of these

Short Answer

Expert verified
The original speed of the man is: (a) \(2(\sqrt{2}-1) \mathrm{m} / \mathrm{s}\).

Step by step solution

01

Understand the Problem

The problem involves the kinetic energy of the man doubling when he increases his speed by 2 m/s. We need to find the original speed of the man.
02

Recall the Kinetic Energy Formula

The kinetic energy (KE) of an object with mass \( m \) moving at speed \( v \) is given by \( KE = \frac{1}{2}mv^2 \).
03

Set Up the Equations

Let's denote the original speed of the man as \( v \). After increasing his speed by 2 m/s, his new speed is \( v + 2 \). The original kinetic energy is \( KE_1 = \frac{1}{2}mv^2 \). The new kinetic energy after increasing speed is \( KE_2 = \frac{1}{2}m(v+2)^2 \). According to the problem, \( KE_2 = 2 \cdot KE_1 \).
04

Substitute and Simplify the Equations

Substitute the kinetic energy expressions into the equation: \[ \frac{1}{2}m(v+2)^2 = 2 \cdot \frac{1}{2}mv^2 \]After simplifying, this becomes:\[ (v+2)^2 = 2v^2 \].
05

Solve the Equation for v

Expand and rearrange the equation:\[ v^2 + 4v + 4 = 2v^2 \]Subtract \( v^2 \) from both sides:\[ 4v + 4 = v^2 \]This simplifies to \[ v^2 - 4v - 4 = 0 \].
06

Use the Quadratic Formula

Solve the quadratic equation \( v^2 - 4v - 4 = 0 \) using the quadratic formula \( v = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \), where \( a = 1 \), \( b = -4 \), and \( c = -4 \).Calculate the discriminant: \( b^2 - 4ac = (-4)^2 - 4 \cdot 1 \cdot (-4) = 16 + 16 = 32 \).So, \( v = \frac{4 \pm \sqrt{32}}{2} = \frac{4 \pm 4\sqrt{2}}{2} = 2 \pm 2\sqrt{2} \).
07

Determine the Original Speed

Since speed cannot be negative, we select \( v = 2(\sqrt{2} - 1) \). This matches with option (a).

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.

Quadratic Equation
In this problem, a key step is solving a quadratic equation. Quadratic equations are mathematical expressions that take the form \( ax^2 + bx + c = 0 \), where \( a \), \( b \), and \( c \) are constants. The solution to a quadratic equation often involves finding two values of \( x \) that satisfy the equation.
To solve a quadratic equation like the one in our exercise, we use the quadratic formula:
\[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]
This formula helps us calculate the roots of the equation by substituting the values of \( a \), \( b \), and \( c \).
  • The expression under the square root sign, \( b^2 - 4ac \), is called the discriminant. It determines the nature of the roots.
  • If the discriminant is positive, we get two distinct real roots. If it's zero, the quadratic has one real root (a repeated root). If negative, the roots are complex.
In this lesson, after simplifying our kinetic energy equation, it turned into the quadratic equation: \( v^2 - 4v - 4 = 0 \). The solutions were found using the quadratic formula, resulting in two possible speeds for \( v \): \( 2(\sqrt{2} - 1) \) and \( 2(\sqrt{2} + 1) \). We chose the positive root that made sense for speed.
Speed
Speed, in physics, is defined as how fast an object is moving. It is a scalar quantity, which means it has magnitude but no direction. Speed is typically measured in meters per second (m/s) or kilometers per hour (km/h).
Understanding speed is essential when dealing with problems involving motion and energy, like the current exercise. In this problem, the man's initial speed \( v \) is key to determining how his kinetic energy changes.
  • If the speed increases, as in this problem where the man increases his speed by 2 m/s, it affects his kinetic energy significantly.
  • Changes in speed lead to changes in kinetic energy; thus, understanding the relationship between them helps in solving related physics problems.
By considering the effects of increasing speed, we set the stage for analyzing how kinetic energy changes, which later allows us to solve for the original speed of the man.
Kinetic Energy Formula
Kinetic energy is the energy an object possesses due to its motion. The formula to calculate kinetic energy is:
\[ KE = \frac{1}{2}mv^2 \]
where \( KE \) is the kinetic energy, \( m \) is the mass of the object, and \( v \) is the speed of the object.
This formula shows that kinetic energy is proportional to the square of the speed. This means that even a small increase in speed results in a significant increase in kinetic energy.
In the problem, the man's kinetic energy doubles when his speed is increased by 2 m/s. Mathematically, this situation is represented by equating twice the original kinetic energy to the new kinetic energy at the increased speed. This setup ultimately helps us form a quadratic equation to find the original speed.
  • The initial kinetic energy \( KE_1 = \frac{1}{2}mv^2 \) is doubled, yielding \( KE_2 = 2 \times KE_1 \).
  • The increase of speed involves using this relationship to form the equation \( (v+2)^2 = 2v^2 \). This transition is crucial to derive the useful formula and solve for \( v \).
Understanding how to utilize the kinetic energy formula in problems like this is key to mastering motion and energy concepts.

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 potential energy of a particle of mass \(5 \mathrm{~kg}\) moving in the \(x-y\) plane is given by \(U=-7 x+24 y\) joule, \(x\) and \(y\) being in metre. Initially at \(t=0\) the particle is at the origin. \((0,0)\) moving with a velocity of \(6[2.4 \hat{\mathrm{i}}+0.7 \hat{\mathrm{j}}]\) \(\mathrm{m} / \mathrm{s}\). The magnitude of force on the particle is: (a) 25 units (b) 24 units (c) 7 units (d) none of these

A chain of mass \(M\) is placed on a smooth table with \(1 / 3\) of its length \(L\) hanging over the edge. The work done in pulling the chain back to the surface of the table is: (a) \(\frac{\mathrm{Mg} L}{3}\) (b) \(\frac{M g L}{6}\) (c) \(\frac{\mathrm{MgL}}{9}\) (d) \(\frac{M g L}{18}\)

A particle of mass \(m\) is moving in a circular path of constant radius \(r\) such that its centripetal acceleration \(a_{c}\) is varying with time \(t\) as \(a_{c}=k^{2} r t^{2}\). The power is:(a) \(2 \pi m k^{2} r^{2} t\) (b) \(m k^{2} r^{2} t\) (c) \(\frac{m k^{4} r^{2} t^{5}}{3}\) (d) zero

An object of mass \(m\) is tiec to a string of length \(L\) anc a variable horizontal forc is applied on it which starts at zero an gradually increases unti the string makes an angl \theta with the vertical. Worl done by the force \(F\) is:(a) \(m g L(1-\sin \theta)\) (b) \(m g L\) (c) \(m g L(1-\cos \theta)\) (d) \(m g L(1+\cos \theta)\)

The potential energy as a function of 'the force between two atoms in a diatomic molecule is given by \(U(x)=\frac{a}{x^{12}}-\frac{b}{6}\) where \(a\) and \(b\) are positive constants and \(x\) is the distance between the atoms. The position of stable equilibrium for the system of the two atoms is given by: (a) \(x=\frac{d}{b}\) (b) \(x=\sqrt{\frac{a}{b}}\) (c) \(x=\frac{\sqrt{3 a}}{b}\) (d) \(x=\sqrt[6]{\left(\frac{2 a}{b}\right)}\)
See all solutions

Recommended explanations on Physics Textbooks

Magnetism

Read Explanation

Electric Charge Field and Potential

Read Explanation

Nuclear Physics

Read Explanation

Wave Optics

Read Explanation

Torque and Rotational Motion

Read Explanation

Rotational 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.