/*! 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 308 A boy of mass \(m\) is standing ... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

A boy of mass \(m\) is standing initially at rest relative to the moving walkway, which has a constant horizontal speed \(u .\) He decides to accelerate his progress and starts to walk from point \(A\) with a steadily increasing speed and reaches point \(B\) with a speed \(\dot{x}=v\) relative to the walkway. During his acceleration he generates an average horizontal force \(F\) between his shoes and the walkway. Write the work-energy equations for his absolute and relative motions and explain the meaning of the term muv.

Short Answer

Expert verified
The term \( muv \) links absolute and relative motion, representing the product of initial speed \( u \) and relative speed \( v \) in the expanded absolute kinetic energy.

Step by step solution

01

Understand the Problem Setup

The boy is initially at rest on a moving walkway that moves with a horizontal speed \( u \). He starts walking, gaining speed until reaching speed \( v \) relative to the walkway. The force he applies to accelerate is \( F \). We need to find work-energy equations for his absolute and relative motion.
02

Establish Definitions

Define the absolute speed of the boy as \( x' = u + v \), which is the sum of the walkway speed \( u \) and his relative speed \( v \). The initial absolute speed is \( x'= u \) because he is initially at rest relative to the walkway, and his final absolute speed is \( x' = u + v \).
03

Work-Energy Equation for Absolute Motion

Using the work-energy principle for absolute motion, the work done by force \( F \) is equal to the change in kinetic energy: \[ W_{absolute} = \Delta KE = \frac{1}{2} m (u+v)^2 - \frac{1}{2} m u^2\] The work done accounts for the increase in speed from \( u \) to \( u+v \).
04

Work-Energy Equation for Relative Motion

For relative motion, consider the kinetic energy relative to the walkway. The relative work-energy equation is:\[ W_{relative} = \Delta KE = \frac{1}{2} m v^2 - 0\] The initial kinetic energy is zero because he starts from rest relative to the walkway.
05

Explain the Term \( muv \)

The term \( muv \) appears when expanding the absolute kinetic energy term:- The initial absolute kinetic energy is \( \frac{1}{2} mu^2 \).- The final absolute kinetic energy is \( \frac{1}{2} m(u+v)^2 \), which expands to \( \frac{1}{2} mu^2 + muv + \frac{1}{2} mv^2 \).- The term \( muv \) represents the cross term from the product of the initial speed \( u \) and the relative speed \( v \), linking absolute and relative motion.

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.

Kinetic Energy
Kinetic energy is the energy a body possesses due to its motion. It depends on the mass and velocity of the object. The formula for kinetic energy (KE) is given by \( KE = \frac{1}{2} mv^2 \), where \( m \) is the mass and \( v \) is the velocity.
In this exercise, the boy starts at rest relative to the moving walkway but increases in speed as he walks. His initial kinetic energy with respect to the walkway is zero, as he is initially standing still on it.
As he begins to walk, his speed increases to \( v \), giving him kinetic energy relative to the walkway. The work done by the force \( F \) results in this increase in his kinetic energy. The work-energy principle helps us understand how energy input, through force, translates into kinetic energy and motion.
Relative Motion
Relative motion refers to the movement of an object with respect to another moving reference frame. In this situation, the boy's relative motion is his movement concerning the moving walkway. The speed \( v \) mentioned in the problem is his speed relative to this walkway and not to a fixed point on the ground.
Calculating relative motion is crucial when two objects are moving. In this case, the boy experiences a speed \( v \) relative to the walkway. His kinetic energy relative to the walkway is given by \(\frac{1}{2} m v^2 \). This type of calculation is often used in real-world scenarios where both objects are motionful, like in transportation where passengers move inside a moving vehicle.
Absolute Motion
Absolute motion considers an object's movement relative to a fixed, non-moving point. Here, the boy's absolute motion is his movement relative to the ground, not just the moving walkway. If the walkway moves with speed \( u \) and he moves relative to it with speed \( v \), then his absolute speed or velocity, \( x' \), is \( u + v \).
Understanding absolute motion is essential because it lets us compute the total actual movement. Whereas relative speed can change depending on the reference point, absolute speed remains constant when referred to a fixed base. It provides context for situations involving moving reference frames. In terms of energy, absolute kinetic energy involves the ground and is expressed as \( \frac{1}{2} m(u+v)^2 \).
Force and Acceleration
Force and acceleration are closely related concepts in physics, described by Newton's second law. The boy applies a force \( F \) that results in his acceleration along the walkway. The relationship is given by:
  • Force (\( F \)) = mass (\( m \)) \( \times \) acceleration (\( a \))
As the boy increases his velocity from rest to \( v \) relative to the walkway, he achieves acceleration. This force also contributes directly to the work done, which increases his kinetic energy. The average horizontal force exerted interacts with friction between the shoes and the walkway.
And why this matters? Because understanding force and acceleration aids in calculating changes in speed and direction, essential for analyzing motion. The work-energy equation combines force into how quickly a body's kinetic energy changes, offering a comprehensive view of 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

The slider of mass \(m_{1}=0.4 \mathrm{kg}\) moves along the smooth support surface with velocity \(v_{1}=5 \mathrm{m} / \mathrm{s}\) when in the position shown. After negotiating the curved portion, it moves onto the inclined face of an initially stationary block of mass \(m_{2}=2 \mathrm{kg}\) The coefficient of kinetic friction between the slider and the block is \(\mu_{k}=0.30 .\) Determine the velocity \(v^{\prime}\) of the system after the slider has come to rest relative to the block. Neglect friction at the small wheels, and neglect any effects associated with the transition.

The 200 -kg glider \(B\) is being towed by airplane \(A\) which is flying horizontally with a constant speed of \(220 \mathrm{km} / \mathrm{h}\). The tow cable has a length \(r=60 \mathrm{m}\) and may be assumed to form a straight line. The glider is gaining altitude and when \(\theta\) reaches \(15^{\circ}\) the angle is increasing at the constant rate \(\dot{\theta}=\) 5 deg/s. At the same time the tension in the tow cable is \(1520 \mathrm{N}\) for this position. Calculate the aerodynamic lift \(L\) and \(\operatorname{drag} D\) acting on the glider.

Determine the coefficient of restitution \(e\) for a steel ball dropped from rest at a height \(h\) above a heavy horizontal steel plate if the height of the second rebound is \(h_{2}\).

A projectile is launched from the north pole with an initial vertical velocity \(v_{0} .\) What value of \(v_{0}\) will result in a maximum altitude of \(R / 3 ?\) Neglect aerodynamic drag and use \(g=9.825 \mathrm{m} / \mathrm{s}^{2}\) as the surface-level acceleration due to gravity.

The car is moving with a speed \(v_{0}=65 \mathrm{mi} / \mathrm{hr}\) up the 6 -percent grade, and the driver applies the brakes at point \(A,\) causing all wheels to skid. The coefficient of kinetic friction for the rain-slicked road is \(\mu_{k}=0.60 .\) Determine the stopping distance \(s_{A B} .\) Repeat your calculations for the case when the car is moving downhill from \(B\) to \(A\).

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.