/*! 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 4 A 75.0-kg man is riding an escal... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 6: Problem 4

A 75.0-kg man is riding an escalator in a shopping mall. The escalator moves the man at a constant velocity from ground level to the floor above, a vertical height of 4.60 m. What is the work done on the man by (a) the gravitational force and (b) the escalator?

Short Answer

Expert verified
(a) The work done by gravity is -3381 J, (b) the work done by the escalator is 3381 J.

Step by step solution

01

Identify Known Variables

First, let's identify the known variables. The mass of the man, \( m = 75.0 \) kg, and the height he is moved to, \( h = 4.60 \) m. The gravitational force can be calculated using the gravitational acceleration, \( g = 9.80 \text{ m/s}^2 \).
02

Calculate Gravitational Force

The gravitational force can be calculated using the formula \( F = m \times g \). Substitute \( m = 75.0 \) kg and \( g = 9.80 \text{ m/s}^2 \), giving \( F = 75.0 \times 9.80 = 735 \) N.
03

Calculate Work Done by Gravitational Force

Work done by the gravitational force is given by \( W_g = F \times h \times \cos(\theta) \). As the displacement is upwards and the gravitational force is downwards (\( \theta = 180^\circ \)), \( \cos(180^\circ) = -1 \), which leads to \( W_g = 735 \times 4.60 \times (-1) = -3381 \text{ J} \).
04

Calculate Work Done by Escalator

Work done by the escalator is equal in magnitude but opposite in direction to the work done by the gravitational force since it lifts the man upwards. Therefore, \( W_e = 3381 \text{ J} \).

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.

Gravitational Force
Gravitational force is a fundamental concept in physics, describing the force that attracts two bodies towards each other. In this particular problem, it acts between the Earth and the man on the escalator.
  • It is calculated using the formula \( F = m \times g \), where \( m \) is the mass of the object and \( g \) is the acceleration due to gravity.
  • Here, the gravitational force acting on the man is determined by multiplying his mass (75 kg) by the gravitational acceleration (9.80 m/s虏).
Therefore, the force exerted by gravity on the man is 735 N. This force acts downwards toward the Earth. Every object with mass experiences this force, making it one of the essential principles in physics, affecting everything from simple machines to planetary orbits.
Work Done Calculation
Work done in physics refers to the amount of energy transferred by a force acting over a distance. In this scenario, we are interested in the work done by gravitational force and the escalator as the man moves to a higher floor.
  • The formula for calculating work is \( W = F \times d \times \cos(\theta) \), where \( F \) is the force, \( d \) is the distance, and \( \theta \) is the angle between the force and the direction of motion.
  • For the gravitational force, the angle is 180掳 because gravity acts downward while the man moves upward, which makes \( \cos(180掳) = -1 \).
Thus, the work done by gravitational force is \(-3381 \text{ J}\). The negative sign indicates that the direction of the gravitational force opposes the movement of the man, showing that gravity is doing negative work as he is lifted upward by the escalator.
Escalator Physics
The physics of an escalator involves understanding how work is done to move objects working against forces like gravity. Escalators are ingeniously designed to lift individuals from one level to another with no apparent effort on their part.
  • When a man rides an escalator, it does work against the gravitational force, moving him upward at a constant height (4.60 m in this example).
  • The key role of the escalator is to provide the energy needed to overcome gravitational forces, facilitating the man's ascent without his input.
  • The direction of the escalator's work is opposite to the gravitational force, which explains why the work done by the escalator is positive \((W_e = 3381 \text{ J})\).
In essence, an escalator transfers mechanical energy to the passengers, lifting them efficiently across floors with minimal energy expenditure for the individual riding it.
Constant Velocity
Constant velocity means that an object travels in a straight line at the same speed over time, experiencing no acceleration. In the context of the escalator and the man, this concept plays a vital role in the physics calculations.
  • When the man moves at a constant velocity, it implies that the forces affecting him, like gravitational force and the force from the escalator, are balanced.
  • Because there are no net forces acting (aside from gravity being countered by the escalator), the man maintains constant speed upward.
  • This is why the calculated work done by the escalator is effectively counterbalancing the gravitational work, showing that no additional energy is being used to change the motion of the man beyond lifting him.
Constant velocity eliminates the complications of acceleration in this problem, simplifying the calculations and focusing solely on the forces in action and the work done against those forces.

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 0.60-kg basketball is dropped out of a window that is 6.1 m above the ground. The ball is caught by a person whose hands are 1.5 m above the ground. (a) How much work is done on the ball by its weight? What is the gravitational potential energy of the basketball, relative to the ground, when it is (b) released and (c) caught? (d) How is the change \(\left(PE_{\mathrm{f}}-PE_{0}\right)\) in the ball's gravitational potential energy related to the work done by its weight?

A slingshot fires a pebble from the top of a building at a speed of 14.0 m/s. The building is 31.0 m tall. Ignoring air resistance, find the speed with which the pebble strikes the ground when the pebble is fired (a) horizontally, (b) vertically straight up, and (c) vertically straight down.

As a sailboat sails 52 m due north, a breeze exerts a constant force \(\overrightarrow{\mathbf{F}}_{1}\) on the boat's sails. This force is directed at an angle west of due north. A force \(\overrightarrow{\mathbf{F}}_{2}\) of the same magnitude directed due north would do the same amount of work on the sailboat over a distance of just 47 \(\mathrm{m}\) . What is the angle between the direction of the force \(\overrightarrow{\mathbf{F}}_{1}\) and due north?

A fighter jet is launched from an aircraft carrier with the aid of its own engines and a steam-powered catapult. The thrust of its engines is \(2.3 \times 10^{5} \mathrm{N}\) . In being launched from rest it moves through a distance of 87 \(m\) and has a kinetic energy of \(4.5 \times 10^{7} J\) at lift-off. What is the work done on the jet by the catapult?

A car accelerates uniformly from rest to 20.0 m/s in 5.6 s along a level stretch of road. Ignoring friction, determine the average power required to accelerate the car if \((a)\) the weight of the car is \(9.0 \times 10^{3} N\) and \((b)\) the weight of the car is \(1.4 \times 10^{4} N .\)
See all solutions

Recommended explanations on Physics Textbooks

Rotational Dynamics

Read Explanation

Electricity and Magnetism

Read Explanation

Wave Optics

Read Explanation

Electromagnetism

Read Explanation

Force

Read Explanation

Magnetism and Electromagnetic Induction

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.