/*! 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 70 A 75.0 -kg person stands on a sc... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

A 75.0 -kg person stands on a scale in an elevator. What does the scale read (in \(N\) and in kg when \((a)\) the elevator is at rest, \((b)\) the elevator is climbing at a constant speed of \(3.0 \mathrm{m} / \mathrm{s},(c)\) the elevator is descending at \(3.0 \mathrm{m} / \mathrm{s},(d)\) the elevator is accelerating upward at \(3.0 \mathrm{m} / \mathrm{s}^{2},(e)\) the elevator is accelerating downward at 3.0 \(\mathrm{m} / \mathrm{s}^{2} ?\)

Short Answer

Expert verified
(a) 735.75 N (75.0 kg), (b) 735.75 N, (c) 735.75 N, (d) 960.75 N, (e) 510.75 N.

Step by step solution

01

Understanding the Force at Rest

When the elevator is at rest, the only force acting on the person is gravity. The force the scale measures is the gravitational force, which is equal to the weight of the person. This force can be calculated using the formula:\[ F = m \times g \]where \( F \) is the force (weight), \( m = 75.0 \) kg is the mass of the person, and \( g = 9.81 \) m/s² is the acceleration due to gravity.
02

Calculating Force at Rest

Substitute the values of mass \( m \) and gravity \( g \) into the equation:\[ F = 75.0 \times 9.81 = 735.75 \] N.
03

Calculating Scale Reading in kg at Rest

Since the scale measures force divided by gravity to give mass, we use \\[ \,\text{mass equivalent reading} = \frac{F}{g} = \frac{735.75}{9.81} \approx 75.0 \, \text{kg} \]
04

Analyzing Force with Constant Speed Upward

When the elevator moves at a constant speed, the acceleration \( a = 0 \), so the force on the scale remains equal to the person's weight. Thus, the reading is still 735.75 N.
05

Recalculating Scale Reading for Constant Speed Downward

Similar to moving upward with constant speed, moving downward at constant speed doesn't change the reading. It remains the same, 735.75 N and 75.0 kg.
06

Analyzing Force when Accelerating Upward

When accelerating upward, the net force equation becomes:\[ F_{ ext{net}} = m(a + g) = 75.0(3.0 + 9.81) \].This increases the felt force, and thus the weight read on the scale.
07

Calculating Scale Reading Accelerating Upward

Substituting the values, \\[ F = 75.0(3.0 + 9.81) = 75.0 \times 12.81 = 960.75 \, \text{N} \].This new force value presents a higher weight than at rest.
08

Analyzing Force when Accelerating Downward

When accelerating downward, the net force equation becomes:\[ F_{ ext{net}} = m(g - a) = 75.0(9.81 - 3.0) \]This decreases the felt force and weight the scale measures.
09

Calculating Scale Reading Accelerating Downward

Substituting the values, \( \,\text{force} = 75.0(9.81 - 3.0) = 75.0 \times 6.81 = 510.75 \, \text{N} \). This is lower than the rest weight.

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.

Understanding Elevator Physics
In the world of physics, elevators provide intriguing opportunities to explore how forces interact. Specifically, when a person stands on a scale in an elevator, their apparent weight (the reading on the scale) changes depending on the motion of the elevator. This can be explained using Newton's Laws of Motion.

When the elevator is at rest or moving at a constant speed, no net external force acts in the vertical direction apart from gravity. Hence, the scale reads the real weight of the person, which is the gravitational force acting downward.

However, when the elevator accelerates upward, an additional force is exerted on the person, increasing their apparent weight. The scale reading is greater than when the elevator is at rest. Conversely, during downward acceleration, the person feels lighter because the effective gravity that pulls them down is reduced, leading to a decreased scale reading. Understanding these variations requires knowledge of basic motion equations and how forces contribute to movement.
Gravitational Force Calculation
Gravitational force is a key player in understanding weight and motion inside an elevator. This force is the product of mass (m) and the acceleration due to the Earth's gravity (g), described by the equation \( F = m \times g \). Here, \( F \) represents gravitational force measured in Newtons (N).

This equation helps us understand why a person's weight remains unchanged whether an elevator is at rest or moving at constant speed. The acceleration due to gravity, \( g \), has a standard value of 9.81 m/s². Thus, for a person weighing 75 kg, the gravitational force is 735.75 N when the elevator is either static or in uniform motion.

Variations occur when the elevator accelerates or decelerates. Here, the net acceleration includes both the elevator's and gravity's influences, altering the reading on the scale. Hence, when calculating gravitational force in different elevator scenarios, it's crucial to adjust for accelerative effects.
Weight Measurement Inside an Elevator
The concept of weight measurement in an elevator can be fascinating and sometimes surprising. A scale in an elevator doesn't just measure weight but the net force on a person due to their mass and the elevator’s acceleration.

At rest, the scale simply reads the gravitational force or actual weight, converting it to kilograms by dividing by the gravitational acceleration. However, when accelerating, the scale captures the changes in motion:
  • If the elevator accelerates upward, the additional force required to accelerate is added to the gravitational pull, making the scale read heavier.
  • When descending or accelerating downward, the opposite occurs. A portion of gravitational force is used to slow the descent, and the scale reads lighter.

Each scenario reveals how weight is not a fixed attribute but is subject to variations in force as external conditions change. Thus, studying weight measurements in elevators delivers insights into how physics rules over daily movements simply and profoundly.

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 1.5 -kg block rests on top of a \(7.5-\mathrm{kg}\) block (Fig, 63\()\) . The cord and pulley have negligible mass, and there is no significant friction anywhere, \((a)\) What force \(F\) must be applied to the bottom block so the top block accelerates to the right at 2.5 \(\mathrm{m} / \mathrm{s}^{2} ?(b)\) What is the tension in the connecting cord?

(II) The cable supporting a 2125 -kg elevator has a maximum strength of \(21,750 \mathrm{N}\) . What maximum upward acceleration can it give the elevator without breaking?

(II) What average force is required to stop a \(950-\mathrm{kg}\) car in 8.0 \(\mathrm{s}\) if the car is traveling at 95 \(\mathrm{km} / \mathrm{h} ?\)

(III) A small block of mass \(m\) rests on the sloping side of a triangular block of mass \(M\) which itself rests on a hori- zontal table as shown in Fig. \(47 .\) Assuming all surfaces are frictionless, determine the magnitude of the force \(\vec{\mathbf{F}}\) that must be applied to \(M\) so that \(m\) remains in a fixed position relative to \(M\) (that is, \(m\) doesn't move on the incline). [Hint: Take \(x\) and \(y\) axes horizontal and vertical.

A "doomsday" asteroid with a mass of \(1.0 \times 10^{10} \mathrm{~kg}\) is hurtling through space. Unless the asteroid's speed is changed by about \(0.20 \mathrm{~cm} / \mathrm{s},\) it will collide with Earth and cause tremendous damage. Researchers suggest that a small "space tug" sent to the asteroid's surface could exert a gentle constant force of \(2.5 \mathrm{~N}\). For how long must this force act?

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.