/*! 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 8 You walk into an elevator, step ... [FREE SOLUTION] | 91Ó°ÊÓ

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You walk into an elevator, step onto a scale, and push the "up" button. You also recall that your normal weight is 625 \(\mathrm{N}\) . Start answering each of the following questions by drawing a free-body diagram. (a) If the elevator has an acceleration of magnitude \(2.50 \mathrm{m} / \mathrm{s}^{2},\) what does the scale read? (b) If you start holding a 3.85 -kg package by a light vertical string, what will be the tension in this string once the elevator begins accelerating?

Short Answer

Expert verified
(a) The scale reads approximately 780.51 N. (b) The tension in the string is approximately 47.40 N.

Step by step solution

01

Understand the Setup and Draw Free-Body Diagram

Visualize and draw the forces acting on you as you stand on the scale in the accelerating elevator: gravitational force (your weight) acting downwards and the normal force from the scale acting upwards.
02

Determine Your Mass

Convert your given weight into mass using the formula: \[ m = \frac{W}{g} \]where \(W = 625 \text{ N}\) and \(g = 9.81 \text{ m/s}^2\). Calculate your mass: \[ m = \frac{625}{9.81} \approx 63.7 \text{ kg} \]
03

Calculate Apparent Weight During Acceleration

When accelerating upwards, your apparent weight is the normal force \( F_N \), which can be calculated using the formula: \[ F_N = m(g + a) \]where \( a = 2.50 \text{ m/s}^2 \) is the upward acceleration. Plug in the values:\[ F_N = 63.7(9.81 + 2.50) \approx 780.51 \text{ N} \]
04

Consider the Package's Free-Body Diagram

The forces acting on the package are its weight acting downwards and the tension in the string acting upwards. Since the whole system is accelerating upward, this affects the tension.
05

Calculate the Weight of the Package

Determine the weight of the package using: \[ W_{pkg} = mg \]where \( m = 3.85 \text{ kg} \) and \( g = 9.81 \text{ m/s}^2 \). Calculate:\[ W_{pkg} = 3.85 \times 9.81 \approx 37.76 \text{ N} \]
06

Calculate the Tension in the String

Use Newton's second law to find the tension \( T \) in the string, considering the upward acceleration:\[ T = m(g + a) \]Plug in the values for the package:\[ T = 3.85(9.81 + 2.50) \approx 47.40 \text{ N} \]

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Free-Body Diagram
A free-body diagram is a simple way to visualize the forces acting on an object. When solving physics problems, especially those involving forces, drawing a free-body diagram is a crucial first step. In this scenario, imagine you are standing on a scale inside an elevator. When you press the "up" button, the elevator begins to accelerate vertically. Here's what you do:
  • Draw the person inside the elevator as a dot or box.
  • Identify and draw arrows representing the forces: the gravitational force (weight) pointing downwards and the normal force (force from the scale) pointing upwards.
This visual representation helps you identify how these forces interact, making it easier to apply Newton's Second Law later. In this setup, note the direction and relative size of the arrows -- the normal force should be greater when the elevator is accelerating upwards.
Newton's Second Law
Newton's Second Law is a fundamental principle in physics that describes how the velocity of an object changes when it is subjected to an external force. The law states that the force acting on an object is equal to the mass of the object multiplied by its acceleration: \( F = ma \).
This equation can be rearranged to solve for different variables such as force, mass, or acceleration, depending on what's needed. In the context of our elevator problem, when the elevator accelerates upwards, Newton's Second Law helps us calculate the apparent weight. The normal force from the scale is adjusted due to this acceleration, effectively altering the reading you see. This change is calculated using the modified equation: \( F_N = m(g + a) \).
Here, \( g \) is the acceleration due to gravity and \( a \) is the elevator's acceleration. The added acceleration component increases the normal force, thus increasing your apparent weight.
Apparent Weight
Apparent weight is the perception of your weight as displayed by a scale, influenced by the forces acting upon you. In a non-moving elevator, your apparent weight matches your actual weight. However, when the elevator accelerates upward, the scale must exert more force against you to overcome the additional downward acceleration from gravity.
To calculate your apparent weight in this scenario, we use \( F_N = m(g + a) \).
Imagine you normally weigh 625 N. Inside the accelerating elevator, your mass calculated from that weight would stay the same (approximately 63.7 kg), but your apparent weight measured by the scale changes to about 780.51 N. This increase occurs because the elevator's upward acceleration adds to gravitational acceleration, making you feel 'heavier' momentarily.
Elevator Acceleration
Elevator acceleration impacts how both weight and tension in any suspended objects (e.g., a package) are perceived. In this elevator scenario, consider how moving upwards changes things. With the elevator accelerating at \( 2.50 \text{ m/s}^2 \), both you and the package are being acted upon by an additional "push" upwards.
For the package held by a string, the tension is calculated to counteract both gravity and the elevator's acceleration. Here: \( T = m(g + a) \) for the package. Given a package mass of 3.85 kg, this results in an upward tension of approximately 47.40 N in the string.
Thus, the formula adapts to these new conditions, demonstrating how motion affects forces. Understanding this helps illustrate Newtonian physics in everyday experiences, like riding an elevator.

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Most popular questions from this chapter

A small car (mass 380 \(\mathrm{kg}\) ) is pushing a large truck (mass900 \(\mathrm{kg}\) ) due east on a level road. The car exerts a horizontal force of 1200 \(\mathrm{N}\) on the truck. What is the magnitude of the force that the truck exerts on the car?

Two crates, \(A\) and \(B,\) sit at rest side by side on a friction- less horizontal surface. The crates have masses \(m_{A}\) and \(m_{B}\) . A horizontal force \(\vec{\boldsymbol{F}}\) is applied to crate \(A\) and the two crates move off to the right. (a) Draw clearly labeled free-body diagrams for crate A and for crate \(B\) . Indicate which pairs, if any, are third- law action- reaction pairs. (b) If the magnitude of force \(\vec{F}\) is less than the total weight of the two crates, will it cause the crates to move? Explain.

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A small 8.00 -kg rocket burns fuel that exerts a time-varying upward force on the rocket as the rocket moves upward from the launch pad. This force obeys the equation \(F=A+B t^{2}\) . Measurements show that at \(t=0,\) the force is \(100.0 \mathrm{N},\) and at the end of the first 2.00 s, it is 150.0 \(\mathrm{N}\) . (a) Find the constants \(A\) and \(B\) , including their SI units. (b) Find the net force on this rocket and its acceleration (i) the instant after the fuel ignites and (ii) 3.00 s after fuel ignition. (c) Suppose you were using this rocket in outer space, far from all gravity. What would its acceleration be 3.00 s after fuel ignition?

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