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Chapter 5: Problem 19

It takes the elevator in a skyscraper \(4.0 \mathrm{s}\) to reach its cruising speed of \(10 \mathrm{m} / \mathrm{s}\). A \(60 \mathrm{kg}\) passenger gets aboard on the ground floor. What is the passenger's apparent weight a. Before the elevator starts moving? b. While the elevator is speeding up? c. After the elevator reaches its cruising speed?

Short Answer

Expert verified
a. The passenger's apparent weight before the elevator starts moving is \(588 \mathrm{N}\). b. The passenger's apparent weight while the elevator is speeding up is \(738 \mathrm{N}\). c. The passenger's apparent weight after the elevator reaches its cruising speed is \(588 \mathrm{N}\).

Step by step solution

01

Calculate Normal Weight

First, calculate the normal weight using the formula: weight = mass * gravity. Given the mass of the person is \(60 \mathrm{kg}\), and acceleration due to gravity \(g = 9.8 \mathrm{m/s^2}\), the weight is \(60 \mathrm{kg} * 9.8 \mathrm{m/s^2} = 588 \mathrm{N}\).
02

Calculate Apparent Weight During Acceleration

The apparent weight of the person while the elevator is accelerating can be calculated using the formula: Apparent weight = mass * (gravity + acceleration). The acceleration of the elevator can be calculated by speed/time, which is \(10 \mathrm{m/s} / 4 \mathrm{s} = 2.5 \mathrm{m/s^2}\). Therefore, the apparent weight is \(60 \mathrm{kg} * (9.8 \mathrm{m/s^2} + 2.5 \mathrm{m/s^2}) = 738\, \mathrm{N}\).
03

Determine Apparent Weight at Cruising Speed

At the cruising speed, the acceleration becomes 0, so the apparent weight will be the same as the normal weight, which is \(588 \mathrm{N}\).

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Key Concepts

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

Physics Problem Solving
In physics, approaching a problem systematically is crucial for finding a solution. When solving a problem like the one concerning the apparent weight of a person in an elevator, it鈥檚 important to break the problem into smaller parts. This makes it easier to analyze and solve each aspect accurately.

Here is a useful methodology for tackling physics problems:
  • **Identify the known values.** Start by reviewing the information provided. In the elevator problem, known values include the mass of the passenger and the gravitational force. Mark these details clearly in your notes.
  • **Understand what is being asked.** Questions about apparent weight before the elevator starts, during acceleration, and at cruising speed need different approaches. Take note of these multiple states.
  • **Apply relevant physics principles and equations.** Ensure you use the correct physics formulas. For the elevator problem, this involves Newton's second law and the equation for apparent weight.
  • **Calculate as required.** Perform the necessary calculations for each part of the question. Check your calculations to prevent errors.
  • **Review the result.** Ensure that your results align logically with the situation described in the problem. For instance, does the apparent weight increase or decrease based on elevator activity?
Elevator Physics
Elevator physics involves understanding how weight and acceleration interact. In an elevator, the movement affects the weight felt by passengers, referred to as the "apparent weight."

**States of Elevator Motion**:
  • **Stationary:** When the elevator is not moving, a passenger's apparent weight is the same as their actual weight, calculated as mass times the acceleration due to gravity.
  • **Accelerating:** When the elevator accelerates (up or down), the apparent weight changes. If the elevator accelerates upward, the apparent weight increases. Conversely, it decreases when moving downward. This is because the force exerted by the elevator floor changes, causing variations in apparent weight.
  • **Constant Speed:** At cruising speed, the elevator moves at a constant speed, meaning no acceleration. Hence, the apparent weight reverts to normal, just like when the elevator is stationary.
Understanding these states helps in predicting how the apparent weight changes throughout the elevator ride.
Newton's Second Law
Newton's second law is central to understanding the concept of apparent weight in the elevator problem. This law states that the force on an object is equal to its mass multiplied by its acceleration:

\( F = m imes a \)

Apparent weight is perceived through the normal force, which is the force exerted by a surface to support the weight of an object resting on it. In the context of an elevator:
  • **When stationary,** the normal force equals the gravitational force acting on the person. This gives a net force of zero.
  • **During upward acceleration,** the normal force increases because the elevator needs to exert additional force. Hence, using Newton's second law, the calculation for apparent weight becomes: \( m imes (g + a) \), where \( a \) is positive.
  • **At cruising speed,** acceleration is zero, making the apparent weight equal to the normal weight, using \( m imes g \).
By applying this law, you can accurately predict changes in apparent weight as an elevator moves.

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Most popular questions from this chapter

Two workers are sliding a \(300 \mathrm{kg}\) crate across the floor. One worker pushes forward on the crate with a force of \(380 \mathrm{N}\) while the other pulls in the same direction with a force of \(350 \mathrm{N}\) using a rope connected to the crate. Both forces are horizontal, and the crate slides with a constant speed. What is the crate's coefficient of kinetic friction on the floor?

Seat belts and air bags save lives by reducing the forces exerted on the driver and passengers in an automobile collision. Cars are designed with a "crumple zone" in the front of the car. In the event of an impact, the passenger compartment decelerates over a distance of about \(1 \mathrm{m}\) as the front of the car crumples. An occupant restrained by seat belts and air bags decelerates with the car. By contrast, an unrestrained occupant keeps moving forward with no loss of speed (Newton's first law!) until hitting the dashboard or windshield, as we saw in Figure \(4.2 .\) These are unyielding surfaces, and the unfortunate occupant then decelerates over a distance of only about \(5 \mathrm{mm}\) a. A \(60 \mathrm{kg}\) person is in a head-on collision. The car's speed at impact is \(15 \mathrm{m} / \mathrm{s}\). Estimate the net force on the person if he or she is wearing a seat belt and if the air bag deploys. b. Estimate the net force that ultimately stops the person if he or she is not restrained by a seat belt or air bag. c. How do these two forces compare to the person's weight?

The acceleration of the spacecraft in which the Apollo astronauts took off from the moon was \(3.4 \mathrm{m} / \mathrm{s}^{2} .\) On the moon, \(g=1.6 \mathrm{m} / \mathrm{s}^{2} .\) What was the apparent weight of a \(75 \mathrm{kg}\) astronaut during takeoff?

A \(10 \mathrm{kg}\) crate is placed on a horizontal conveyor belt. The materials are such that \(\mu_{\mathrm{s}}=0.50\) and \(\mu_{\mathrm{k}}=0.30\) a. Draw a free-body diagram showing all the forces on the crate if the conveyer belt runs at constant speed. b. Draw a free-body diagram showing all the forces on the crate if the conveyer belt is speeding up. c. What is the maximum acceleration the belt can have without the crate slipping? If the acceleration of the belt exceeds the value determined in part c, what is the acceleration of the crate?

Josh starts his sled at the top of a \(3.0-\mathrm{m}\) -high hill that has a constant slope of \(25^{\circ} .\) After reaching the bottom, he slides across a horizontal patch of snow. Ignore friction on the hill, but assume that the coefficient of kinetic friction between his sled and the horizontal patch of snow is \(0.050 .\) How far from the base of the hill does he end up?
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