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Zach, whose mass is \(80 \mathrm{kg}\), is in an elevator descending at \(10 \mathrm{m} / \mathrm{s}\). The elevator takes 3.0 s to brake to a stop at the first floor. a. What is Zach's apparent weight before the elevator starts braking? b. What is Zach's apparent weight while the elevator is braking?

Short Answer

Expert verified
Before the elevator starts braking, Zach's apparent weight is 784 N. While the elevator is braking, his apparent weight is 517.6 N.

Step by step solution

01

Calculate Zach's apparent weight before the elevator starts braking.

Let's start by calculating Zach's weight before the elevator slows down. As there's no acceleration during this phase, only the force of gravity is acting on him. The formula to calculate weight is \(F = mg\), where \(m\) = 80 kg is the Zach's mass and \(g\) = 9.8 m/s² is acceleration due to gravity. Substituting numbers in the formula we get:\( F = 80 \times 9.8 = 784 N\). Therefore, Zach's apparent weight before the elevator starts braking is 784 N.
02

Calculate the acceleration of the elevator

The elevator takes 3.0 s to brake to a stop from 10 m/s. The acceleration (negative because it's a deceleration) can be calculated by the formula \(a = (v_f - v_i)/t\), with \(v_i\) as initial velocity equals 10 m/s, \(v_f\) as final velocity equals 0 m/s (it comes to a stop) and \(t\) as time it takes to stop, equals 3 s. Substituting numbers, we get \(a = (0 - 10)/3 = - 3.33 m/s² \).
03

Calculate Zach's apparent weight while the elevator is braking.

Now that we have the deceleration (negative acceleration), we can calculate Zach's apparent weight while the elevator is braking using the formula \(F = mg + ma\). This time, \(a = -3.33 m/s²) (the negative sign indicates downward direction). Substituting numbers in the formula, we get: \(F = 80 \times 9.8 + 80 \times -3.33 = 784 -266.4 = 517.6 N\). Therefore, Zach's apparent weight while the elevator is braking is 517.6 N

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Key Concepts

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

Elevator Physics and its Effects
Elevator physics is all about how forces act on objects inside an elevator, affecting the perception of weight. When you're in an elevator that's either moving vertically upward or downward, you experience changes in your apparent weight. Your apparent weight differs from your actual weight—the force that gravity exerts on your mass. This is due to the presence of additional forces acting on you. For instance, when the elevator moves at a constant speed, your apparent weight equals your real weight, since only gravity is acting on your body.
  • When the elevator accelerates downward (like when it starts descending from a higher floor), your apparent weight feels lighter.
  • If it accelerates upwards (as when stopping at a higher floor), your apparent weight feels heavier.
  • Apparent weight depends on the net acceleration you feel—so it can change as the elevator accelerates, decelerates, or moves between floors.
Understanding these concepts helps us better grasp real-world situations we might encounter, such as in the original exercise, where Zach's apparent weight differs depending on whether the elevator is braking or moving at a constant speed.
Newton's Second Law - The Force Baseline
Newton's Second Law of Motion is crucial for understanding how forces affect motion. According to this law, the force \( F \) exerted on an object is equal to its mass \( m \) multiplied by its acceleration \( a \): \[ F = m \cdot a \]. This equation helps us to relate the forces acting on objects inside elevators, such as in the example with Zach. When Zach is in an elevator descending at a constant speed, the force acting on him is purely gravitational. However, once the elevator starts braking, an additional force comes into play due to the deceleration. Therefore, the total force experienced by Zach becomes a combination of gravitational force and the deceleration force from the elevator slowing down:\[ F = m \cdot g + m \cdot a \] (where \( g \) is gravity).
  • Newton’s second law helps to calculate apparent weight changes as elevators accelerate or decelerate.
  • The exercise demonstrates Newton's Law by computing different apparent weights depending on elevator speed changes.
  • This change provides insight into how net forces alter perceived weight.
Gravity - The Ever-present Force
Gravity is an invisible force pulling objects towards the center of the Earth. It is constant and, for practical calculations on Earth's surface, has a value of approximately \( 9.8 \mathrm{m/s}^2 \). While gravity is a constant force, its effect on perceived weight can vary depending on other forces at play.
  • Gravity is responsible for defining what we call 'weight'. Your true weight is a product of your mass and the gravitational acceleration (\( F = m \times g \)).
  • Any other force—like acceleration in an elevator—can change your apparent weight, though your actual weight remains consistent.
  • Apparent weight is affected by the gravitational pull plus any additional forces acting in the system.
In everyday terms, while you remain of the same mass and the gravitational force acting on you remains constant, your sensation of weight can vary depending on other external factors.

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Most popular questions from this chapter

Dana loads luggage into an airplane using a conveyor belt tilted at an angle of \(20^{\circ} .\) She places a few pieces of luggage on the belt before it starts to move, then she turns the belt on. It takes the belt \(0.70 \mathrm{s}\) to reach its top speed of \(1.2 \mathrm{m} / \mathrm{s}\). Does the luggage slip? Assume \(\mu_{\mathrm{s}}=0.50\) between the luggage and the belt.

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?

Each of 100 identical blocks sitting on a frictionless surface is connected to the next block by a massless string. The first block is pulled with a force of \(100 \mathrm{N}\). a. What is the tension in the string connecting block 100 to block 99? b. What is the tension in the string connecting block 50 to block \(51 ?\)

A simple model shows how drawing a bow across a violin string causes the string to vibrate. As the bow moves across the string, static friction between the bow and the string pulls the string along with the bow. At some point, the tension pulling the string back exceeds the maximum static friction force and the string snaps back. This process repeats cyclically, causing the string's vibration. Assume the tension in a 0.33 -m-long violin string is \(50 \mathrm{N}\), and the coefficient of static friction between the bow and the string is \(\mu_{\mathrm{s}}=0.80 .\) If the normal force of the bow on the string is \(0.75 \mathrm{N},\) how far can the string be pulled before it slips if the string is bowed at its center?

You probably think of wet surfaces as being slippery. Surprisingly, the opposite is true for human skin, as you can demonstrate by sliding a dry versus a slightly damp fingertip along a smooth surface such as a desktop. Researchers have found that the static coefficient of friction between dry skin and steel is \(0.27,\) while that between damp skin and steel can be as high as \(1.4 .\) Suppose a man holds a steel rod vertically in his hand, exerting a \(400 \mathrm{N}\) grip force on the rod. What is the heaviest rod he can hold without slipping if a. His hands are dry? b. His hands are wet?

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