### Understanding Elevator Safety Design using SpringsAn engineer is designing a spring to be placed at the bottom of an elevator shaft. If the elevator cable should break when the elevator is at a height of 103 meters above the top of the spring, the objective is to calculate the value that the spring stiffness constant should have so that passengers undergo an acceleration of no more than 3.9 times the acceleration due to gravity (3.9g) when brought to rest. Given:- Height above the spring: \(103 \text{ m}\)- Total mass of the elevator and passengers: \(924 \text{ kg}\)- Maximum allowed acceleration: \(3.9g\) (where \(g \approx 9.8 \, \text{m/s}^2\))#### Explanation:1. **Energy Considerations**: - When the elevator cable fails, the elevator falls freely under gravity. - The potential energy at height \(h\) (103 m) is converted into kinetic energy just before contact with the spring. - The spring then needs to absorb this kinetic energy and decelerate the elevator gently without exceeding the specified acceleration.2. **Calculating Potential Energy**: \[ E_p = mgh \] where \(m = 924 \, \text{kg}\), \(g = 9.8 \, \text{m/s}^2\), and \(h = 103 \, \text{m}\).3. **Kinetic Energy to Spring Compression**: The kinetic energy \(E_k\) at the moment of potential energy conversion is given by: \[ E_k = \frac{1}{2} k x^2 \] where \(k\) is the spring stiffness constant, and \(x\) is the compression of the spring.4. **Maximum Compression and Acceleration**: Ensure that the deceleration provided by the spring does not exceed \(3.9g\). This involves setting up an equation that includes the maximum force the spring exerts in terms of the maximum allowable acceleration.5. **Solving for Spring Constant \(k\)**: Utilize energy conservation and force equations to derive the necessary spring constant \(k\) such that the deceleration condition is satisfied.These steps help ensure that the design protects the passengers by limiting the maximum shock experienced. Students and engineers

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An engineer is designing a spring to be placed at the bottom of an elevator shaft. If the elevator cable should break when the elevator is at a height of 103m above the top of the spring, calculate the value that the spring stiffness constant should have so that passengers undergo an acceleration of no more than 3.9g when brought to rest. Let 924kg be the total mass of the elevator and passengers.

An engineer is designing a spring to be placed at the bottom of an elevator shaft. If the elevator cable should break when the elevator is at a height of 103m above the top of the spring, calculate the value that the spring stiffness constant should have so that passengers undergo an acceleration of no more than 3.9g when brought to rest. Let 924kg be the total mass of the elevator and passengers.

College Physics

11th Edition

ISBN:9781305952300

Author:Raymond A. Serway, Chris Vuille

Publisher:Raymond A. Serway, Chris Vuille

Chapter1: Units, Trigonometry. And Vectors

Section: Chapter Questions

Problem 1CQ: Estimate the order of magnitude of the length, in meters, of each of the following; (a) a mouse, (b)...

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