### Spring Design for Elevator Safety**Problem Statement:**An engineer is tasked with designing a spring to be placed at the bottom of an elevator shaft. In the event that the elevator cable breaks when the elevator is at a height of 273 meters above the top of the spring, we need to calculate the spring stiffness constant (k) that will ensure passengers experience an acceleration of no more than 5.1g when the elevator is brought to rest. Assume that the total mass of the elevator and passengers is 2185 kg.---**Detailed Explanation:**To solve this problem, we need to apply the principles of physics, specifically the dynamics of motion and Hooke's Law, which relates the force exerted by a spring to its displacement and stiffness constant.**1. Understanding the Parameters:** - Height (h): 273 meters - Maximum acceleration (a_max): 5.1 times the acceleration due to gravity (g), where g = 9.81 m/s² - Total mass (m): 2185 kg **2. Assessing the Maximum Allowable Acceleration:** \[ a_{max} = 5.1g = 5.1 \times 9.81 \, m/s^2 \] \[ a_{max} = 49.931 \, m/s^2 \]**3. Calculating the Force Due to Maximum Acceleration:** Using Newton's second law ( \( F = ma \) ): \[ F_{max} = m \cdot a_{max} = 2185 \, kg \times 49.931 \, m/s^2 \] \[ F_{max} = 109054.735 \, N \]**4. Energy Considerations:** The potential energy (PE) due to height when the elevator is at 273 meters needs to be considered. It’s converted to kinetic energy (KE) and then to the elastic potential energy (EPE) in the spring: \[ PE = mgh = 2185 \, kg \times 9.81 \, m/s^2 \times 273 \, m \] \[ PE = 5828204.65 \, J \]**5. Spring Energy Equation:** At maximum

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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 273m 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 5.1g when brought to rest. Let 2185kg 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 273m 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 5.1g when brought to rest. Let 2185kg 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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