Finding spring constant in elevator

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Homework Help Overview

The problem involves a 2000-kg elevator that falls at 25 m/s before contacting a spring designed to stop it. The spring compresses 3m, and a constant frictional force of 17,000N is applied during the motion. The goal is to determine the spring constant required for this scenario.

Discussion Character

  • Exploratory, Assumption checking, Problem interpretation

Approaches and Questions Raised

  • Participants discuss the setup of the energy equation, questioning the assumptions about the initial conditions, including the height and distance over which friction acts. There is also debate about the interpretation of kinetic and gravitational potential energy in the context of the spring's compression.

Discussion Status

The discussion is active, with participants providing insights into the energy components involved and clarifying the conditions at the moment the elevator contacts the spring. Some guidance has been offered regarding the energy equations, but questions remain about the assumptions made regarding distances and heights.

Contextual Notes

There is uncertainty regarding the initial height of the elevator and the total distance over which the frictional force acts, as well as the interpretation of the energy conservation principles in this scenario.

Maiia
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Homework Statement


In a 'worst-case' design scenario a 2000-kg elevator with broken cables is falling at 25m/s when it first contacts a cushioning spring at the bottom of the shaft. The spring is supposed to stop the elevator, compressing 3m as it does so. During the motion, a safety clamp applies a constant 17,000N frictional force to the elevator. As an energy consultant, you are asked to determine what the force constant of the spring should be.

The Attempt at a Solution


The way I set up the problem was like this:

Work done by noncons force + GPE initial+ KE initial+ EPE initial= GPE final + KE final+ EPE final
EPE initial is 0. I set my zero as the point when the mass is coming to rest on the spring for a slight instant before compressing the spring. I assume the 25m/s is a constant velocity...? If so, then KE inital and KE final cancel. So I am left with:
-fd + mgy0= mgyf + .5kx^2
thing is, then I have two unknowns, because I don't know the spring constant and I don't know what height it starts off at...If someone could point me in the right direction, I would greatly appreciate it:)
 
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No, 25 m/s is not constant. It was 25 m/s at the point where the elevator first contacts the spring, then, at full compression, the speed is zero.

The energy going into the problem is KE+GPE. The height for GPE is the 3m it will compress.

The dissipated energy is-Fd, ans you noted, where d is the same 3m. The remaining energy is in the spring.
 
Why is d 3m for -Fd? I thought the d would be the entire distance the elevator was falling? Because isn't Work done the total distance over which the friction acts?
 
in the same vein, why would the h for the GPE be 3m as well? assuming that this is the initial GPE, wouldn't h be the height at which the elevator starts falling at? How do we know it starts at a height of 3m?
 
Your equation should reflect both terms.

KE of the initial 1/2mv² = 1/2kx² + 17000 * x

Looks like everything is known but the spring constant k, which happily is what they asked you to find.
 
Last edited:
Maiia said:
Why is d 3m for -Fd? I thought the d would be the entire distance the elevator was falling? Because isn't Work done the total distance over which the friction acts?

The problem "starts" at the moment the elevator contacts the spring. At this point it is moving at 25 m/s. The energy of the system is assessed at this moment. Whatever happened before is irrelevant. At this moment, it has KE (due to 25 m/s) and GPE (due to 3m above final position).
 

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