Finding spring constant in elevator

In summary: As the spring compresses, the elevator slows down and the KE decreases. At the same time, the spring gains EPE as it compresses. The total energy of the system stays the same, but the form of it changes. In summary, in a 'worst-case' design scenario, a 2000-kg elevator falling at 25 m/s first contacts a cushioning spring at the bottom of the shaft. The spring is meant to stop the elevator, compressing 3m while a safety clamp applies a constant 17,000N frictional force. As an energy consultant, you are tasked with determining the force constant of the spring, which can be found by setting up an equation that takes into account the
  • #1
Maiia
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0

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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  • #2
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.
 
  • #3
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?
 
  • #4
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?
 
  • #5
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:
  • #6
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).
 

1. How do you find the spring constant in an elevator?

The spring constant in an elevator can be found by measuring the displacement of the elevator when a known weight is placed on it. The force applied by the weight can be calculated using Newton's second law (F = ma), and the displacement can be measured using a ruler or measuring tape. The spring constant can then be calculated using the equation k = F/x, where k is the spring constant, F is the force applied, and x is the displacement.

2. Why is it important to know the spring constant in an elevator?

The spring constant is an important factor in understanding the behavior of the elevator. It determines how much force is required to compress or stretch the spring, and therefore affects how the elevator responds to external forces. It is also used in safety calculations to ensure that the elevator can safely support its maximum weight capacity.

3. What factors can affect the spring constant in an elevator?

The material and design of the spring itself can affect the spring constant, as well as any external factors such as temperature and humidity. The weight and distribution of weight in the elevator can also impact the spring constant, as well as the condition and maintenance of the elevator.

4. How can the spring constant be altered in an elevator?

The spring constant can be altered by changing the material or design of the spring, adjusting the weight and distribution of weight in the elevator, or by changing the condition or maintenance of the elevator. However, altering the spring constant may also affect the overall function and safety of the elevator, so it should only be done by trained professionals.

5. Can the spring constant in an elevator change over time?

Yes, the spring constant in an elevator may change over time due to wear and tear on the spring or changes in the weight or distribution of weight in the elevator. Regular maintenance and inspections can help identify any changes in the spring constant and make necessary adjustments.

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