Orientation on Calculus Work Problem. Hooke's Law.

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Homework Help Overview

The discussion revolves around a work problem related to Hooke's Law, specifically focusing on the work done to compress a spring from its natural length of 13 inches to a compressed state. The original poster attempts to calculate the work done during the first inch of compression and raises questions about the limits of integration and the nature of the force involved.

Discussion Character

  • Exploratory, Assumption checking, Conceptual clarification

Approaches and Questions Raised

  • Participants explore the correct limits for integration when calculating work, questioning whether it should be from 0 to 1 inch or from 12 to 13 inches. There is also discussion about the units of force and the interpretation of the compression distance.

Discussion Status

Some participants have pointed out potential errors in units and the setup of the problem, while others seek clarification on the initial and final lengths of the spring during the compression process. The conversation indicates a lack of consensus on the correct approach to defining the limits of integration.

Contextual Notes

There are uncertainties regarding the units of force and the interpretation of the compression distance, as well as the distinction between a spring and a string. Participants are also addressing the implications of the work calculation based on different assumptions about the compression limits.

Dan350
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A force of 16,000lb compresses a string from its natural length of 13 inch to 8 inch. Find the work done to compress it to the first inch

W=$$\int F dx$$


F=kx
16000=K(5)
3200=KW=$$\int F dx$$

W=$$\int_1^{13}\!\ 3200xdx$$

[1600x^2] from 1 to 13

w= 268800ftlb

Am I right?
I think the trick here is the limits.
If they would be from 0 to 1,, aint that to little work?
I mean, as you compress down, it's harder to do it. The work has to increase as you reach 0 inch
Am I right?
If not, please explain

Or is it from 12 to 13?

how I viewed is that they want to know the work as you compress it to the first inch.. in this case 1

I need a little orientation

| 16000Lb |
/----*------------------------*----------------------/
0inch 1inch 8inch 13inc
Thanks!

 
Last edited:
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Dan350 said:
A force of 16,000ft-lb
Wrong units for a force.
160000=K(5)
Extra 0 crept in. What units do you want the work in? What units is the 5 in?
W=$$\int_1^{13}\!\ 3200xdx$$
What are the initial and final lengths of the spring as it goes through "the first inch" of its compression from 13 inches to 8 inches?
 
haruspex said:
Wrong units for a force.

Extra 0 crept in. What units do you want the work in? What units is the 5 in?

What are the initial and final lengths of the spring as it goes through "the first inch" of its compression from 13 inches to 8 inches?

Sorry about that
force is 16000lb
using hooke's law I got the K
16000lb=K(5) 5 as the inches that the Force compressed

The length of the string is 13 inches
It was compressed 5 inches by a 16000lb force
I need to find the work of the first inch

Thanks
 
Dan350 said:
The length of the string is 13 inches
A spring, presumably, not a string.
It was compressed 5 inches by a 16000lb force
I need to find the work of the first inch
So answer my question:
"What are the initial and final lengths of the spring as it goes through "the first inch" of its compression from 13 inches to 8 inches? "​
 
haruspex said:
A spring, presumably, not a string.

So answer my question:
"What are the initial and final lengths of the spring as it goes through "the first inch" of its compression from 13 inches to 8 inches? "​


1 and 13
the spring is 13 inch long. from 0 to 13 there's no work, now, whether it compresses or streches, a work can be calculated
 
Dan350 said:
1 and 13
No.
The x in your integral is the extent of compression. The spring starts at its relaxed length of 13 inches. That's x = 0, no compression. It is compressed from 13 inches to 8 inches. That's a compression of 5 inches, x = 5. What is the value of x when the spring has been compressed by only the first of those five inches of compression?
 

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