Orientation on Calculus Work Problem. Hooke's Law.

AI Thread Summary
The discussion focuses on calculating the work done to compress a spring using Hooke's Law, with a force of 16,000 lb compressing it from 13 inches to 8 inches. The work formula, W = ∫F dx, is applied, but there is confusion regarding the limits of integration and the correct interpretation of the compression distance. Participants clarify that the spring's initial length is 13 inches, and the compression of interest is the first inch, leading to a need for precise limits in the integral. The conversation emphasizes the importance of correctly identifying the extent of compression to accurately calculate the work done.
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!

 
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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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