Orientation on Calculus Work Problem. Hooke's Law.

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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:
on Phys.org
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?