- #1

- 44

- 0

**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=$$\int F dx$$

W=$$\int F dx$$

F=kx

16000=K(5)

3200=K

W=$$\int F dx$$

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!

F=kx

16000=K(5)

3200=K

W=$$\int F dx$$

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: