- #1
Dan350
- 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=$$\int F dx$$
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!
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: