## Unam Group Title Any help? I'm not so familiar with the pounds and feet...would be easier for me to have N and cm A force of 7 pounds is required to hold a spring stretched 0.2 feet beyond its natural length. How much work is done in stretching the spring from its natural length to 1 feet beyond its natural length? Don't forget to enter the correct units. (You may enter lbf or lb*ft for ft-lb.) one year ago one year ago

1. mark_o. Group Title

it will be the same thing, what is the formula for work here?

2. mark_o. Group Title

@ Unam

3. Unam Group Title

@mark_o. W, F=kx

4. mark_o. Group Title

hi my pc got froze.... :D

5. mark_o. Group Title

if F=-kx $W=\int\limits_{}^{}F(x) dx$

6. mark_o. Group Title

$W=\int\limits_{x _{i}}^{x _{f}} -Kx dx$ $W=-K[\frac{ x ^{2} }{ 2 }]_{x _{i}}^{x _{f}}$ $W=-\frac{ K }{ 2 }[x ^{2}-x _{f}^{2}]$

7. mark_o. Group Title

if the initial x=0 then $W=-\frac{ K }{ 2 }(x _{f})^{2}$

8. mark_o. Group Title

can you try to do it ?

9. mark_o. Group Title

@ Unam .... :D take care now :D

10. Unam Group Title

oh got it

11. Unam Group Title

sorry internet was out

12. Unam Group Title

just got back..but thanks alot for the help

13. Unam Group Title

i have the formula but just cos its in pounds and ft so i'm not familiar with the stuff

14. mark_o. Group Title

F=kx pound F= K x (in ft) then k=lb/ft = pound per foot now work w= -(k/2)x^2 W= (lb / ft)(ft)^2=lb ft

15. mark_o. Group Title

did solve W=? is it 7/4 ft lbs ?

16. Unam Group Title

@mark_o. nope :)