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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.)

Mathematics
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it will be the same thing, what is the formula for work here?
@ Unam
@mark_o. W, F=kx

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Other answers:

hi my pc got froze.... :D
if F=-kx \[W=\int\limits_{}^{}F(x) dx\]
\[W=\int\limits_{x _{i}}^{x _{f}} -Kx dx\] \[W=-K[\frac{ x ^{2} }{ 2 }]_{x _{i}}^{x _{f}}\] \[W=-\frac{ K }{ 2 }[x ^{2}-x _{f}^{2}]\]
if the initial x=0 then \[W=-\frac{ K }{ 2 }(x _{f})^{2}\]
can you try to do it ?
@ Unam .... :D take care now :D
oh got it
sorry internet was out
just got back..but thanks alot for the help
i have the formula but just cos its in pounds and ft so i'm not familiar with the stuff
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
did solve W=? is it 7/4 ft lbs ?
@mark_o. nope :)

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