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Unam
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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 ftlb.)
 one year ago
 one year ago
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 ftlb.)
 one year ago
 one year ago

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mark_o. Group TitleBest ResponseYou've already chosen the best response.1
it will be the same thing, what is the formula for work here?
 one year ago

Unam Group TitleBest ResponseYou've already chosen the best response.0
@mark_o. W, F=kx
 one year ago

mark_o. Group TitleBest ResponseYou've already chosen the best response.1
hi my pc got froze.... :D
 one year ago

mark_o. Group TitleBest ResponseYou've already chosen the best response.1
if F=kx \[W=\int\limits_{}^{}F(x) dx\]
 one year ago

mark_o. Group TitleBest ResponseYou've already chosen the best response.1
\[W=\int\limits_{x _{i}}^{x _{f}} Kx dx\] \[W=K[\frac{ x ^{2} }{ 2 }]_{x _{i}}^{x _{f}}\] \[W=\frac{ K }{ 2 }[x ^{2}x _{f}^{2}]\]
 one year ago

mark_o. Group TitleBest ResponseYou've already chosen the best response.1
if the initial x=0 then \[W=\frac{ K }{ 2 }(x _{f})^{2}\]
 one year ago

mark_o. Group TitleBest ResponseYou've already chosen the best response.1
can you try to do it ?
 one year ago

mark_o. Group TitleBest ResponseYou've already chosen the best response.1
@ Unam .... :D take care now :D
 one year ago

Unam Group TitleBest ResponseYou've already chosen the best response.0
sorry internet was out
 one year ago

Unam Group TitleBest ResponseYou've already chosen the best response.0
just got back..but thanks alot for the help
 one year ago

Unam Group TitleBest ResponseYou've already chosen the best response.0
i have the formula but just cos its in pounds and ft so i'm not familiar with the stuff
 one year ago

mark_o. Group TitleBest ResponseYou've already chosen the best response.1
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
 one year ago

mark_o. Group TitleBest ResponseYou've already chosen the best response.1
did solve W=? is it 7/4 ft lbs ?
 one year ago

Unam Group TitleBest ResponseYou've already chosen the best response.0
@mark_o. nope :)
 one year ago
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