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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.)
 2 years ago
 2 years 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.)
 2 years ago
 2 years 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?
 2 years ago

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

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

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

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

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

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

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

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

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

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

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