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anonymous
 3 years ago
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.)
anonymous
 3 years ago
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.)

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anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0it will be the same thing, what is the formula for work here?

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0hi my pc got froze.... :D

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0if F=kx \[W=\int\limits_{}^{}F(x) dx\]

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0\[W=\int\limits_{x _{i}}^{x _{f}} Kx dx\] \[W=K[\frac{ x ^{2} }{ 2 }]_{x _{i}}^{x _{f}}\] \[W=\frac{ K }{ 2 }[x ^{2}x _{f}^{2}]\]

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

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0can you try to do it ?

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0@ Unam .... :D take care now :D

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0sorry internet was out

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0just got back..but thanks alot for the help

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0i have the formula but just cos its in pounds and ft so i'm not familiar with the stuff

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0F=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

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0did solve W=? is it 7/4 ft lbs ?
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