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