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Unam

  • 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 ft-lb.)

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  1. mark_o.
    • 3 years ago
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    it will be the same thing, what is the formula for work here?

  2. mark_o.
    • 3 years ago
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    @ Unam

  3. Unam
    • 3 years ago
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    @mark_o. W, F=kx

  4. mark_o.
    • 3 years ago
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    hi my pc got froze.... :D

  5. mark_o.
    • 3 years ago
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    if F=-kx \[W=\int\limits_{}^{}F(x) dx\]

  6. mark_o.
    • 3 years ago
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    \[W=\int\limits_{x _{i}}^{x _{f}} -Kx dx\] \[W=-K[\frac{ x ^{2} }{ 2 }]_{x _{i}}^{x _{f}}\] \[W=-\frac{ K }{ 2 }[x ^{2}-x _{f}^{2}]\]

  7. mark_o.
    • 3 years ago
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    if the initial x=0 then \[W=-\frac{ K }{ 2 }(x _{f})^{2}\]

  8. mark_o.
    • 3 years ago
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    can you try to do it ?

  9. mark_o.
    • 3 years ago
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    @ Unam .... :D take care now :D

  10. Unam
    • 3 years ago
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    oh got it

  11. Unam
    • 3 years ago
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    sorry internet was out

  12. Unam
    • 3 years ago
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    just got back..but thanks alot for the help

  13. Unam
    • 3 years ago
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    i have the formula but just cos its in pounds and ft so i'm not familiar with the stuff

  14. mark_o.
    • 3 years ago
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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

  15. mark_o.
    • 3 years ago
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    did solve W=? is it 7/4 ft lbs ?

  16. Unam
    • 3 years ago
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    @mark_o. nope :)

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