A community for students.
Here's the question you clicked on:
 0 viewing
anonymous
 4 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
 4 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.)

This Question is Closed

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

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

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

anonymous
 4 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
 4 years ago
Best ResponseYou've already chosen the best response.0if the initial x=0 then \[W=\frac{ K }{ 2 }(x _{f})^{2}\]

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

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

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

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

anonymous
 4 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
 4 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
 4 years ago
Best ResponseYou've already chosen the best response.0did solve W=? is it 7/4 ft lbs ?
Ask your own question
Sign UpFind more explanations on OpenStudy
Your question is ready. Sign up for free to start getting answers.
spraguer
(Moderator)
5
→ View Detailed Profile
is replying to Can someone tell me what button the professor is hitting...
23
 Teamwork 19 Teammate
 Problem Solving 19 Hero
 Engagement 19 Mad Hatter
 You have blocked this person.
 ✔ You're a fan Checking fan status...
Thanks for being so helpful in mathematics. If you are getting quality help, make sure you spread the word about OpenStudy.