Question

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.)

Answer 1

it will be the same thing, what is the formula for work here?

Answer 2

@ Unam

Answer 3

@mark_o. W, F=kx

Answer 4

hi my pc got froze.... :D

Answer 5

if F=-kx
\[W=\int\limits_{}^{}F(x) dx\]

Answer 6

\[W=\int\limits_{x _{i}}^{x _{f}} -Kx dx\]
\[W=-K[\frac{ x ^{2} }{ 2 }]_{x _{i}}^{x _{f}}\]
\[W=-\frac{ K }{ 2 }[x ^{2}-x _{f}^{2}]\]

Answer 7

if the initial x=0 then
\[W=-\frac{ K }{ 2 }(x _{f})^{2}\]

Answer 8

can you try to do it ?

Answer 9

@ Unam .... :D
take care now :D

Answer 10

oh got it

Answer 11

sorry internet was out

Answer 12

just got back..but thanks alot for the help

Answer 13

i have the formula but just cos its in pounds and ft so i'm not familiar with the stuff

Answer 14

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

Answer 15

did solve W=? is it 7/4 ft lbs ?

Answer 16

@mark_o. nope :)