application of integral calculus: Work (spring)

[makovx]

hi there! i'm having some troubles regarding this question:

"if a force of 5 pounds produces a stretch of 1/10 of the natural length, L , of the spring, how much work is done in stretching the spring to double its natural length?"

i tried answering this but i'm not sure if my answer is correct.
will the answer be in terms of L? thanks a lot!

[Doc Al]

will the answer be in terms of L?
Yes.

[makovx]

i answered W= 75 L is it correct?

[Doc Al]

i answered W= 75 L is it correct?
No. How did you get that answer?

[makovx]

my solution is this:

F(x)=k(x)
F(x)= 5 pounds
x=1/10 of L = L/10

5 = k(L/10)
k= 50/L

then F(x)= 50/L x

my lower limit is L and my upper limit is 2L.

then, i did the usual integration in the equation W= integral of F(x) dx (with the upper limit and the lower limit)

[Doc Al]

my solution is this:

F(x)=k(x)
F(x)= 5 pounds
x=1/10 of L = L/10

5 = k(L/10)
k= 50/L

then F(x)= 50/L x
So far, so good.
my lower limit is L and my upper limit is 2L.

then, i did the usual integration in the equation W= integral of F(x) dx (with the upper limit and the lower limit)
Careful. In F = kx, x is the amount of stretch, not the total length. (Fix the limits of your integration.)

[makovx]

yes. x is the amount of stretch. if the length is L, and the amount of stretch is 1/10, wouldn't be x= 1/10 of L?

did i miss something there or is it just my limits of integration that made my answer wrong?

[Doc Al]

yes. x is the amount of stretch. if the length is L, and the amount of stretch is 1/10, wouldn't be x= 1/10 of L?
That part, where you calculated the spring constant, was just fine.
did i miss something there or is it just my limits of integration that made my answer wrong?
It's your limits of integration. When you stretch the spring from L (unstretched) to 2L, how does x vary?

[makovx]

I finally got your point. Thanks a lot! ^^,