Calculating Work Needed to Stretch a Spring: 100J to 0.75m

[iPhotonHQ]

"It takes 100J of work to stretch a spring 0.5m from its equilibrium position. How much work is needed to stretch it an additional 0.75m."
Attempt: w = ⌠_a^bF(x)dx
work = F x D
100J = F x 0.5m
F = 200J

0.75 + 0.5 = 1.25
w = ⌠_0.5^1.25 200dx
w = 150 J

The correct answer: w = 525 J

what did I do wrong? Thanks!

Nevermind, I found the exact problem online, so sorry!

 

[Arman777]

Work is not constant in spring.You can't use W=Fd.You should take an integral
(Carefull for the signs when you take an integral).
Another thing that Questions ask "The work done by our force".

 

[iPhotonHQ]

Work is not constant in spring.You can't use W=Fd.You should take an integral
(Carefull for the signs when u take an integral).
Another thing that Questions ask "The work done by our force".

My problem was that I forgot hooke's law.
F = kx
I can apply it to the given information to find the target solution.

 

[Arman777]

Any ideas How ?

 

[iPhotonHQ]

Any ideas How ?

w = ⌠_.5^1.25 kx dx
you can solve for k by plugging in known integral...
100 = ⌠₀^.5kxdx

you should get k = 800

 

[Arman777]

great