# Thread: one more hookes law question (work and springs)

1. ## one more hookes law question (work and springs)

suppose that 2 J of work is needed to stretch a string from its natural length of 30 cm to a length of 42 cm. how much work is needed to stretch it from 35 cm to 40 cm?

just when i thought i had this down, i got to this question (i got an exam tomorow, and the teacher just went over one easy example today, so i dont really have anything to reference)... this seems like im supposed to work backwards from the basic questions but im not at all sure how to tackle this.. preliminarily, i set up an integral from 0 to 12 that equals 2, but i dont know where to go at all from there... i would desperately appreciate it if someone can walk me through this one...

thanks so much!

2. Originally Posted by twostep08
suppose that 2 J of work is needed to stretch a string from its natural length of 30 cm to a length of 42 cm. how much work is needed to stretch it from 35 cm to 40 cm?

just when i thought i had this down, i got to this question (i got an exam tomorow, and the teacher just went over one easy example today, so i dont really have anything to reference)... this seems like im supposed to work backwards from the basic questions but im not at all sure how to tackle this.. preliminarily, i set up an integral from 0 to 12 that equals 2, but i dont know where to go at all from there... i would desperately appreciate it if someone can walk me through this one...

thanks so much!
this is it, bud ... I start charging by the keystroke after this

$\displaystyle \int_0^{.12} kx \, dx = 2
$

evaluate the above definite integral and solve for the spring constant, k.

after you find k, then determine the integral ...

$\displaystyle \int_{.05}^{.1} kx \, dx$

3. lol i think thats it for now... i totally get this now.. i wouldve failed this exam without you!