If 27 J of work are needed to stretch a spring from 16 cm to 22 cm and 45 J are needed to stretch it from 22 cm to 28 cm, what is the natural length of the spring?

Question

dan815 · Answer

do u know the equation for energy stored in a spring

dan815 · Answer

F=kx

anonymous · Answer

E = (1/2) K * x^2, where K is the spring constant and x is the displacement

dan815 · Answer

integrate to find work
so w=int(kx dx)

dan815 · Answer

plug in your values and solve

mony01 · Answer

do i subtract 22 from 16 first?

dan815 · Answer

http://answers.yahoo.com/question/index?qid=20100130230220AAt40TO

dan815 · Answer

http://www.sosmath.com/CBB/viewtopic.php?t=987

dan815 · Answer

here u go a full solution http://www.freemathhelp.com/forum/threads/46261-Natural-length-of-a-spring-given-work

dan815 · Answer

we dont know how much it was stretched past the equlibrium position

dan815 · Answer

w=1/2k*(x)^2
where x is the stretch from equilibirum

dan815 · Answer

w=(k/2)*(22-a)^2-[k/2*(16-a)^2]=27

w=(k/2)*(28-a)^2-[k/2*(22-a)^2]=45

dan815 · Answer

you have 2 equations 2 unknows solve them

mony01 · Answer

would the answer be 9?

whpalmer4 · Answer

does 9 give a valid solution if you plug it back in to both equations?

mony01 · Answer

where can i plug in 9 at a?

whpalmer4 · Answer

well, how did you come up with 9?  surely it was on opposite side of an equals sign from a letter, right?  which letter? :-)

whpalmer4 · Answer

$$\frac{k}{2}(22-a)^2-\frac{k}{2}(16-a)^2=27$$$$\frac{k}2(28-a)^2-\frac{k}2(22-a)^2=45$$

dan815 · Answer

have learnt how to solve equations

dan815 · Answer

think of it like balancing wieghts

whpalmer4 · Answer

$$(16-a)^2 = a^2-32a+256$$$$(22-a)^2=a^2-44a+484$$$$(28-a)^2=a^2-56a+784$$
Multiply both equations by $(2/k)$:

$$(22-a)^2-(16-a)^2 = 54/k$$$$(28-a)^2-(22-a)^2 = 90/k$$$$a^2-44a+484-a^2+32a-256 = 54/k$$$$a^2-56a+784-a^2+44a-484=90/k$$
$$-12a+228=54/k$$$$-12a+300=90/k$$Multiply one row by -1 and add together
$$72=36/k$$You should be able to find $a$...