1. ## Work Problem!

What is the natural length of a spring (in cm) which needs 6 Joules of work to stretch the length from 10cm to 12cm and another 10 Joules to stretch it from 12cm to 14cm?

I believe I solved it but a friend said the answer was wrong but I'm pretty sure mine is right. He says my technique is wrong but I don't know why it would be, it makes sense to me. Anyways, here my work:

$6 = \int^{12-y}_{10-y} kx\ dx$
$12 = \int^{14-y}_{12-y} kx\ dx$

Where k is the spring constant, x is force, and y is natural length.

I solve for both and since k is a constant make each equation equal which results in $y=9$. My friend claims the answer is 8cm and he said he needed to use linear algebra or something silly like that. Hope you can clear things up for me!

2. Originally Posted by jfinkbei
What is the natural length of a spring (in cm) which needs 6 Joules of work to stretch the length from 10cm to 12cm and another 10 Joules to stretch it from 12cm to 14cm?

I believe I solved it but a friend said the answer was wrong but I'm pretty sure mine is right. He says my technique is wrong but I don't know why it would be, it makes sense to me. Anyways, here my work:

$6 = \int^{12-y}_{10-y} kx\ dx$
$\textcolor{red}{12} = \int^{14-y}_{12-y} kx\ dx$

Where k is the spring constant, x is force, and y is natural length.

I solve for both and since k is a constant make each equation equal which results in $y=9$. My friend claims the answer is 8cm and he said he needed to use linear algebra or something silly like that. Hope you can clear things up for me!
is it 10J or 12J ? your method is sound. if it's 12J, I get y = 9 cm.

3. Thank you! This is why this forum is so helpful, seeing dumb mistakes. It is actually 10J and with 10J the natural length is 8cm. Weird how it managed to work out to a whole number both ways...