springs 2

[jeca86]

If 6J of work is needed to stretch a spring from 10cm to 12cm and another 10J is needed to stretch it from 12cm to 14cm, what is the natural length of the spring?

I dont even know how to begin with this one.

Maybe:
W=F times x
F= k times x
integrate k times x times dx?

but what numbers go where?
I'm sorry im asking for so much, but these problems just keep getting harder. (for me anyways)

 

[galactus]

No takers, I may as well give it a shot.

This is a little trickier than most spring-work problems.

We need the length of the spring, but we know work.

I am going to use meters. Let y equal the length of the spring.


\(\displaystyle \L\\6J=\int_{.10-y}^{.12-y}{kx}dx\)

Integrating gives:

\(\displaystyle \L\\6=\frac{kx^{2}}{2}\rightarrow\6=\frac{k((.12-y)^{2}-(.10-y)^{2})}{2}\)

\(\displaystyle \L\\12=k((.12-y)^{2}-(.10-y)^{2})\)

\(\displaystyle \L\\12=k(.0044-.04y)\)

Now, we still need to stretch the spring from 12 to 14 cm:

\(\displaystyle \L\\10J=\int_{.12-y}^{.14-y}{kx}dx\)

\(\displaystyle \L\\20=k((.14-y)^{2}-(.12-y)^{2})\)

\(\displaystyle \L\\20=k(.0052-.04y)\)

\(\displaystyle \L\\12-k(.0044-.04y)=20-k(.0052-.04y)\)

Solving for k, we get k=10,000

\(\displaystyle \L\\12=(10000)(.0044-.04y)\)

Solve for y(length of spring). Multiply by 100 to convert ot cm.

You could actually use either equation to solve. Try the other one and see.

I hope I didn't lead you down the primrose path. This seems correct.