force and work

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hgaon001

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So the problem reads : A force of 40 pounds compresses a 12-inch spring by 3 inches. How much work (in foot-pounds) is required to compress the spring from a length of 8 inches to a length of 2 inches?

My main problem is im not sure if when i find k it's like
F=kd, im not sure if its 40=k3 or 40=k4

and then i would just set up an integral F(x)=kx with that from 4 to 10 right?
 
The limits of integration must be the displacement from the natural length, NOT the length of the spring.

So, yes, you are correct.
 
k is using the 3. \(\displaystyle 40=3k\Rightarrow k=\frac{40}{3}\)

May I ask, from where did you get the 4?.
 
because the question asked to compress it from 8.. which means it was already compressed 4 inches..
 
Initially, we are told it takes 40 foot-pounds to compress the spring 3 inches.

\(\displaystyle 40=3k\)

\(\displaystyle k=\frac{40}{3}\)

But then compressing it from 8 to 2 gives the spring a displacement, as you said, from 4 to 10.

\(\displaystyle \int_{4}^{10}\frac{40}{3}xdx\)
 
hgaon001 said:
So the problem reads : A force of 40 pounds compresses a 12-inch spring by 3 inches. How much work (in foot-pounds) is required to compress the spring from a length of 8 inches to a length of 2 inches?

My main problem is im not sure if when i find k it's like
F=kd, im not sure if its 40=k3 or 40=k4

and then i would just set up an integral F(x)=kx with that from 4 to 10 right?
Click to expand...

I don't know whether you are being taught problems with large displacement and non-linear spring or not!

The displacements in this problem are ridiculously large - specially in the compressive range ( 12" spring compressed to 2"!!!).

Given everything else (material, coil diameter, wire diameter of the spring, etc.) held constant - the spring constant of a spring is inversely proportional to its length (that is it is a function of 'l').

40 = k/12 * 3

k = 160

\(\displaystyle W \, = \, \int_4^{10}F\cdot dx \, = \,\int_4^{10} \frac{k}{12-x}x \cdot dx\)

Similar situation comes up in Hooke's law when you consider "true strain" of the structure.

However, there is a argument to be made where it is postulated:

The spring constant of a spring is inversely proportional to the number of coils.

So when you physically cut a spring - it's number of coil is reduced - and hence spring constant would go up proportionately. But it will not change during small displacement (since the number of coil would remain constant).

Now you can state your assumptions and do the problem accordingly.
 
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