Work Required to Stretch Spring 4-7 ft: 8.4 ft-lbs

[blessedcurse]

Homework Statement

The amount of WORK to stretch a spring 4 feet beyond its natural length of 2 feet is 10 ft-lbs. Find the work required to stretch the spring from 4 feet to 7 feet.

Homework Equations

W=$$\int^{b}_{a}Fdx=\int^{b}_{a}kxdx=[kx^{2}/2]^{b}_{a}$$

The Attempt at a Solution

10=$$[kx^{2}/2]^{4}_{0}$$
10=k(16)/2-k(0)/2
10=8k
k=4/5

W=$$[kx^{2}/2]^{b}_{a}$$
W=4/5$$[x^{2}/2]^{5}_{2}$$
W=4(25)/10-4(4)/10
W=42/5

W=8.4 ft*lb

 

[Vuldoraq]

Could you explain your limits of integration? The spring is either being stretched from 0 to 6 ft or from 2 to 6 ft, I don't see where you got 0 to 4 from.

 

[blessedcurse]

It's being stretched 4 feet beyond its natural length of 2 feet, so the natural length is the 0 and then 4 feet beyond that is 4.

At first I had 2 to 6, but then I realized that that meant stretching it from 2 feet beyond its natural length to 6 feet beyond it because the limits refer to change in distance... so the change in distance is 0 to 4.

 

[Vuldoraq]

Oh, I see. Sorry, that is a badly worded question (or more likely I read it badly). I thought it meant it's being stretched four feet beyond it's elestic limit! My bad.

Given that is true I think your answer is correct, at least in it's principle (Ihaven't checked the numerics).