# Thread: Applications of integration: Work

1. ## Applications of integration: Work

A force of 10 lb is required to hold a spring stretched 4 in. beyond its natural length. How much work is done in stretching it from its natural length to 6 in. beyond its natural length?

This is what I did...
F=10
x=4
4 going to 6

F=kx
10=k(4)
k=2.5

f=2.5x

I then took the integral of 2.5x from 4 to 6 and my answer was 25 ft-lbs. The correct answer is 15/4 ft-lbs.
Can someone explain?

2. Originally Posted by abclarinetuvwxyz
A force of 10 lb is required to hold a spring stretched 4 in. beyond its natural length. How much work is done in stretching it from its natural length to 6 in. beyond its natural length?

This is what I did...
F=10
x=4
4 going to 6

F=kx
10=k(4)
k=2.5

f=2.5x

I then took the integral of 2.5x from 4 to 6 and my answer was 25 ft-lbs. The correct answer is 15/4 ft-lbs.
Can someone explain?

The Hooke's constant is actually 30 (with units of lbs/ft)

Then, the work done going from natural length (x=0) to x=1/2 (6 inches is half a foot) is:
$\displaystyle \int_0^{1/2} 30x dx =15/4$

But who the heck uses lbs and ft as their units? Everyone else uses SI units, with kg and m...lol

3. [QUOTE=abclarinetuvwxyz;244519]A force of 10 lb is required to hold a spring stretched 4 in. beyond its natural length.
So F= kx and 10= 4k. k= 10/4= 5/2= 2.5

How much work is done in stretching it from its natural length to 6 in. beyond its natural length?

This is what I did...
F=10
x=4
4 going to 6

F=kx
10=k(4)
k=2.5
so k= 2.5 pounds/inch, not "pounds/foot"!

f=2.5x

I then took the integral of 2.5x from 4 to 6 and my answer was 25 ft-lbs. The correct answer is 15/4 ft-lbs.
Can someone explain?
$\displaystyle \int_4^6 2.5 x dx= c\left| \frac{2.5}{2}x^2\right|_4^6$
$\displaystyle = \frac{2.5}{2} (6^2- 4^2)= \frac{2.5}{2}(36-16)$
$\displaystyle = \frac{2.5}{2}(20)= 25$inch-lbs. That is the correct answer but since there are 12 inches/foot, that is 25/12 foot-pounds.

4. ## How to actually do it.

Originally Posted by abclarinetuvwxyz
A force of 10 lb is required to hold a spring stretched 4 in. beyond its natural length. How much work is done in stretching it from its natural length to 6 in. beyond its natural length?

This is what I did...
F=10
x=4
4 going to 6

F=kx
10=k(4)
k=2.5

f=2.5x

I then took the integral of 2.5x from 4 to 6 and my answer was 25 ft-lbs. The correct answer is 15/4 ft-lbs.
Can someone explain?
- OK! You just screwed up one thing. You took it from 4 inches past its natural length to 6 inches past its natural length. You were supposed to take it from 0 to 6. Your K will be the same form 0 to 6 just like 4 to 6 inches.

- This would give you 45 inches per pound. (12 inches per foot)!

- THIS GIVES YOU (45/12) OR 15/4 foot per pounds!!!