# Integral, Work

1. Feb 12, 2008

### Enjoicube

1. The problem statement, all variables and given/known data
A uniform cable hanging over the edge of a tall building is 40 ft long and weighs 60 lb. How much work is required to pull 10 ft of the cable to the top?

2. Relevant equations
Density=3/2 ft/lb

3. The attempt at a solution
Ok, the problem is this: I can solve this problem one way, but the way the teacher explains in class (the bounds in particular) is confussing.

Here is my solution: Since you are only going to pull 10 ft of the rope up, the 30 odd feet below this 10 feet will remain a constant, therefore, no integral required for this part.
30ft*3/2 lb/ft*10 ft=450 ft*lbs

So lets now deal with the ten feet you will have to pull up, first divide the 10 ft. into sub intervals dy thick, and lets say that at any given point y on the rope (the top of the building being at height 0), we have y ft to pull up till that segment reaches the top of the building. Therefore, the work to pull one of these slices becomes:
W:3/2 ft/lb * dy * y

so lets integrate 3y/2, from 0 to 10.
10
$$\int(3y/2dy)$$=75 ft*lbs
0
Now, my teachers solution: she says forget the first part of my solution and deal with the problem in one swoop by integrating:

40
$$\int(3y/2dy)$$=525ft*lbs
30

My question: WHERE did she get these bounds? and why?

Last edited: Feb 12, 2008
2. Feb 12, 2008

### NateTG

I would guess those come from the amount of rope that's hanging. (Just like you're going from 0 to 10, she's going from 30 to 40.)