We will compute the work required to pull up both the cable and the crate separately.

1. Work Done in Pulling Up the Cable

Since the cable has a uniform weight density, we use integration to sum up the work required to lift each small segment.

• Length of cable: 40 ft
• Total weight of cable: 120 lbs
• Weight density:
  $\frac{120 \text{ lbs}}{40 \text{ ft}} = 3 \text{ lbs/ft}$

Consider a small segment of cable at height $x$ above the bottom. This segment has weight $3 \, dx$ and must be lifted a distance $x$. The total work is:

$W_{\text{cable}} = \int_0^{40} 3x \, dx$

Computing the integral:

$W_{\text{cable}} = 3 \int_0^{40} x \, dx = 3 \left[ \frac{x^2}{2} \right]_0^{40}$

$= 3 \times \frac{40^2}{2} = 3 \times \frac{1600}{2} = 3 \times 800 = 2400 \text{ ft-lbs}$

2. Work Done in Pulling Up the Crate

The crate starts at 200 lbs and ends at 100 lbs, meaning its weight decreases linearly.

• Weight function: Since weight decreases uniformly, we define the weight function as:

  $W(x) = 200 - \frac{100}{40}x = 200 - 2.5x$

Each part of the crate must be lifted a distance $40$, so the total work is:

$W_{\text{crate}} = \int_0^{40} (200 - 2.5x) \, dx$

Computing the integral:

$W_{\text{crate}} = \int_0^{40} 200 \, dx - \int_0^{40} 2.5x \, dx$

[ = \left[ 200x \right]_0^{40} - 2.5 \left[ \frac{x^2}{2} \right]_0^{40} ]

$= (200 \times 40) - 2.5 \times \frac{40^2}{2}$

$= 8000 - 2.5 \times \frac{1600}{2}$

$= 8000 - 2.5 \times 800$

$= 8000 - 2000 = 6000 \text{ ft-lbs}$

Final Answers

1. Work done in pulling up the cable: 2400 ft-lbs
2. Work done in pulling up the crate: 6000 ft-lbs

Would you like me to explain any part in more detail?

Related Questions

1. How would the work change if the cable’s weight density were different?
2. What if the crate’s weight decreased at a non-uniform rate instead?
3. How would the work change if the cable were not uniform but had varying density?
4. What if the crate were lifted only halfway instead of fully?
5. Can we use a different method (like energy principles) to solve this problem?

Tip:

When dealing with varying weight distributions, always define a weight function $W(x)$ before integrating to ensure proper calculations!