A bucket that weighs 4 lb is lifted vertically at a constant rate of 2 ft/sec by means of a rope of negligible weight. As the bucket rises, water leaks out at a

Question

A bucket that weighs 4 lb is lifted vertically at a constant rate of 2 ft/sec by means of
a rope of negligible weight. As the bucket rises, water leaks out at a rate of 0.2 lb/sec. If the
bucket contained 40 lb of water at the instant the lifting began, determine the work done in raising
the bucket 80 feet.

Answer 1

We compute work as the integral of a force over a distance:

 

w=∫_x1^x2[f(x) dx]. 

 

For this problem we are interested in the distance the bucket is moved, so the integral will be from 0 to 80.

 

The force we're interested in is the leaking bucket's weight. We're going to try to express the weight as a function of position (x) so we can integrate it. 

 

The bucket leaks at 0.2 lb/sec and is lifted at 2 ft/sec. That means we could also say the bucket leaks at (0.2 lb/sec)/(2 ft/sec) = 0.1 lb/ft. We went from a time-based rate of leakage to a position-based rate using the bucket's velocity. **Make sure you understand how the units of these quantities cancel out.

 

If the bucket leaks at 0.1 lb/ft and moves a height X ft, that means it lost (0.1 lb/ft)*(X ft)=0.1X lbs of water. So at any given point, the bucket should weigh

 

f(x)=4+(40-0.1x) lbs = 44-0.1x

 

Integrating from x=0 to x=80 yields the work in units of ft*lbs.