A bucket that weighs 6 lb and a rope of negligible weight are used to draw water from a well that is 50 ft deep. The bucket is filled with 40 lb of water and is pulled up at a rate of 2 ft/s, but water leaks out of a hole in the bucket at a rate of 0.2 lb/s. Find the work done in pulling the bucket to the top of the well.

Therefore total work done =2050 ft-lb

Step-by-step explanation:

Given that, a bucket are used to drawn water from a well that is 50 ft deep.

(1)

Work done = Force×displacement.

Work done to pull the bucket is

= Weight of the bucket × displacement

=(6×50) ft-lb

=300 ft-lb

(2)

The bucket fill with 40 lb of water and is pulled up at a rate of 2ft/s, but leaks out of a hole in the bucket at a rate of 0.2 lb/s.

Since the well is 50 ft deep.

It takes =25 s to pull the bucket at the top of the well.

After 25 s it lost (25×0.2)lb =5 lb water.

So the remaining water is = (40-5) lb= 35 lb

Let y is distance above the original depth of 50 ft.

When y=50 ft, weight of the bucket= 40 lb,when y=0 weight of the bucket= 35 lb

The slope of the of water leakage is

The bucket holds lb of water when it is y ft above the original depth.

Work done to pull the water in the bucket is

=1750 ft-lb

Therefore total work done

=Work done to pull the bucket+Work done to pull the water in the bucket

=(300+1750) ft-lb

=2050 ft-lb

correct answers below:

no

step-by-step explanation: