Answer:
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
![=\frac{40-35}{50-0}](https://tex.z-dn.net/?f=%3D%5Cfrac%7B40-35%7D%7B50-0%7D)
![=\frac{1}{10}](https://tex.z-dn.net/?f=%3D%5Cfrac%7B1%7D%7B10%7D)
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
![=\int _0^{50}(40-\frac1{10}y)dy](https://tex.z-dn.net/?f=%3D%5Cint%20_0%5E%7B50%7D%2840-%5Cfrac1%7B10%7Dy%29dy)
![=[40y- \frac1{10}y^2]_0^{50}](https://tex.z-dn.net/?f=%3D%5B40y-%20%5Cfrac1%7B10%7Dy%5E2%5D_0%5E%7B50%7D)
![=[(40\times 50)-\frac1{10}(50)^2]-[(40\times 0)-\frac1{10}(0)^2]](https://tex.z-dn.net/?f=%3D%5B%2840%5Ctimes%2050%29-%5Cfrac1%7B10%7D%2850%29%5E2%5D-%5B%2840%5Ctimes%200%29-%5Cfrac1%7B10%7D%280%29%5E2%5D)
=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