A 50-lb bucket is at the bottom of a 100-ft well. A 200 lb rope (also 100 ft long) is tied securely to the bucket. We will use rope to lift this bucket out of the wall, at a rate of 1 foot every second.
(a) How much does this rope weigh per foot? How much would a 50 foot length of this rope weigh?
(b) The force that is applied to the system, F(t) at a given time t is equal to the total weight that you are holding in the well: the weight of the rope remaining in the well, plus the weight of the bucket. What is F(0) F(50), F(100)? Find a formula F(t) that describes the force at time t.
(c) Find the total work done to lift the bucket to the top of the well.
I think I know how to do the first part but not sure so I'm just gonna say what I did:
Since the rope is 100 ft long and it weighs 200 lb's, shouldnt its weight per foot just be 200/100 = 2 lb/ft? If thats right then it should be 4 lb/ft for a 50 ft length rope.
For part b im lost and I'm pretty sure I need part b done to be able to do part c.
Help is required,
THanks
You are correct that the rope weighs 2 lb/ft, so 50 ft of rope would weigh 100 lb.
After t seconds, the rope has been lifted t ft, so the length of rope being held will be 100-t ft. Therefore $F(t)=2(100-t)+50$, since the rope weighs 2(100-t) lb and the bucket weighs 50 lb.
To find the total work done, you have to integrate $F(t)$ from $t=0$ to $t=100$.