Question

A bucket that weighs 4 lb and a rope of negligible weight are used to draw water from a well that is 80 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.

Answer 1

The bucket is outside again in 20s, at a cost of 4 lb water. Which is an average amount of water of 38 lb water. \[38+4= 42\] lb. Work in this case is \[42\div20=2,1 lb per second\]

Answer 2

wouldn't work be measured in ft-lbs? What I'm looking at defines Work as force x distance. And force= mass x acceleration. In this case, is mass=weight? Or weight=force?

Answer 3

both, you see.

You need to understand that mass=/= force, which implies that the power you need is not equal to the weight of the bucket+water.

Answer 4

I understand that but I'm just trying to look at this with the factor of speed. How exactly is speed related to work? I'm still confused sorry.

Answer 5

wouldn't it take 40 seconds for the bucket to be outside again? considering the bucket is moving at a constant rate of 2ft/s and well is 80ft deep?

Answer 6

uhh... msayer, you're right.

Answer 7

also the mass/weight would be a function of time since it starts with 44lbs initial and loses 0.2lb/s therefore, mass/weight would be
\[m(t) = 44-(0.2 \times t)\]

Answer 8

I'm following...

Answer 9

just multiply m(t) x 80 ft (depth of well) = work?

Answer 10

not quite. First the units wouldn't work out. If we "assume" that the lb is in lbm rather than lbf, we would have to convert that.

Answer 11

even though initially the rate of speed was in ft/s already?

Answer 12

if the lb is in lbf, the units work out since it would be force x distance

Answer 13

Would I have to factor in the force of gravity as well?

Answer 14

making an assumption that the 44lb of water and bucket are weight rather than mass, you wouldn't have to add gravity to your calculation - this would be saying they are in units of lbf rather than lbm

Answer 15

ah so the solution should be as simple as 36lb x 80 ft? (Subtracting m(t) from 44 for weight)

Answer 16

not quite - what you stated indicates that you are pulling 36lb weight up 80ft. In this problem, we are pulling an initial weight of 44lbs which becomes 36lbs at the end. This means that you will have to integrate.

Answer 17

oh yeah.. duh. The Equation module is out of view as far as correct symbols so, Would the integral look like (0-40) (44-.2x)dx

Answer 18

b=initial time of 40 sec