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anonymous
 5 years ago
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.
anonymous
 5 years ago
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.

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anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0The 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 \[42\div20=2,1 lb per second\]

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0wouldn't work be measured in ftlbs? 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?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0both, 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.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0I 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.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0wouldn'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?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0uhh... msayer, you're right.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0also 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 \[m(t) = 44(0.2 \times t)\]

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0just multiply m(t) x 80 ft (depth of well) = work?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0not 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.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0even though initially the rate of speed was in ft/s already?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0if the lb is in lbf, the units work out since it would be force x distance

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Would I have to factor in the force of gravity as well?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0making 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

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0ah so the solution should be as simple as 36lb x 80 ft? (Subtracting m(t) from 44 for weight)

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0not 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.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0oh yeah.. duh. The Equation module is out of view as far as correct symbols so, Would the integral look like (040) (44.2x)dx

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0b=initial time of 40 sec
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