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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 \[42\div20=2,1 lb per second\]
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?
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.
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.
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?
uhh... msayer, you're right.
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 \[m(t) = 44-(0.2 \times t)\]
just multiply m(t) x 80 ft (depth of well) = work?
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.
even though initially the rate of speed was in ft/s already?
if the lb is in lbf, the units work out since it would be force x distance
Would I have to factor in the force of gravity as well?
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
ah so the solution should be as simple as 36lb x 80 ft? (Subtracting m(t) from 44 for weight)
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.
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
b=initial time of 40 sec