Work Needed to Lift a Leaky BucketDate: 08/01/99 at 17:03:03 From: Allison Subject: Calculus, physics applications Here's the problem: A cylindrical bucket has height of 1 ft and a radius of .5 ft, and weighs 4 pounds. It is attached to a rope of negligible weight. It is thrown down a well 100 feet deep and filled up with water to the top. There is a hole in the bucket, causing water to leak out at a rate that is proportional to the height of the water in the bucket. The bucket is being pulled up at a rate of 2ft/sec, and the height of the water in the bucket is .9 feet when the bucket reaches the top of the well. How much work is being done in lifting the bucket and the water to the top of the well? Assume the density of the water is 62 pounds per cubic ft. Here is what I have done: The bucket holds pi/4 ft^3 of water when full rho = density of water in lbs/ft^3 density = 16/pi It takes 50 sec to haul up the bucket, and the rate of leakage is proportional to the height of the water dh/dt = -kh dh/h = -kdt Separate the variables: h(t) = ce^-kt when t = 0 h(t) = 1, thus c = 1 when t = 50 h(t) = .9, so k = .002107 x(t) = height of the bucket above the bottom of the well at time t. If the bucket is brought up at a rate of 2ft/sec then x(t) = 2t. Integrate weight: 4 + (16/pi * e^(-kx/2))dx from 0 to 100. My final answer is 883.38. Is this correct? Are my units right? I have done a lot of work, please help me, and if I am wrong tell me how and where to correct it. Thanks, Allison Date: 08/09/99 at 17:02:02 From: Doctor Fwg Subject: Re: Calculus, physics applications Dear Allison, The first part of your calculation is okay, but I probably would have written H (as a function of t or x) like this: 1) H = Ho Exp[-kt] = Ho Exp[-kx/v], where H is a function of t or x (where t = x/v), v is the lifting velocity, and Ho is the water level when bucket is full. In your work above you let t = x/2 which is fine also as long as x is measured in feet. For an incremental amount of work (i.e., dW) to lift the bucket through a distance of dx, you can write: 2) dW = F dx = M g dx, where F (note: F = M g) is the incremental force exerted to lift the pail plus water through a small distance of "dx," M is the mass of the pail plus water as a function of time, and g is the gravitational constant (about 32 ft/sec^2 or 980 cm/sec^2). But, noting that the mass of the pail alone is Mp (or 4.0 lb = 1818 g), and using Eq. 1, M can be rewritten as: 3) M = Mp + Rho A H = Mp + Rho A {Ho Exp[-kx/v]} = Mp + Rho A Ho Exp[-kx/v], where Rho is the water's density (which is, predominantly, a function of the water's temperature, but here can probably be assumed to be a constant equal to about 62 lb/ft^3 or about 1.0 g/cm^3), A is the cross sectional area of the water in the bucket (note: A = 3.14159 R^2), H is the height of the water in the bucket at time "t" (about 1.0 ft or about 30.5 cm when t = 0.0 sec), k is a constant, and x/v (equal to t) is time. Note: (A H) is the volume of water in the pail at time t or height x, and (Rho A H) means density times volume which gives the mass of water in the pail at that height or time. Now, plugging M in Eq. 3, into Eq. 2: 4) dW = F dx = M g dx = {Mp + Rho A Ho Exp[-kx/v]} g dx. Eq. 4 can be integrated with W varying from 0 to W and x varying from 0 to x. Because v is constant, the result is: 5) W = Mp g {x - 0} - (v/k)(Rho A Ho g){Exp[-kx/v] - Exp[0]}, or: 6) W = Mp g x - (v/k)(Rho A Ho g){Exp[-kx/v] - 1.0}. You need to be careful about the units here. It might be best to convert to all cgs units. Namely: grams for mass, g = 980 cm/sec^2, velocity in cm/sec, k in sec^(-1), Rho in gm/cm^3, and x (and Ho) in cm. If you use cgs units, the units of work will be in ergs, where: 7) 1.0 erg = 1.0 g cm^2/sec^2. The total work required will be a little less than the work needed to lift the mass of the bucket and the pail full of water because only a little water leaks out. In other words, the second term in Eq. 6 (i.e. -(v/k)(Rho A Ho g){Exp[-kx/v] - 1.0} must be positive and only a little smaller than the term: 8) Ww = Mw g x, where, Ww is the work needed to lift only the water, Mw is the mass of only the water in the full bucket, g is the acceleration due to gravity, and x is height the water is lifted through. Note: The term Exp[-kx/v] will be less than one (but not negative) for all positive values of x. Note also that I am using the convention in this problem that x is zero at the bottom of the well and increases in the "plus" direction as the pail is lifted. There is at least one other way to solve this problem using t as the integration variable, but the way illustrated above may seem more straightforward to you. I hope so and I hope this illustration has answered your question. With Best Wishes, - Doctor Fwg, The Math Forum http://mathforum.org/dr.math/ |
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