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Math Help - Work problem

  1. #1
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    Work problem

    A bucket that weighs 3kg and a rope of negligible weight are used to draw water from a well that is 100m deep. The bucket is filled with 40 liters of water and is pulled up at a rate of 2m/s, but water leaks out of a hole in the bucket at a rate of 0.2 m/s (I think the question is supposed to say 0.2 liters/s ?). Find the work done in pulling the bucket to the top of the well.

    I came across a more basic physics type problem recently and it made me interested so I have been looking into it as a side project lately. I came across this random problem which I don't have a solution for and gave it a go. Sorry if I am missing gaps in my knowledge, I just started with them

    My attempt:

    I assume that 40 liters means 40kg of weight/force.

    Weight K of the bucket at any given time = 3+(40-0.2x)

    Work \Delta W = x\times K = 3x+40x-0.2x^2 \Deltax

    W = \int^{50}_{0} 43x-0.2x^2~dx

    = 45416.67

    Am I way off?

    Any help/direction is much appreciated.
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  2. #2
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    Re: Work problem

    Work is distance times force, not time times force. Also, to get work in Joules, you need to multiply the mass in kilograms by g = 9.8 m/s^2.
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  3. #3
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    Re: Work problem

    I was working in meters, not seconds. Every 2 meters, the bucket leaks 0.2 liters, so each slice is 2m. Also that's why I integrated between 0 and 50 not 0 and 100.
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  4. #4
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    Re: Work problem

    Quote Originally Posted by terrorsquid View Post
    I was working in meters, not seconds.
    Hmm.

    Quote Originally Posted by terrorsquid View Post
    Weight K of the bucket at any given time = 3+(40-0.2x)
    I agree with the formula (modulo the factor g) if x denotes time. If x denotes height, then K(x) should be 43 - 0.1x.

    Quote Originally Posted by terrorsquid View Post
    Also that's why I integrated between 0 and 50 not 0 and 100.
    So, you are saying the well depth is 50m?

    Let t denote time and x denote height. Then x(t)=2t. As was said above, K(t)=g(43-0.2t), so K(x)=g(43-0.1x). By definition of work, W=\int_0^{100} xK(x)\,dx=\int_{0}^{100}gx(43-0.1x)\,dx. Since x(0)=0 and x(50)=100, using integration by substitution, this is the same as \int_0^{50}x(t)K(x(t))\frac{dx}{dt}dt=\int_0^{50}4  gt(43-0.2t)\,dt.

    Edit: corrected formula for work.
    Last edited by emakarov; November 22nd 2011 at 01:02 PM.
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  5. #5
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    Re: Work problem

    1 liter of water has a mass of 1 kg

    initial mass = 43 kg

    initial weight = 43g = 421.4 Newtons

    lifting the bucket at a rate of 2 m/s takes 50 seconds. in that time, (50 sec)(0.2 L/sec) = 10 L leak out, ending with a final weight of 33g = 323.4 N

    Work done is the area of the trapezoid in the Force vs. displacement graph shown.
    Attached Thumbnails Attached Thumbnails Work problem-workgraph.jpg  
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  6. #6
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    Re: Work problem

    Quote Originally Posted by terrorsquid View Post

    Weight K of the bucket at any given time = 3+(40-0.2x)
    Yah, sorry maybe I should have said at any point x is a multiple of 2.

    Quote Originally Posted by emakarov View Post
    So, you are saying the well depth is 50m?
    The way I was thinking is that every two meters, the bucket loses some weight, and this happens 50 times along its 100m journey. If you make the formula 0.1 every meter like you have done wont that change the total work? as lifting a weight 2 meters requires more work than lifting the same weight 1 meter then dropping 0.1 and lifting it another meter right?

    To be honest, I wasn't sure why the timing was even given for the question initially.
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    Re: Work problem

    Quote Originally Posted by terrorsquid View Post
    Yah, sorry maybe I should have said at any point x is a multiple of 2.



    The way I was thinking is that every two meters, the bucket loses some weight, and this happens 50 times along its 100m journey. If you make the formula 0.1 every meter like you have done wont that change the total work? as lifting a weight 2 meters requires more work than lifting the same weight 1 meter then dropping 0.1 and lifting it another meter right?

    To be honest, I wasn't sure why the timing was even given for the question initially.
    F = 43g - (0.2g)t

    x = 100 - 2t

    t = \frac{100-x}{2}

    F(x) = 43g - (0.2g) \cdot \frac{100-x}{2} = g \left(33 + \frac{x}{10}\right)

    W = g \int_0^{100} 33 + \frac{x}{10} \, dx = 37240 \, J
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