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Purpose of computing work

  1. Oct 7, 2004 #1
    Can someone tell me what im doing wrong? "A glass is a cylinder, 15cm tall and with a radius of 3cm. It is filled 12.5cm high with water d=1000kg/m^3.
    A straw is placed in the glass, with the top of the straw 5cm above the top of the glass. How much work does it take to drink through the straw? Note..for the purpose of computing work, it is safe to assume that you are always drinking from the top most "layer" of water, no matter where the straw is placed."

    I started using a system that the bottom of the cup is 0 height..and the top of the cup 15....I think some people do it the other way around :confused:
    By the way I converted from cm to m randomly in the problem whenever the units came up in the problem.
    This is my work...




    using the equations form above [tex]m=.9\pi,kg/m[/tex]





    so far so good?

    after integrating...i said that it takes [tex]2182.1J [/tex] to do the work im assuming thats wrong...thats alot of Joules just to drink some water ay?
    Last edited: Oct 7, 2004
  2. jcsd
  3. Oct 7, 2004 #2


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    Imagine a thin (thickness dx) "layer" of water a height x (meters) above the base. That layer is a disk of radius 0.03 m in radius so its volume is 0.0009πdx cubic meters and so its mass is .9 πdx kg and its weight is 8.829 π dx Newtons.
    That appears to be what you have but then you say
    Where is that "+ 05" from?

    That layer has to be lifted a distance .15- x meters which requires work of
    8.82π(0.15- x) dx Joules. The total is the integral of that as x goes from 0 to 0.15.
  4. Oct 7, 2004 #3

    Doc Al

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    I have a hard time understanding what you did. Try it this way. First define a slice of water of thickness "dy". Its mass is [itex]dm = \rho \pi r^2 dy[/itex]. The amount of work you need to do to lift each slice from its position y to the final height of h = .2 m is [itex]dW = g(h - y)dm = \rho \pi r^2 g(h - y)dy[/itex]. Integrate this from y = 0, 0.125m.
  5. Oct 7, 2004 #4
    Okay let me try again from mass, [tex]m=.9h,kg\pi[/tex] and then i plug

    that into [tex]\vec{F}=m\vec{a}[/tex] and [tex]\vec{a}=9.8m/s^2[/tex] get [tex]\vec{F}=8.82h,N\pi[/tex]

    so then I plug that into [tex]\vec{W}=\vec{F}d[/tex] which would give me

    [tex]\vec{W}=(8.82h)(.20-h)J\pi[/tex]..... I used this [tex].20-h[/tex] because

    the drink is going to be moving through the straw which is .05m above the top

    of the glass(top of glass=.15m) and then to find the work I

    [tex]\pi\int_{0}^{.125}(8.82h)(.20-h)dh[/tex] and this time I got

    [tex].0252554597J[/tex].....how does that look I think the only place we got

    different results is the the height. So is this correct? And would

    [tex].05+x[/tex] be okay?
    Last edited: Oct 7, 2004
  6. Oct 7, 2004 #5

    Doc Al

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    Again, it is difficult to follow what you are doing. It seems you are using "h" both as a constant (the height of the fluid?) and as your variable of integration. See my previous post and see if you can follow what I did.

    Also, most of the time it's easier if you wait until the last minute before plugging in numbers and grinding out the arithmetic. That also minimizes errors.
  7. Oct 7, 2004 #6




    And then integrate away? Is that correct? I got [tex].476J[/tex]
  8. Oct 8, 2004 #7

    Doc Al

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    Staff: Mentor

    Looks right to me.
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