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1 : sam 2 #DESCRIPTION
2 :     # Integration
3 :     # Application-based
4 :     # Physics--Work.
5 :     #ENDDESCRIPTION
6 :    
7 :     #KEYWORDS('Integration', 'Physics', 'Applications')
8 :    
9 :     DOCUMENT(); # This should be the first executable line in the problem.
10 :    
11 :     loadMacros(
12 :     "PG.pl",
13 :     "PGbasicmacros.pl",
14 :     "PGchoicemacros.pl",
15 :     "PGanswermacros.pl",
16 :     "PGauxiliaryFunctions.pl"
17 :     );
18 :    
19 :     TEXT(&beginproblem);
20 :     $showPartialCorrectAnswers = 1;
21 :    
22 :     #Here we ensure that the height of the pool is always greater than the depth of
23 :     #the water
24 :     $c1 = random(5,12,.5);
25 :     $c2 = random(1,11,.5);
26 :     @cs =($c1,$c2);
27 :     @sortedcs = num_sort(@cs);
28 :     $d = $sortedcs[0];
29 :     $h = $sortedcs[1];
30 :     $r = random(8,20,.5) ;
31 :     $w = random(63,66,.1);
32 :     $pi = 4*arctan(1);
33 :    
34 :     BEGIN_TEXT
35 :     You are visiting your friend Fabio's house. You find that, as a joke, he filled
36 :     his swimming pool with Kool-Aid, which dissolved perfectly into the water.
37 :     However, now that you want to swim, you must remove all of the Kool-Aid
38 :     contaminated water. The swimming pool is round, with a $r foot radius. It is $h
39 :     feet tall and has $d feet of water in it. $BR
40 :     How much work is required to remove all of the water by pumping it over the
41 :     side?
42 :     Use the physical definition of work, and the fact that the weight of the
43 :     Kool-Aid contaminated water is \( \sigma = $w lbs/ft^3 \) $BR
44 :     \{ans_rule(45)\}
45 :     END_TEXT
46 :    
47 :     &HINT(EV3(<<'EOT'));
48 :     $HINT $BR
49 :     The formula for work is: $BR
50 :    
51 :     \[\int_{a}^{b} Force * distance \] $BR
52 :    
53 :     Where distance is the distance over which the force is exerted.
54 :     EOT
55 :    
56 :     &SOLUTION(EV3(<<'EOF'));
57 :     $SOL $BR
58 :     Consider a horizontal cross-section of the pool, with thickness \(dx\) if we
59 :     consider the x-axis to be vertical, in the center of the pool. This is
60 :     just a very short cylinder, so its volume is: $BR
61 :    
62 :     \( dV = \pi r^2 dx \). $BR
63 :    
64 :     We know r, the radius of the pool, is a constant, \(r= $r\). Now that we have
65 :     the volume of an arbitrary cross-section of the water, we need to find the
66 :     force which is exerted on the volume. That force is nothing more than the
67 :     weight. The constant \( \sigma \) gives us weight-per-volume of the liquid.
68 :     Therefore, by multipling the volume of the slice by \( \sigma \), we find: $BR
69 :    
70 :     \( dF = \sigma \pi r^2 dx \) $BR
71 :    
72 :     Since Work (W) is given by: $BR
73 :    
74 :     \( W = Fx = F\int_{a}^{b}xdx \) $BR
75 :    
76 :     in the case of a constant force F, all that remains is to find an expression
77 :     for \( D\), the distance each slice of water is lifted. If we consider the top
78 :     of the pool as x=0, $h-$d is the distance to the surface of the water, since the
79 :     height of the pool is $h, and the depth is $d. So we have the distance
80 :     x from x=$h-$d until x=$h. This results in the integral: $BR
81 :    
82 :     \[ W = \int_{$h-$d}^{$h} \sigma \pi r^2 xdx \] $BR
83 :    
84 :     which is simple to evaluate.
85 :     EOF
86 :    
87 :     $answer = (.5*$w*$pi*$r**2)*(2*$d*$h - $d**2) ;
88 :     &ANS(std_num_cmp($answer));
89 :     ENDDOCUMENT(); # This should be the last executable line in the problem.

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