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# View of /trunk/NationalProblemLibrary/Rochester/setDiffEQ11ModelingWith2ndOrder/ur_de_11_3.pg

Fri Apr 28 13:46:21 2006 UTC (7 years ago) by jjholt
File size: 3457 byte(s)
Added tags.  --JH


    1 ## DESCRIPTION
2 ## Calculus
3 ## ENDDESCRIPTION
4
5 ## KEYWORDS('differential equation' 'second order' 'linear' 'nonhomogeneous')
6 ## Tagged by tda2d
7
8 ## DBsubject('Calculus')
9 ## DBchapter('Second-Order Differential Equations')
10 ## DBsection('Applications of Second-Order Differential Equations')
11 ## Date('')
12 ## Author('')
13 ## Institution('Rochester')
14 ## TitleText1('')
15 ## EditionText1('')
16 ## AuthorText1('')
17 ## Section1('')
18 ## Problem1('')
19
20 DOCUMENT() ;
21
23 "PG.pl",
24 "PGbasicmacros.pl",
25 "PGchoicemacros.pl",
27 "PGauxiliaryFunctions.pl",
28 "PGdiffeqmacros.pl",
29 ) ;
30 #######
31 $ss = random(2,7,1) ; 32$n = random (2,9,1) ;
33
34 #######
35 $ans = ivy (1,32,32*$ss,0,$n) ; 36 37 TEXT(beginproblem()) ; 38$showPartialCorrectAnswers = 1 ;
39
40 BEGIN_TEXT
41
42 A hollow steel ball weighing 4 pounds is suspended from a spring.
43 This stretches the spring $$\frac{1}{ss}$$ feet.
44
45 The ball is started in motion from the
46 equilibrium position
47 with a downward velocity of $$n$$ feet per second.
48 The air resistance (in pounds) of the  moving ball
49 numerically equals 4 times its
50 velocity (in feet per second) .
51 $PAR 52 Suppose that after t seconds the ball is y feet below its rest position. 53 Find y in terms of t. (Note that the positive direction is down.) 54 55$PAR
56 Take as the gravitational acceleration 32 feet
57 per second per second.
58 $BR$BR
59 $$y=$$ \{ans_box(2,80)\}
60
61 END_TEXT
62
63 HINT(EV3(<<'EOF'));
64 When using English units (lb, ft, etc.) you need to be a bit careful with equations involving
65 mass. This is causing some confusion --- I hope this hint helps:
66 $PAR 67 Pounds (lb) is a unit of force, not mass. Using mg=F and g=32 ft/sec^2 we see that 68 an object at the surface of the earth which weighs 32 lbs (i.e. the force on it is 32 lbs) will 69 have a mass of 1 (the unit of mass in the English units is called the slug -- really!) So one 70 slug weighs 32 lbs at the surface of the earth ( or lb = (1/32)*slug*ft/sec^2 ). 71$PAR
72 When using metric units, kilogram is a unit of mass not force or weight.  A 1 kilogram mass will weigh
73 1 * 9.8  newtons on the surface of the earth.  (g= 9.8 m/sec^2 and newton = kg*m/sec^2 ).
74 $PAR 75 Saying that a mass "weighs" 1 kilogram is technically incorrect useage, but it is often used. 76 What one really means is that it has 1 kilogram of mass and therefore weighs 9.8 newtons. 77$PAR
78 EOF
79
80 ANS(fun_cmp($ans , var=>"t",reltol=>1 )) ; 81 82 ENDDOCUMENT() ; 83 84 85 86 87 88 89 90 ################################################## 91 my$XML_INFORMATION = <<'END_OF_XML_TRAILER_INFO';
92 <?xml version="1.0"?>
93 <metaPGdata>
94         <author>David Prill</author>
95         <course>MTH163</course>
96         <description>spring problem, underdamping</description>
97         <fullPath>setDESOLinear/15.pg</fullPath>
98         <institution>University of Rochester</institution>
99         <keywords>spring problem, underdamping</keywords>
100         <libraryPath>setDESOLinear/15.pg</libraryPath>
102         <modified><dateTime.iso8601>20000718T13:05:40</dateTime.iso8601></modified>
103         <msgNum>404</msgNum>
104         <pgProblem>true</pgProblem>
105         <preface></preface>
106         <problemVariants></problemVariants>
107         <probNum></probNum>
108         <psvn></psvn>
109         <revisedVersions></revisedVersions>
110         <setName>DESOLinear</setName>
111         <titleRoot>15</titleRoot>
112         </metaPGdata>
113
114 END_OF_XML_TRAILER_INFO
115 ##################################################
116