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{SECT 0 {EXCHG {PARA 0 "" 0 "" {TEXT -1 29 "Rebekah Boyd, Melinda Chau
vin" }}{PARA 0 "" 0 "" {TEXT -1 19 "Page 97, problem 18" }}{PARA 0 "" 
0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 776 "When an object move
s on a surface, it encounters a resistance force of friction.  This fo
rce has a magnitude of uN where u is the coefficient of friction and N
 is the magnitude of the normal force that the surface applies to the \+
object.  Friction acts in opposition to motion, but cannot initiate mo
tion.  Suppose that an object of mass 30 kg is released from the top o
f an inclined plane that iss inclined 30 degrees to the horizontal. (S
ee picture in book on page 98).  Assume that the gravitational force i
s constant, air resistance is negligible, and the coefficient of frict
ion u=0.2.  Determine the equation of motion for the object as it slid
es down the plane.  If the top surface of the plane is 5 m long, what \+
is the velocity of the object when it reaches the bottom?" }}{PARA 0 "
> " 0 "" {MPLTEXT 1 0 8 "restart:" }}}{EXCHG {PARA 0 "> " 0 "" 
{MPLTEXT 1 0 14 "with(student):" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 
473 "Since we are starting at the top of incline plane, we are having \+
to consider the forces that flow with the object as it slides down the
 plane and the forces that act on the object.  Thus we have the normal
 force of N1+u*N2=m*dv/dt.  N1 is represented by m*g*sin(Pi/6), where \+
m = mass, g = gravity , and Pi/6 is 30 degrees, and N2 is represented \+
by m*g*cos(Pi.6).  As you can see the m's cancel and we are left with \+
the following equation after putting 0.2=u, g = 9.81m/s^2." }}{PARA 0 
"> " 0 "" {MPLTEXT 1 0 51 "V:=(v,t)->(9.81*(sin(Pi/6))-(.2*9.81*(cos(P
i/6))));" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%\"VGR6$%\"vG%\"tG6\"6$%)
operatorG%&arrowGF),&-%$sinG6#,$%#PiG#\"\"\"\"\"'$\"$\")*!\"#-%$cosGF0
$!%i>!\"$F)F)F)" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 284 "In this case,
 V represents the statement that velocity equals the sum of the forces
 on an object.  Since the plane is inclined, the usual gravitational f
orce is altered by the degree of incline.  Since all of the above are \+
known, we found the numeric answer to simplify for use later." }}
{PARA 0 "> " 0 "" {MPLTEXT 1 0 49 "simplify(9.81*(sin(Pi/6))-(.2*9.81*
(cos(Pi/6))));" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#$\"+d\"ee?$!\"*" }}}
{EXCHG {PARA 0 "" 0 "" {TEXT -1 46 "(expression evaluated to get a num
eric answer)" }}{PARA 0 "> " 0 "" {MPLTEXT 1 0 15 "diff(v(t),t)=%;" }}
{PARA 11 "" 1 "" {XPPMATH 20 "6#/-%%diffG6$-%\"vG6#%\"tGF*$\"+d\"ee?$!
\"*" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 116 "Using dsolve, we find the
 equation of velocity.  This represents the velocity of the object as \+
is follows the plane." }}{PARA 0 "> " 0 "" {MPLTEXT 1 0 15 "A:=(dsolve
(%));" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%\"AG/-%\"vG6#%\"tG,&F)$\"+d
\"ee?$!\"*%$_C1G\"\"\"" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 18 "F
:=eval(A, _C1=0);" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%\"FG/-%\"vG6#%
\"tG,$F)$\"+d\"ee?$!\"*" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 261 "C1 is
 set to 0 because the object has no initial velocity; therefore the on
ly forces acting on the object at this time will be gravity (in this c
ase, \"pulling\" the block down the incline) and the opposing force of
 friction.  Thus at t = 0, all forces are equal." }}{PARA 0 "> " 0 "" 
{MPLTEXT 1 0 33 "X:=dsolve((diff(x(t),t)=rhs(%)));" }}{PARA 11 "" 1 "
" {XPPMATH 20 "6#>%\"XG/-%\"xG6#%\"tG,&*$)F)\"\"#\"\"\"$\"+z!HHg\"!\"*
%$_C1G\"\"\"" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 214 "X represents the
 distance the object travels down the incline in terms of time squared
.  This number is found by integrating the velocity function.  In the \+
case of our plane, the distance is known to be 5 meters.  " }}{PARA 0 
"> " 0 "" {MPLTEXT 1 0 15 "eval(X, _C1=0);" }}{PARA 11 "" 1 "" 
{XPPMATH 20 "6#/-%\"xG6#%\"tG,$*$)F'\"\"#\"\"\"$\"+z!HHg\"!\"*" }}}
{EXCHG {PARA 0 "" 0 "" {TEXT -1 114 "Again, the constant is set to 0 b
ecause the object has at t = 0, the distance the object has traveled i
s 0 meters." }}{PARA 0 "> " 0 "" {MPLTEXT 1 0 15 "eval(%,x(t)=5);" }}
{PARA 11 "" 1 "" {XPPMATH 20 "6#/\"\"&,$*$)%\"tG\"\"#\"\"\"$\"+z!HHg\"
!\"*" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 12 "solve(%, t);" }}
{PARA 11 "" 1 "" {XPPMATH 20 "6$$!+o5:m<!\"*$\"+o5:m<F%" }}}{EXCHG 
{PARA 0 "" 0 "" {TEXT -1 123 "Since time cannot be negative in this ca
se, it will take 1.766 seconds for the block to slide to the bottom of
 the incline." }}{PARA 0 "" 0 "" {TEXT -1 129 "Velocity was evaluated \+
(above) as 3.206 meters per second.  Multiplying that velocity by the \+
time it takes the block to reach the" }}{PARA 0 "" 0 "" {TEXT -1 72 "b
ase of the incline gives an ending velocity of 5.662 meters per second
." }}{PARA 0 "> " 0 "" {MPLTEXT 1 0 22 "eval(F,t=1.766151068);" }}
{PARA 11 "" 1 "" {XPPMATH 20 "6#/-%\"vG6#$\"+o5:m<!\"*$\"+3)H?m&F)" }}
}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}}}{MARK "11 0 0" 114 }
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