Final Exam
Richland College
1. In the picture θ=25.0°, mass of the object is 7 kg and coefficient of kinetic friction is
0.19. Find the acceleration of the object when moves downward. (15 points)
2. Find the resultant of the three vectors in the picture. (20 points)
3. A student applied 6.0 J work to stretch a spring 2 cm from its relaxed position. How
much work her friend should apply to stretch the spring 4 cm in addition. (10 points)
4. James had an accident when driving at the speed of 95 km/h. The car travels 0.8 m after
hitting an object and before coming to a complete stop. Find the average acceleration of
the car. (15 points)
5. In the picture, the mass of dart is 0.100 kg and the constant of the spring is 250 N/m
(spring has negligible mass). The spring is compressed 6cm (First picture). What would
be the velocity of the dart in the second picture when it leaves the gun while the spring
is at equilibrium. (15 points)
6. A 10 kg box rests on a horizontal floor. Find the force of friction when the applied force
is 20 N, 38 N, 40 N. The coefficient of static friction is 0.4 and the coefficient of kinetic
friction is 0.3
(15 points)
7. Find the displacement from 1.5 seconds to 3.1 seconds if v(t)=25+18t (v is in m/s and t is
in seconds) (10 points)
Final Exam
Richland College
1. In the picture θ=25.0°, mass of the object is 7 kg and coefficient of kinetic friction is
0.19. Find the acceleration of the object when moves downward. (15 points)
2. Find the resultant of the three vectors in the picture. (20 points)
3. A student applied 6.0 J work to stretch a spring 2 cm from its relaxed position. How
much work her friend should apply to stretch the spring 4 cm in addition. (10 points)
4. James had an accident when driving at the speed of 95 km/h. The car travels 0.8 m after
hitting an object and before coming to a complete stop. Find the average acceleration of
the car. (15 points)
5. In the picture, the mass of dart is 0.100 kg and the constant of the spring is 250 N/m
(spring has negligible mass). The spring is compressed 6cm (First picture). What would
be the velocity of the dart in the second picture when it leaves the gun while the spring
is at equilibrium. (15 points)
6. A 10 kg box rests on a horizontal floor. Find the force of friction when the applied force
is 20 N, 38 N, 40 N. The coefficient of static friction is 0.4 and the coefficient of kinetic
friction is 0.3
(15 points)
7. Find the displacement from 1.5 seconds to 3.1 seconds if v(t)=25+18t (v is in m/s and t is
in seconds) (10 points)
Formula Sheet
.

Final
Onedimensionalkinematics
r
displacemqfi: Ax = xI  xi
.
averagesp"*: totaldistancetraveled
total time
instantanous speed =fnf (mafritude of the instantaneous velocity)
xf  xi
averagevelocity: u* =+ =
o
.
Lt tI t,
o
instantaneous
.
av€rage acceleration:
o
instantaneous accel leration: o
o
velocity
,= li:or*=;Trf
aou:
vi
at =vr
tfti
L!
= ,yoo*=ITr#
Onedimmsional motion with constant acceleration (4 facts):
(l)
v
=vo+ at
(2) x=xo+)00+u)t
(3) x=xo+vst+Lat2
(4)
o
Free
'r
,2 =ri,
*Zo(*ro)
fall (positive direction fory taken tobe upword)
x
1!
and a
r g
in the above 4 facts:
(1) v=vogt
(z) y=yo+)Qo+u)t
(3) y= yo+vot)stz
(4) u'=u3zsbY)
r
Vectors
The resultant vector fot several vectors is the vector sum. For examplg if you have three displacements
the resultant displacerirent ,E is given by:
, and
ir, i,
ir,
fr.=ir+ir+ir.
Ifa vector 7
o
is unitten in component"for as tr =
of
Getting magnitude and directim
ll=,[e?T]
.(,c \
0 tantl
=
(magnitudeof 7 )
? t (aire"tior of i
\e,
;
Ari+
Ari that:
7 from the components:
\
 Final
Formula Sheet
r
'
Page 2
Getting components from magrritude and direction:
4 =lllcosl
(xcomponent
e, =lilsine
@ component
of 7 )
\ d understood to be the angte that
of 7 ) I tr mateswith the positivex axis
Projectile Motion
e
.r direction (motion with constant velocity):
a ar=0
t
o
o
v, =vr!
X=XO*VxOt
y direction (free fall
(l) vy = vyo  gt
... positive
direction fory taken tobe upward):
*bro *rr)t
(2)
y = yo
(3)
y = yo +vrot )&2
$) "?,=ujozsj n)
.
Newton's Laws of Motion
o
Twokindsofforces: contaclforces (objects incontactwithoneanother)andfieldforces (objects
not in contact with one another). Gravity is the only field force we will deal with in this course.
.
t F' = zi (Newton's 2od law)
l/
ximplies two statementr,
)4
=
ma, atdZr,
=
*,
o Weighf W =mg
o Normal forces: in general, whenever two surfaces
are in contact, each strface exerts a force on
the other in a direction that is perpendicular to the two surfaces.
.
Friction forces
o
.
Twokinds: theforceof staAcfriction, f, andtheforceofkineticfriction, fr:
. f,3lt,N (force of static friction)
. .f* = ltrN (force ofkinetic friction)
l, = " the coeffi cient of static fiiction" i p * = " the coeffi cient of kinetic friction"
Strings and Springs
o
o
Strings
the tension in a sfrng olways pulls m the o$ects attached to the ends of the shing
Springs
o
o
o
Hooke's
fa*'
lFl =
rt
= "the spring constant" or "the force constanf'.
Work required to sfietch or compress a spring
W
o
*r.
a
distancex from its equilibrium length:
=!62
2
Potential $ergystored in a spring; U"prns
=l*'
Formula Sheet. Final
Page 3
Translational Equilibrium
. IF=6 + d=d.
o
The statement immediately abovg really implies two sets of statements:
!A
.
=oand)^$
=o =
a,=oa11da,  e.
Circular Motion
o
a* =,2
'r
Centipaalacceleration:
Work and Kinetic Energy
e
o
r
.
Work: W =Fd e,os?
Kinetic energy: X
!*u,
2
Workenergy theorem: Wno = [,y
Work required to stretch or conpress a spring a distance x from its equilibrium leirgth:
w=l k'
2
Potential Energy and Conservative Forces
o
o
Work done by a conservative force: W"  LU , for some potential energy U
Gravitational potential en€rgy (near surface of Earth): (J
= mg
o
o
Potential €n€rgy stored in a spring:
n*
Total mechanical €xrergy: E =
U**, =)U'
K+U
Linear Momentum
o
o
Linearmomeirtum: F=
Newton's second
ni
9p
^r
(Yr=nm€'
\4J
u_>O
N _./
Implies the
following:
(I
\La r)
tqt, 
a'P'
N
(arteragenet force equals averagerate of change of linear momentum).
7=E F)*u
a
lmpulse:
a
Impulsemomentum theorem:
I
= AF
Formula Sheet
.

Final
Collisions
o
o
Two broadcategories: heqdonand glancing
For each category three classes:
l.
2.
3.
o
elastic:
I
inelastic:
andr(conserved
f
conserved,r(not
completely inelastic:
f
conserved, Knot, objects stick together
Headon collisions:
l.
elastic:
' frtvti + m2vzi = mtvrf + m2v2l (pconservatior)
t rti * vry =v2; + v2y (bthed' equation..' derived in class)
2. inelastic:
€vrenl t
. ntvti+m2v2i=mltf +m2vzf (p*rrrouutLi )' ;A"/l
3. completelyinelastic:
. ntvti * m2v2i = (ry + mzb t (pconservation)
o
Glancing collisions:
'
t
.
Page 4
Pxi=P4
Pi=Pvl
RotationalKinematics
o
o
angle
d inradians; 6=! 6=lengthofarcsweptout)
angulardispiacernent
.
ayerageangular
r
L0=0t 0i
velocity:
r* =X =#
o
instantaneousangularvelocity
.
avsrage angular accelerat io,r:
o
instantaneotrs angular acceleration:
*
o
o
d understood
at= tim{
At+o Ll
do,
'
Lal
 @i
Lt @l
tI ti
, = l\#
to be in radiansin all of the above
Period (timerequired for one complete revolution);
,a f
=4
Rotational motion with constant angular acceleration (4 facts):
(l)
@=@o+dt
e)
o=on+f,@o+ait
(3) 0=eo+aht+!ai
(4) at' = ai
+2a(0
0)
L
+
Formula Sheet
.

Final
Page 5
Rotational and Linear Quantities
' lt = rct)
a clr=rd
o Centipetal accelerationt a"o  7412
o igid obiect: Everypoint in the object
has the same angular velocity as ev€ry other point in the
object. Everypoint in the object has the same angular acceleration as every otler point in the
object.
.
Rolling Without Slipping
.
v=ra
Ypoint
,
(v:translational
ofcontac =
0
speed ofcenter
ofrolling object)
at each instant
Rotational Kinetic Energy and Moment of tnertia
'
.
General Formula for Moment of Inertia for Collection
of.l/Point Masses:
=f*,r!
i=t
e
o
Rotational Kinetic Energy: X
o
Total Kinetic Energy: Kaut = Ko*o + K,
Moments of Inertia for Distributed Objects: Table 10l will be given.
*, : !2 tat2
=*.rr'", *f,t"*r'
(Note: Here vs* isthehanslationalvelocity ofthecenta ofmass md
inqtia about the center of mass.)
r
t
Ig, isthemomentof
Torgue and Angular Acceleration
o
.
Torque:
t = rF
(r is the moment arm)
Relation Between Torque and Angular Acceleration (Newton's Second Law for Rotational
Motion):
to", = Ia
.
Equilibrium
Three conditions required for true equilibrium (translational and rotational equilibrium):
l. Zr, =o (i.e., a, =o)
2. Zr,
3.
=o
trr, =0
(i.e., ay
=o)
fot any axis (i.e.,
a=0
about
anyais)
Formula Sheet
'

Final
Page 6
Angular Momentum
.
o
Angular momentum: 7, = Ia .
For point particle of mass rz moving with speed v at an angle
o
0 with respect to the radial
L=rmvsial
direction:
Newton's Second [.aw for Rotational Motion;
tr",
Note here that
,ro =*
.
isthe average net uternal tuqre.
t.tiu:.1111.'11.J ..l.i
r
f
rJi
i
l' i
il
rR:
'.t ,...
t  .._;t: __
IIIr;".
i
l[ i
i,i"f
/t:{Stir.
'{"{;in
lft"
.tt
ii
+ri
ti
r,l
1i,:, '/
Disk or
solid cvlinder
1i
r
I
r=+riRz
(axis at rim)
i
Ir[il'ff*^
1
l**u'
Longihin rod
Disk or
solid cylincler
I
i\L
ii \
ut
'l::i
il
ilr
\
"iIL
I { jt.a,ri
1. ..r':.

r
...,,,,,
Long thin ro
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