Physics 221 2005S Exam 1-solutions
PHYSICS 221
Spring 2005
EXAM 1: Feb 17 2005 8:00pm—10:00pm
Name (printed): ____________________________________________
ID Number: ______________________________________________
Section Number: __________________________________________
INSTRUCTIONS:
Each question is of equal weight, answer all questions. All questions are multiple choice.
Choose the best answer to each question.
Before turning over this page, put away all materials except for pens, pencils, erasers,
rulers, your calculator and “aid sheet”. An “aid sheet” is one two sided 8½×11 page of
notes prepared by the student. There is also a list of possibly useful equations at the end
of the exam.
"In general, any calculator, including calculators that perform graphing numerical
analysis functions, is permitted. Electronic devices that can store large amounts of text,
data or equations are NOT permitted." If you are unsure whether or not your calculator
is allowed for the exam ask your TA.
Examples of allowed calculators: Texas Instruments TI-30XII/83/83+/89, 92+
Casio FX115/250HCS/260/7400G/FX7400GPlus/FX9750 Sharp EL9900C.
Examples of electronic devices that are not permitted: Any laptop, palmtop, pocket
computer, PDA or e-book reader.
In marking the multiple choice bubble sheet use a number 2 pencil. Do NOT use ink. If
you did not bring a pencil, ask for one. Fill in your last name, middle initial, and first
name. Your ID is the middle 9 digits on your ISU card. Special codes K to L are your
recitation section, for the Honors section please encode your section number as follows:
H1⇒02; H2⇒13 and H3⇒25.
If you need to change any entry, you must completely erase your previous entry. Also,
circle your answers on this exam. Before handing in your exam, be sure that your
answers on your bubble sheet are what you intend them to be.
It is strongly suggested that you circle your choices on the question sheet. You
may also copy down your answers on the record sheet at the end and take this page
with you for comparison with the answer key to be posted later.
When you are finished with the exam, place all exam materials, including the bubble
sheet, and the exam itself, in your folder and return the folder to your recitation
instructor. No cell phone calls allowed. Either turn off your cell phone or leave it at home.
Anyone answering a cell phone must hand in their work; their exam is over.
Total number of questions is 27.
Best of luck, David Atwood and Paula Herrera-Siklody
Page 1 of 18
Physics 221 2005S Exam 1-solutions
The following applies to questions [1] and
[2]: A 2-kg block slides up an incline which.
The initial speed of the block, at the bottom
of the incline, is 2 m/s. The acceleration of
the block is 7.2 m/s2 down the ramp.
∆x=?
[1] How far along the incline will the block
go before it momentarily stops?
(A) 9cm (B) 13cm
(E) 33cm
(C) 20cm (D) 28cm
ANSWER: D
Using the v2 formula, 2a ∆x=0-v2
Thus ∆x=-v2/(2a)=-(2m/s) 2/(2(-7.2m/s))=0.28m
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[2] If a hand pushes on the sliding block perpendicularly to the incline surface,
(A) The magnitude of weight increases.
(B) The magnitude of the normal force by the incline on the block increases.
(C) The magnitude of the friction force by the incline on the block increases.
(D) Both A and B.
(E) Both B and C.
Answer (E):
The weight of the block is the force of gravity on the block and is not altered by pushing
on it. The normal force must increase because the block does not sink into the ramp. If
the normal force increases, then kinetic friction which is proportional to the normal also
increases
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Page 2 of 18
Physics 221 2005S Exam 1-solutions
[3] A train car moves along a straight track. The graph shows the position as a function
of time for this train. The graphs shows that the train:
Fast
Slow
Position
Fast
Time
(A) Speeds up all the time
(B) Slows down all the time
(C) Speeds up initially but then slows down.
(D) Moves at a constant speed
(E) Slows down initially but then speeds up.
Answer: (E):
The velocity is the slope of the curve. Clearly the slope of the curve is initially large, then
levels off to a lower value and later increases again. The velocity thus follows the same
pattern.
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[4] If John walks 1km north at 5.0km/hr and then runs another 1km north at 15.0km/hr.
What is the magnitude of his average velocity during the 2km trip?
(A)20.0 km/hr (B)10.0 km/hr
(C)7.5 km/hr
(D)6.0 km/hr
(E) 5.0 km/hr
Answer: (C):
The time for first leg=1km/(5km/hr)=0.20hr. Time for second leg is
1km/(15km/hr)=0.066hr. The average velocity=(total distance)/(total time)
=(2km)/(.266hr)=7.5km/hr.
Page 3 of 18
Physics 221 2005S Exam 1-solutions
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[5] Suppose that a particle moving along the x-axis has velocity as a function of time
described by
v(t ) = a + bt 4
where a=3m/s and b=4m/s5. What is the acceleration of the particle at t=1s?
(A) a=4 m/s² (B) a=7 m/s² (C) a=12 m/s² (D) a=16 m/s² (E) a=20 m/s²
_
Answer (D):
a=dv/dt=4bt3. Substituting in t=1s and b=4m/s5 we get a=16m/s2.
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The following applies to questions [6] and [7]: You want to throw a 400-g snowball to
one of your friends who stands unaware of your evil intentions 10-m away from you. The
ball leaves your hand 1 m from the ground at an angle of 50° with the horizontal and hits
him right on the top of his head. Your friend is 1.7-m tall. Neglect air resistance.
[6] Find the work done by gravity while the ball is in the air.
(A) –6.7J
(B)-2.7J
(C)0J
(D)+2.7J
(E)6.7J
Answer: (B)
First, the work done by gravity is negative since in net the snowball moves upwards. If m
is the mass of the snowball, h is the vertical displacement between your hand and your
friends head and A is the horizontal displacement,
Calculating the work:
G G
W = F ⋅ ∆r = (−mgˆj ) ⋅ ( Aiˆ + hˆj ) = − mgh = −(0.400kg )(9.8m / s 2 )(0.7m) = −2.7 J
[7] What is the direction of the acceleration of the snow ball at the top of its trajectory?
(A) Up.
(B) Down.
(C) In the direction of motion.
(D) Opposite to the direction of motion.
(E) The acceleration is instantaneously zero.
Answer (B):
The acceleration of an object in free fall is always downwards and equal to g=9.81m/s² in
magnitude.
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Page 4 of 18
Physics 221 2005S Exam 1-solutions
[8] A rock is thrown straight upwards at a velocity of 9.8m/s from a bridge that is 39.2m
above a lake. What is the magnitude of the velocity of the rock when it strikes the lake?
Neglect air resistance.
(A) 9.8m/s (B) 19.6m/s
Answer: (C)
Using the v² formula:
v 2 = v02 + 2a y ∆y
(C) 29.4m/s
(D) 39.2m/s
(E) 49.0 m/s
∴ v = v02 + 2a y ∆y
= (9.8m / s ) 2 + 2(−9.8m / s 2 )(−39.2m)
= 29.4m / s
You can also obtain the result using the work energy theorem. If the rock has mass m the
initial KE is Kinit=mv02/2. The work done by gravity is thus adds to the KE to give the
final KE: Kfinal=Kinit+W=mv02/2+Fgrav∆y= mv02/2+mg∆y. Setting this equal to
mv2/2=Kfinal and solving for v we obtain the same result as above. Also, you can recast
the solution in terms of the conservation of mechanical energy but we didn’t learn that by
lecture 14.
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[9] Cars B moves along a straight road. Shown below are the snapshots from a
stationary camera of the positions at t = 1, 2, 3 and 4 s. Assume that the motion is smooth
(no unexpected behaviors between consecutive snapshots)
B
1
2
3
4
What is the sign of the net work done on B ?
(A) Positive
(B) Negative
(C) Zero
(D) It cannot be said without knowing the detail of the forces acting on B.
(E) It depends on the choice of axes.
Answer (B):
Apparently car B is slowing down. Its kinetic energy is decreasing so by the work energy
theorem a negative amount of work is being done on car B
Page 5 of 18
Physics 221 2005S Exam 1-solutions
[10] City A is 480km directly west of city B. A plane has an air speed of 260 km/hr. If
there is wind blowing north at 100 km/hr, what is the time it takes to fly from city A to
city B? Assume that the air traffic controllers have routed the plane to fly directly over
the interstate which runs in a straight line from A to B.
(A)1.72hr
(B) 1.84hr
(C)2.00hr
(D)2.50hr
(E)3.00hr
vplane,ground
A 260km/hr
vplane,air
100km/hr
vair,ground
B
Answer(C):
G
G
G
Using the relative velocity formalism, v plane, ground = v plane, air + v air , ground . From the question
G
G
we know that v plane, ground is due east while vair , ground is due north at 100km/hr.Finally the
magnitude of | v plane, air |= 260km / hr . The three vectors form a right triangle as shown
G
above so the magnitude of v plane , ground is . (260km / hr ) 2 − (100km / hr ) 2 = 240km / hr .
Since the cities are 480km apart, the time for the journey is 2hr.
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[11] The figure below shows the elliptical path along which a particle moves at a
constant speed clockwise. At
which of the indicated points
is the acceleration of the
particle the greatest in
magnitude?
(A) point P
(B) point Q
(C) point R
(D) both P
and R
(E) The acceleration is the
same at all points
Answer (A):
Acceleration is the rate of change in velocity. The place at which the velocity changes
most quickly (in direction) is the region where the curve is the tightest. The acceleration
is thus greatest at point P.
Page 6 of 18
Physics 221 2005S Exam 1-solutions
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a=3m/s²
[12] A watermelon of mass 10kg is hanging from the ceiling
of an elevator by a massless string. If the elevator is
accelerating upwards at a rate of 3m/s², what is the
magnitude of the net force on the watermelon?
10kg
(A) 128N
(B) 98N
(C) 78N
(D)30N (E)0N
Answer (D):
By Newton’s second law,
Fnet=ma=(10kg)(3m/s2)=30N
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[13] A 4-kg book is kept fixed against a vertical wall by a hand
that applies a 200-N force perpendicular to the surface of the
book. The coefficient of static friction between the book and the
wall is µS = 0.3. What is the magnitude of the friction force?
(A) 12N
(B) 20N
(C) 39N
(D) 59N
m=4kg
(E)200N
Answer (C):
If the book is really pinned against the wall then the force of static friction must be equal
in magnitude to the weight of the box, mg=(4kg)(9.8m/s2)=39N. Now we should check
that the force is within the static friction limit. The normal force of the wall on the book
is 200N so the static friction limit is Nm=(200N)(0.3)=60N>39N so the situation is
consistent.
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[14] A 4.0-kg flower pot slips from a window sill and hits the street at 20 m/s.
What is the kinetic energy of the flower pot right before hitting the ground?
(A) 0J
(B) 80J
(C) 400J
(D) 600J
Answer (E):
The kinetic energy is K=mv2/2=(4.0kg)(20m/s)2/2=800J
Page 7 of 18
(E) 800J
Physics 221 2005S Exam 1-solutions
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[15] In the figure below, in which case will the dot product of the two vectors be the
largest (including sign). All of the vectors are of the same length.
Case I
(A) Case I
Case II
(B)Case II
Case III
(C)Case III
(D)Case IV
Case IV
(E)All the same
ANSWER: A
The dot product is proportional to the cosine of the angle between the vectors. The case
with the smallest angle will therefore have the largest dot product. Case I therefore has
the largest dot product.
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The following applies to questions [16] and [17]: Pete applies a horizontal force F=20N
on the system depicted below, made of two boxes A and B placed on a frictionless
horizontal floor. The masses of the boxes are mA = 20 kg and mB = 40 kg.
B
A
[16] What is the magnitude of the acceleration of box B?
(A) 0, you cannot move an object whose weight is 400 N with a 20-N force.
(C) 0.5 m/s2
(D) 1 m/s2
(E)1.2 m/s2
(B) 0.3 m/s2
Answer (B):
The system consisting of the two boxes has a total mass of 60kg. The force acting on this
system is 20N so the acceleration of the system which is therefore the acceleration of box
B is (20N)/(60kg)=0.3 m/s2
_______________________________________________________________________
[17] The magnitude of the force on A by B is _______________ the magnitude of the
force on B by A.
(A) Four times
(B)Twice
(C)The same as
(D) Half
(E)One fourth
Answer (C):
By Newton’s third law!
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Page 8 of 18
Physics 221 2005S Exam 1-solutions
[18] Two people pull their suitcases along the same horizontal airport hallway. They
both exert the same force on the suitcases, but in case 1, the force is perfectly horizontal,
whereas in case 2, the force makes an angle of 25° with the floor.
What is the ratio (work in case 1):(work in case 2)?
(A) 1:1
(B) 1:sin25°
(C) sin25°:1
(D) 1:cos25°
(E)cos25°:1
Answer (D):
G G
Work is W = F ⋅ ∆r . The dot product is proportional to the cosine of the angle between
the vectors hence the ratio of the work is 1:cos25°.
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[19] Consider the system shown below made of three identical blocks of mass m, two
ideal, massless strings and an ideal, massless pulley. The friction between the blocks and
the table is negligible.
m
T1
m
T2
m
Let T1 and T2 be the tension on each string, as shown in the figure. Which of the
following is true?
(B)T1 < T2 = mg
(C)T1 < T2 < mg
(A) T1 = T2 = mg
(E)T1 = T2 > mg
(D)T1 = T2 < mg
Answer (C):
The hanging block should accelerate downwards so the net force is downwards. Thus the
tension in string 2 must be less than this weight so T2<mg. The two blocks on the table
have the same acceleration, a. The net force of the two block system is T2 so T2=2ma.
The net force on block 1 is the T1 so T1=ma<2ma=T2 and T2<mg.
Page 9 of 18
Physics 221 2005S Exam 1-solutions
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[20] The graph below shows a position dependent net force acting on a 1kg block
moving along the x-axis. Initially the block is at x=0m, moving in the +x direction and
has a kinetic energy of 12J. When the block gets to x=8m, what is its kinetic energy?
(A) 12J
(B)18J
(C)24J
(D)36J
(E)48J
Net Force
3N
2N
6J
1N
3J
1m
2m
3m
3J
4m
5m
6m
7m
8m
x
Answer (C):
By the work energy theorem, the change in kinetic energy is the work done on the block.
The work is given by
W = ∫ F dx = 12 J
since this is the area under the graph. This is the increase in kinetic energy. Since it starts
with 12J, the final kinetic energy is 24J.
Page 10 of 18
Physics 221 2005S Exam 1-solutions
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10m
30kg
[21] How much work does Bob do on a 30kg box by pushing it across a level floor
through a distance of 10m at a constant velocity. The coefficient of kinetic friction
between the floor and the box is µk=0.2?
(A)588J
(B)294J
(C)147J
(D)60N
(E)0J
Answer (A):
The net force on the box is 0 since it is moving at a constant velocity. The force that Bob
exerts must therefore be equal in magnitude to the kinetic friction which
is fbob=fk=µkmg=58.8N. The work that bob does is thus fbob ∆x=588J.
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[22] Robin Hood fires an arrow at velocity v0=30m/s at an angle of θ=30o above the
horizontal from the top of a cliff of height h. The sheriff of Nottingham standing at
distance d from the base of the cliff notices that the arrow impacts him with a speed of
36m/s. What is the height h of the cliff? Neglect air resistance.
(A) h=10m
Answer (B):
(B) h=20m
(C) h=30m
(D) h=40m
Using the v² formula:
G G
v 2 − v02 = 2a ⋅ ∆r
G
G
G G
where a = − ˆjg and ∆r = −hˆj + diˆ so a ⋅ ∆r = hg thus,
G G
v 2 − v02 = 2a ⋅ ∆r = 2 gh
∴ h = (v 2 − v02 ) /(2 g )
= ((36m / s ) 2 − (30m / s ) 2 ) /(2(9.8m / s 2 ))
= 20m
Page 11 of 18
(E) h=50m
Physics 221 2005S Exam 1-solutions
Robin
Hood
v0=30m/s
θ=30o
h
v=36m/s
d
Page 12 of 18
Sheriff of
Nottingham
Physics 221 2005S Exam 1-solutions
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The following applies to questions [23] and [24]: An 1kg particle is constrained to
move along the x-axis. At t=0 its velocity is v0 = 2m / s in the +x direction and its initial
position is x = 0m . The acceleration as a function of time is given by the graph below
Acceleration
(m/s²)
2
1
2
4
6
8
Time (s)
[23] Which of the following graphs is the velocity as a function of time for the particle?
(A) Graph I
(B) Graph II
(C) Graph III
(D) None of the graphs.
6
Velocity 4
(m/s)
2
Total
area=31m
Graph I
12m
6
Velocity 4
(m/s)
2
6
Velocity 4
(m/s)
2
10m
3m
6m
2
4
6
8
2
4
6
8
2
4
6
8
Graph II
Graph III
Time (s)
Page 13 of 18
Physics 221 2005S Exam 1-solutions
Answer (A): The slope of the velocity curve should match the acceleration. Thus from
t=0-3s the slope is flat at a value of v=2m/s, the initial value. From t=3-4s the slope is
2m/s² while at from t=4-6s the slope is 1m/s². From t=6-8 the slope is again 0 at a final
velocity of 6m/s. Thus Graph I is the correct graph.
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[24] Continuing from the question [23], what is the particle’s position at t=8s.
(A)x=16m
(B)x=24m
(C)x=28m
(D)x=31m
(E)x=32m
Answer (D): The change in position is the integral of velocity hence the area under the
curve. As shown above, the area under Graph I is 31m and since the particle starts at
x=0m, it winds up at x=31m
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[25] A cannon is fired from a steep slope which makes a
constant 60o angle to the horizontal. It is fired in a direction
perpendicular to the slope and lands further down the slope.
If the cannon ball is fired with an initial velocity of 196m/s
how long after it is fired does it land? Neglect air resistance.
(A) 5s
(B) 10s
(C) 20s
(D) 40s
(E) 80s
60o
Answer (E):
Set up a coordinate system with the x-axis parallel to the
slope as shown. The acceleration of gravity in this frame
G
is a = giˆ sin θ − gˆj cosθ For this problem, we are only
interested in the motion in the y-direction. The net
displacement in the y direction is 0 the average velocity
in the y direction is 0 by eqn 5.3 on the equation sheet.
This implies vfy=-viy. The time aloft is thus
∆vy/ay=(-2viy)/(ay)
=(-2(196m/s))/(-g cos(60°))
=80s
Page 14 of 18
y
x
60o
Physics 221 2005S Exam 1-solutions
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[26] A small stone of mass m = 20 g is placed at the rim (R=27.0cm) of a level pottery
wheel of radius R = 27.0 cm. The wheel is initially at rest and begins moving with a
constant angular acceleration of 0.300 rad/s2. When the system reaches 7.00 rad/s, the
stone flies off. What is the coefficient of static friction between the wheel and the stone?
(A) 0.34
(B) 0.57
(C) 0.81
(D) 1.35
(E) 1.61
Answer(D):
The normal force between the stone and the wheel is N=mg=(0.02kg)(9.8m/s)=0.20N.
The radial acceleration of the stone at the critical point where it slides is
ar = Rω 2 = (0.27m)(7 s −1 ) 2 = 13.2m / s 2
The tangential acceleration is
at = Rα = (0.27m)(0.3s −2 ) = 0.081m / s 2
The magnitude of the total acceleration is thus
a = ar2 + at2 = 13.2m / s 2
By Newton’s 2nd law, the critical value for the static friction is
Fmax=ma=(.020kg)(13.2m/s²)=0.264N
The coefficient of static friction is thus µs=Fmax/N=1.3.
Page 15 of 18
Physics 221 2005S Exam 1-solutions
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[27] Four blocks of masses m1=1kg, m2=2kg, m3=3kg, m4=4kg are on a frictionless
horizontal surface as shown on the figure below. The blocks are connected by ideal
massless strings. A force FL=10N is applied to the left block and is directed to the left.
Another force FR=50N is applied to the right block, and is directed to the right. What is
the magnitude of the tension T in the string between m2 and m3.
FL=10N
T=?
m4=4kg
m3=3kg
m2=2kg
m1=1kg
FR=50N
(A) T=14N
(B)T=20N
(C)T=38N
(D)T=40N
(E)T=44N
Answer (C):
Take right as positive. The net force on the whole train is 40N. The acceleration of the
whole train is (+40N)/(10kg)=+4m/s². This is the acceleration of the first two blocks m1
and m2.which together have has a mass of 3kg so the net force on those blocks is +12N.
The applied force is +50N therefore the force from the center string is −38N. The tension
in the string is therefore 38N.
_______________________________________________________________________
Best of Luck
David Atwood and Paula HerreraSiklody
Page 16 of 18
Physics 221 2005S Exam 1-solutions
Formula Sheet for Exam 1
1. Physical Constants
(numerical value used to derive answers in exam):
1.1) Acceleration of gravity on Earth’s Surface: g=9.8m/s²
1.2) Radius of Earth: Rearth=6.38×106m
1.3) Mass of Proton: mp=1.67×10-27kg
3. Vectors
G G
G G
3.1) Dot Product: A ⋅ B = Ax B x + Ay B y + Az B z =| A || B | cosθ
G
G
where θ is the angle between A and B .
G
3.2) Components: A = Ax iˆ + Ay ˆj + Az kˆ
G
G G
3.3) Magnitude: | V |= V = V x2 + V y2 + V z2 = V ⋅ V
5. One Dimensional Motion
5.1) Average Velocity: v = ∆x / ∆t
5.2) Instantaneous Velocity: v = dx / dt
2. Calculus
2.1)
d
dx
x n = nx n −1
d
dx
sin x = cos x
x n +1
n +1
d
dx cos x = − sin x
n
∫ x dx =
4. Algebra
4.1) The solutions to ax 2 + bx + c = 0
are x =
1
2a
(− b ±
b 2 − 4ac
)
6. Forces
G
G
6.1) Newton’s Second: F = ma
G
G
6.2) Newton’s Third: FAB = − FBA
6.2) Kinetic Friction: f k = µ k N
6.4) Static Friction: f s ≤ µ s N
6.5) Centripetal Force: F =
v x = v0 x + a x t
x = x0 + v0 x t + 12 a x t 2
5.3) For Constant Acceleration only: v 2 − v 2 = 2a ( x − x )
0x
0
x
x
x − x0 1
= 2 (v x + v 0 x )
t
7. Three Dimensional Motion
G
7.1) Position Vector: r = xiˆ + yˆj + zkˆ
G
G
G
G
2 G
7.2) Velocity and Acceleration: v = dtd r
a = dtd v = dtd 2 r
G G G
v = v0 + at
G G G
G
r = r0 + v 0 t + 12 at 2
7.3) Constant Acceleration only: v 2 − v 2 = 2aG ⋅ (rG − rG )
0
G G0
r − r0 1 G G
= 2 (v + v 0 )
t
ω = 2πf
v = Rω
7.4) Circular Motion: f = 1 / T
7.4a) Angular Velocity: ω = dθ / dt
7.5) Centripetal Acceleration: a rad = Rω 2 = v 2 / R = ( 4π 2 R ) / T 2
G
G
G
7.6) Changing Reference Frames: v PA = v PB + v BA
8. Kinetic Energy and Work
8.1) Linear Motion: K = 12 mv 2
8.2) Rotational Motion: K rot = 12 Iω 2
8.3) Work by a constant force
G G
W = F ⋅ s = Fs cosθ
8.4) Work done by a variable force in 1
dim:
x2
W = ∫ Fx dx
x1
8.5) Work in 3D:
P2
G G P2
W = ∫ F ⋅ dl = ∫ F cos φ dl
P1
P1
8.6) Power: P=dW/dt
Page 17 of 18
mv 2
R
G G
P = F ⋅v
Physics 221 2005S Exam 1-solutions
Record Sheet
You may fill in this sheet with your choices, detach it and take it with you after the exam
for comparison with the posted answers
1D
11A
21A
2E
12D
22B
3E
13C
23A
4C
14E
24D
5D
15A
25E
6B
16B
26D
7B
17C
27C
8C
18D
28
9B
19C
29
10C
20C
30
Page 18 of 18