Transcript Answers to Sample exam 2004

Physics 1100
Final Exam
Part A Mulitple Choice Select the best response and indicate your choice on the
computer card provded.
1.
The kilogram is currently defined as:
A
B
C
D
2.
the mass of a platinum-iridium cylinder kept in Scves, France
the mass of 5.9786 x 1026 protons
the mass of one liter of pure water, free of air, at standard temperature and pressure
the mass of a cube of pure water, free of air, 10 cm on each side, at standard temperature
and pressure
How many significant figures are in 120.07 cm?
A
B
C
D
3.
VALUE 20%
6
5
4
3
The Hoover Dam is 726 feet high. What is its height in kilometers? Note 1 km = 0.621 mi and 1 mi =
5280
feet.
A
B
C
D
0.726 km
0.138 km
0.221 km
0.275 km
4.
Figure 3 represents the velocity of a particle as it travels along the axis. In what direction is the acceleration at
t = 0.5 s?
A
B
C
D
5.
toward the positive x-axis
toward the negative x-axis
there is no acceleration at t = 0.5 s
you need an ‘acceleration vs time’ graph to do this problem
At a given instant, the acceleration of a certain particle is zero. Does this mean that:
A
B
C
D
the velocity is constant
the velocity is increasing
the velocity is decreasing
the velocity is not changing at that instant
6.
Figure 2 represents the position of a particle as it travels along the x-axis. What is the magnitude of
the average velocity of the particle between t = 1 s and t = 4 s?
7.
A
B
C
0.57 m/s
0.67 m/s
1.0 m/s
D
1.3 m/s
Vector M = 4.00 points eastward and vector N = 3.00 points northward. The resultant vector M + N is
given by
A 5.00 m at an angle 71.6o north of east.
B 5.00 m at an angle 18.4o north of east
C 5.00 m at an angle 53.1o north of east
D
5.00 m at an angle 36.9o north of east
8.
What is the angle between the vectors A and –A when these vectoes are drawn from a common origin?
A
B
C
D
90o
0o
360o
180o
9.
Mass is a vector quantity.
A True
B
False
10.
A boy kicks a football with an initial velocity of 20 m/s at an angle of 25 o above the horizontal. The
acceleration of the ball while in flight is
A
B
C
D
11.
20 m/s2
-20 m/s2
-9.8 m/s2
0
For which value of θ is the range of a projectile fired from ground level a maximum?
A
B
C
D
30o above the horizontal
45o above the horizontal
600 above the horizontal
90o above the horizontal
12. A rock is thrown from ground level at some angle above the horizontal with a certain velocity. It reaches its
highest point and starts falling down. What is the magnitude of the velocity of the rock just before it hits the
ground?
A
B
C
D
13.
An object moving along a curved path with constant speed does not have a net force acting on it.
A
B
14.
it is equal to its initial vertical velocity
it is equal to its initial horizontal velocity
it is equal to the magnitude of its initial velocity
0
True
False
An object is moving with constant velocity. Which of the following statements is true:
A
B
C
D
a constant force is being applied in the direction of motion
there are no forces acting on the object
the net force on the object is zero
there is no frictional force acting on the object
15. What average net force is required to accelerate a car with a mass of 1200 kg from 0 to 27 m/s in 10 s?
A
B
C
D
444 N
3240 N
4355 N
11760 N
16. A 55-kg box rests on a horizontal surface. The coefficient of static friction between the box and the surface
is 0.30. A horizontal force of 156 N is applied to the box. Does the box move?
A
B
yes
no
17. The SI unit of work is expressed as
A
B
C
D
Watt
Newton
Newton/second
Joule
18. A person carries a mass of 10 kg and walks along the +x-axis for a distance of 100 m with a constant
velocity of 2 m/s. What is the work done by this person?
A
B
C
D
0J
20 J
200 J
none of the other answers given is correct
19. On a force vs position graph, the area under the curve is a representation of
A
B
C
D
Force
Position
Kinetic Energy
Work
20. An object of mass 4.0 kg is moved from a point A to a point C as follows: 0.80 m vertically from A to B, and
then 0.60 m horizontally from B to C. What is the amount of work done as the object is moved from point A
to C?
A 1.3 J from A to B and 16 J from B to C
B
2.2 J from A to B and 32 J from B to C
C
32 J from A to B and 0 J from B to C
D
4.0 J from A to B and 16 J from B to C
21. The center of mass for a continuous, uniform object is located at the geometric center of the object.
A
B
True
False
22. What is the correct expression for torque, in terms of the magnitude of the force , F, the radial distance
from the axis of rotation, r, and the angle θ between the force and the radial line?
A
B
C
D
γ = F r sin θ
γ = F r cos θ
γ = F r tan θ
γ=Fr θ
23. A 55-kg box rests on a horizontal surface. The coefficient of static friction between the box and the surface
is 0.30. A 140-N horizontal force is applied to the box. What is the frictional force on the box?
A
B
C
D
162 N
540 N
140 N
Since the box is not accelerating the frictional force is 0
24. Vector A is initially pointing eastwards. This vector is rotated clockwise by an angle of 1100. Which of the
following statements is correct regarding the final position of vector A?
A
B
C
D
It is pointing 200 west of south
it is pointing 700 south of west
it is pointing 200 south of west
it is pointing 700 east of south
PART B Problems Do ten (10) out of the twelve (12) given problems. Mark a large
X through the two (2) problems that you choose to omit. Your solutions
must be organized and complete for full credit.
VALUE 60%
1.
A 55.0 kg skateboarder starts from rest at the top of a 10.0-m ramp with a height of 2.00 m. The coefficient of
friction is 0.12 on the ramp, and on the horizontal parking lot at the bottom.
A
ANS.
Using energy principles, find the skateboarder’s speed at the bottom of the ramp.
We set the height of parking lot as h =0 for GPE
Ei  E f  Wnc
10.00m
2.00m
Ki U i  K f U f W f
mghi 
Vf 
1
mv 2f  F f d
2
2
(mghi  F f d ) 
m
2
(mghi  umg cos  )d
m
V f  2 g (hi  u cos  )  2(9.81
V f  6.23 m
θ
s2
m
2.00 m
)( 2.00 m  0.12 
)
s2
10 .0m
(4)
B
If the skateboarder continues moving in a straight line across the parking lot until she coasts to a stop,
use energy principles to find how far she travels.
2
ANS.
We have a net force since there’s accelartion. This work on the object changes the mechanical (kinetic) energy of
the object. Work Energy Theorm:
KE  Work
net
KE
d

F
f
KE
 F d
f
1 m v2
 KE
2 i
f
i

F
um g
f
4.0 m
v2
2
i
s
d

2ug 2(0.12)(9.81 m
 6. 9 m
s2
)
2.
A
A ball is thrown horizontally from the roof of a building 45.0 m tall and lands 24.0 m from the base.
How long was the ball in the air?
ANS:
We consider the vertical motion of the Ball
y o  45 .0m
vo  0
y  y o  vo t 
1 2
at
2
0  yo  0 
1
at 2
2
a  g
t ?
v?
y0
t
 2 yo

a
 2(45 .0m)
 9.81 m 2
s
t  3.0s
B
What was the ball’s initial speed?
ANS
We consider the horizontal motion of the ball
vo  ?
v?
xo  0
x  24.0
a0
t  3.0s
x  xo  vo t 
1
at 2
2
x  0  vo t  0
vo 
vo 
x
t
24 .0 m
3 .0 s
v o  8.0
m
s
3.
A book of mass 1.7 kg rests on the rear window shelf of a car that’s rounding a flat curve in the road that has a radius of
curvature
of 98m. The coefficient of static friction between the book and the car is 0.24.
A
ANS
Find the magnitude and give the direction of the frictional force between the shelf and book while the car
is rounding the curve at a speed of 36 km/h.
Centripetal force to keep the book going in a circular motion is supplied by the static friction the book and the shelf. Static
friction varies from 0 to μs mg.
Note that μsfn=μsmg is the maximum static frictional force possible.
Static frictioanl force =
ma cp 
2
mv

r
36 m 2
)
3.6 s
98 m
(1.7 kg)(
 1.7 N
B
What is the maximum speed of the car on the curve before the book begins to slide?
free body diagram for book
Vectors Fx  F f
Newton Fx  ma
FN
 ma  F f
max
Ff
ma cp  u s F N
mg
mv 2
 u s mg
r
v
u s rg 
v  15
(0.24 )(98 m)(9.81
m
s
m
)
s2
4.
A stone is launched straight up by a slingshot. It’s initial speed is 19.6 m/s and the stone is 1.50 m above the
ground when launched.
A
ANS
How high above the ground does the stone rise?
We consider the upward motion of stone
v0
V o  19 .6
yo  0
y ?
y
a  g
m
s
t ?
0  v o2 2a ( y  0)
m 2
)
s
y
m
2(9.81 2 )
s
 v o2
y 
2a
y  19.6m
v  v  2a ( y  y o )
2
2
o
 (19 .6
Total height = 19.6m + 1.5m = 21.1
B
How much time does the stone spend in the air?
We consider upward and downward motion
y
v o  19 .6
v?
m
s
y0
yo  0
a  g
t ?
1
y  y o  v o t  at 2
2
0  0  vo t 
t
 2v o
a
1 2
at
2
m
)
s
m
 9.81 2
s
 2(19 .6
t 
t  2.00 s
5.
The maximum load that can be supported by a rope in an overhead hoist is 400.0 N.
A
ANS
What is the maximum acceleration that can be safely be given to a 25.00-kg object being hoisted
vertically upward?
Freebody diagram of crate
Vectors F y  FT  mg
Newton F y  ma
a
FT
 ma  FT  mg
FT  mg
a
m
mg
a
400 .0 N  25 .00 * 9.81N
25 .00 kg
a  6.190
m
s2
B
What would be the tension in the rope if the load was being lowered at an acceleration of 3.000 m/s 2 ?
FT
ma  FT  mg
FT  ma  mg
a
FT  m(a  g )
mg
FT  25 .00 kg(3.00
FT  170 N
m
m

(

9
.
81
)
s2
s2
6.
A 25.0-kg boy is wind sailing on his 10.0-kg sail-equipped cart. The wind pushes him and his cart at a constant speed up a
25.0- m-long incline that rises 5.00 m.
A
ANS
Find the work done by the wind on the cart.
Since the wind pushes the cart with a constant velocity, we know ∑ Fx = 0 ,Therefore FW = Wx =mg sinθ
Fw  Wx  mg sin 
Fw  (25 .0kg  10 .kg)(9.81
m 5.00 m
)(
)
2
s
25 .0m
Fw  68.7 N
Ww  Fw d  (68.7 N )( 25.0m)  1.72 KJ
B
What is the power developed by the wind if the ride up the incline lasts 11.0 s?
P
W 1.72 KJ

 156W
T
11 .0 s
7.
A
An experimental rocket car starting from rest reaches a speed of 560 km/h after a straight 400 m run on a level roadway
Assuming that acceleration is constant, what was the time of the run?
x
Vo  0
Xo  0
a?
km
V  560
h
X o  420 m
X  Xo 
X  0
1
(V  Vo )t
2
1
(V  V o )t
2
t 
2X
V  Vo
t
2(420 m)
560 m
(0 
)
3.6 s
t ?
t  5.4s
B
What is the magnitude of the acceleration?
V  Vo  at
a
V  Vo
t
560 m
0
3
.
6
s
a
5.4 s
a  28 .8
m
s2
8.
A swimming pool worker uses a uniform 3.45-m pole with a dip net at the far end to retrieve objects that fall in
the pool. He holds it with his right hand at the end, and his left hand at the 50.0 cm mark. If the pole’s mass is
2.84 kg, and there is a 1.25-kg load in the dip net, find the size and direction of the force exerted by each
hand when the pole is horizontal.
y
FL
Note Fw = 2.84 * 9.81N = 27.9N
FD = 1.25 * 9.81N = 12.3N
Distance from center of pole to end is
3.45m/2 = 1.73m
x
FD
FW
FR
Equilrium
cond(i)
  Fy  0
cond(ii)
T  0
FL  FR  FW  FD  0
use left end as pivot
RL FL  Rw FW  RD FD  0
From Eq 2
FL 
…….. …………… Eq 1
…………… Eq 2
RW FW  R D FD (1.73)( 27 .9 N )  (3.45 m)(12 .3N )

RL
0.500 m
FL  181 N
Subst. Into Eq 1
FR  FL  FW  FD  181 N  27.9 N  12.3N  141 N
A soccer player kicks a stationary ball, giving it a speed of 20.0 m/s at an angle of 15.00 to the horizontal
9.
A
What is the maximum height reached by the ball?
Vo Y
We consider upward motion of the ball
Vo=20.0 m/s
θ
y
Vy  0
V o y  5.18
y ?
Yo  0
t ?
a  g
m
s
Vox
Vox  Vo cos
V ox  (20 .0
m
) cos(15 .0  )
s
m
s
Voy  Vo sin 
V ox  19 .3
m
) sin(15 .0  )
s
m
V o y  5.18
s
Voy  (20 .0
V y2  Voy2  2a( y  y o )
0  Voy2  2a( y  0)
y
 (V o2y )
2a
m 2
)
s
y
m
2(9.81 2 )
s
 (5.18
y  1.37m
B
What is the ball’s range?
V o2 sin 2
Range 
g
(20 .0
Range 
m 2
) sin(2 *15 .0  )
s
m
9.81 2
s
Range  20.4m
10
A
.The ferris wheel at a circus has a diameter of 18.2 m and rotates once in 12.0 s.
What is normal force exerted by the seat on a 57.5 kg rider when he is at maximum height?
FN
y
X
mg
Vectors F y  FN  mg
Newton F y  ma cp
 ma cp  FN  mg
FN  ma cp  mg
Note the speed of Rider


dis tan ce D

time
t
 (18 .2m)
12 .0 s
 4.76
m
s
FN  m(a cp  g )
Install a negative sign since
acp is negative when rider is at
the top
V2
F N  m( 
 g)
r
m 2


  (4.76 )

m
s
FN  (57 .5kg)  
 9.81 2 
 18 .2m
s 


2


FN  421 N
B
At what speed in m/s must the wheel rotate so that a rider feels 1.5 times his normal weight at the
lowest position during the ride?
Vectors F y  FN  mg
y
Newton F y  ma cp
x
FN
 ma cp  FN  mg
ma cp  1.5mg  mg
mg
v2
 0.5 g
r
v
0.5 gr
v  (0.5)(9.81
m
v7
s
m 18 .2m
)(
)
2
s
2
11.
The carton in the diagram below lies on a plane inclined at an angle of 22.00 to the horizontal. The
coefficient of kinetic friction is 0.120.
9.30m
θ=22.0∙
y
A
FN
Determine the acceleration of the carton as it slides down the incline
Ff
x
mgy
mg
ma  mg x  mg y
a  g x  g y
a  g sin   g cos
Vectors Fx  mg x  F f
a  g (sin    cos )
Newton Fx  ma
a  (9.81
 ma  mg x  F f
m
)(sin 22 .0   0.120 cos 22 .0  )
2
s
m
a  2.58 2
s
B
If the carton starts from rest 9.30 m up the plane from its base, what will be the carton’s speed when it
reaches the bottom of the incline?
vo= 0
y
xo= 0
v= ?
x = 9.30 m
x
a = 2.58 m/s
v 2  v o2  2a( x  x o )
v 2  0  2 a ( x  0)
v  2ax
v 2(2.58
m
)(9.30 m)
s
m
v  6.93
s
t=?
12.
An aircraft that has an air speed of 225 km/h is to fly to a destination that lies in the direction 10.00 North of
East. A steady wind of speed 45.0 km/h blows from the direction 15.00 East of North.
A
Draw a relative velocity vector diagram.
n
Note: P – plane
A – Air
w
e
E - Earth
Vpa
Vae
s
15.0
10.0
Vpe
B
Use vector components to find the velocity of the plane with respect to the earth during this flight
Using the same
diagram as in part
A. All angles are
in degrees
C  10.0  90.0  15.0  115.0 
B
15.0
15.0
10.0
sin A 
10.0
a sin C
c
A  10.4 
 Direction V PA  10 .0   10 .4   20 .4 
V PE  VPA  V AE
VECTOR
N-COMP
225
VPA
 45 .0
VAE
2
2
V PE  V PEN
 V PEE
km
km
cos15 .0   43 .5
h
h
34 .9
VPE

(34 .9
E-COMP
km
km
sin 20 .4   78 .4
h
h
km
h
km 2
km 2
)  (199
)
h
h
225
km
km
cos 20 .4   211
h
h
 45 .0
km
km
sin 15 .0   11 .6
h
h
199
 202
a sin C
)
c

km 

  45 .0
 sin 115 .0
h 

A  sin 1 
km
225 .0

h

a
c

sin A sin C
c sin A  a sin C
C
A
A  sin 1 (
km
 at 10 .0  N  of  E 
h
km
h






PART 3
1.
Complete Question # 1 and ONE other question.
Each Question Valued 10 for a total value of 20.
A lab cart is set in motion, by hand, on a frictionless incline. Once the cart has been given its motion, only the
force of gravity acts on it, and thus gravity controls its velovity on the incline.
The three `velocity Vs time` curves presented in the graph below result from three different angles for the incline.
The inlcine-angles, above the horizontal, are in random order, 10∙, 20 and 25 degrees.
Note that students are expected to associate each curve with its correct incline whose size is given in
degrees above the horizontal
Question and answer on the next slide. You will have to
flip back and forth to see graph and answer.
A
Interpret the lab results from the previous page and complete the table below.
The initial section of each curve with negative slopes indicate the cart’s motion while being pushed. The nearly straight line sections with
positive slope indicate the cart’s constant acceleration and down the incline. The bottom of the curves that end on the x-axis is where the
cart was stopped and at the bottom of the incline.
m
m
 (0.8 )
s
s
(10  incline)  a 
2.85 s  1.55 s
m
m
1.50  (1.0 )
s
s
(20  incline)  a 
2.75 s  1.95 s
m
m
1.50  (1.10 )
s
s
(25  incline)  a 
1.85 s  1.25 s
1.2
θ
 1 .8
m
s2
 3 .1
m
s2
 4.3
m
s2
Sinθ
Acceleration

0.1736
1. 8
m
s2
20

0.3420
3.1
m
s2
25

4 .3
m
s2
10
0.4226
Mechanical theory proclaims that the acceleration of an object due to gravity on a frictionless incline is found using
the relationship….
a = g sin θ where a - is the acceleration on the incline.
θ - is the size of the incline in degrees
g – is the acceleration due to gravity.
B
Using the information from the table on the previous page, plot a curve a Vs sinθ on the grid
below.
You should get a
line something
like this.
C
Slope 
Determine the slope of the above curve and using the accepted valve of g =9.8m/s2, find the %
discrepancy for the experimental value for g .
y 2  y1
x 2  x1
% Discrepancy 

4.7  1.1
m
 9.0 2
.5  .1
s
9.8  9.0
*100 %
9.8
 8%
2.
A car goes around a curve on a road that is banked at an angle of 30.0°. Even though the road is slick, the car
will stay on the road without any friction between its tires and the road when its speed is 24.0 m/s. Note that we
choose the positive y axis to point vertical and the positive x direction to point toward the center of the circular
path.
Find the equation for r and calculate the radius of the curve for this car?
The equation for r
 Fy  N cos  mg  0
mg
cos 
 Fx  N sin  macp
N
acp 
N sin 
m
 mg  sin  



 cos  m 
v2
acp  g tan  
r
2
v
r
g tan 
(24 .0

(9.81
m 2
)
s
m
) tan 30 .0 
2
s
The radius of the
curve
 102 m
3.
When you arrive at Duke’s Dude Ranch, you are greeted by the large wooden
sign shown in the figure below. The left end of the sign is held in place by a bolt,
the right end is tied to a rope that makes an angle of 20.0° with the horizontal.
If the sign is uniform, 3.20 m long, and has a mass of 16.0 kg,
What is :
(a) the tension
in the rope
(b) the horizontal and vertical components of the force, F, exerted by the bolt?
(c) the Force on the bolt F?
note that from the sign’s point of view the situation is symmetric: it has a movable support at each end and doesn’t “know” whether the
support is a wall or a wire. So, the force at the wall bolt is the same as it would be if the wall were instead a wire running up and to the
right at 20.0° above horizontal. By symmetry, f = T and  Fy  2T s in  mg  0.

(16.0 kg) 9.81
mg
T

2sin 
2 sin20.0
m
2
s

229 N
Fx  F cos  T cos  (229.5 N) cos20.0  216 N
Fy  F sin  T sin 
mg

2

(16.0 kg) 9.81
2
m
s2
 78.5 N
4.
A basketball is thrown horizontally with an initial speed of 4.20 m/s . A straight line drawn from the release point
to the landing point makes an angle of 30.0° with the horizontal. What was the release height?
h
The horizontal distance the ball travels is given by x  v0 2h and x  tan  .
g
h
 v0
tan 
2h
g
2v0 2 h

g
tan 2 
2v0 2
h
tan 2 
g
h2

2 4.20 m
s

m
9.81 2
s
 1.20 m

2
tan 2 (–30.0)