Spring 2015
INSTRUCTIONS: a) No books or notes are permitted.
Final Exam
April 27, 2015 b) You may use a calculator. c) You must solve all problems beginning with the equations on the “Information Sheet” provided with this test. Please write down the equation you are using, or the principle you are applying to a given situation. d) You must show your work to receive credit!!!! If a problem requires only a simple
answer, you must explain your answer. Please show all of your work on the sheets
provided (use the back if necessary). d) Please draw a box or a circle around each of your answers.
1

1. A farmer, fleeing from an enraged bull, seeks safety by going through an open gate in a fence and closing the gate behind him. The gate is initially standing open at an angle of 90
0
to the fence. The gate has a moment of inertia of 2500 kg-m
2
. The farmer exerts a force of 220 N at a point 4 m from the hinge. As the gate swings shut, the farmer always applies the force in the most advantageous way.
(a) (10 pts.) Find the angular acceleration of the gate as it closes.
(b) (10 pts.) What is the corresponding linear acceleration and what is its direction?
(c) (10 pts.) How long does it take to shut the gate?
2

2.
Assume that the moon travels around the earth in a circular orbit.
(15 pts.) Find the period in days of the moon in its orbit?
(M
10
6 earth
= 6 x 10
24
kg; M moon
= 7.4 x 10
22
kg; radius of earth = 6.37 x 10
m; distance from earth to moon = 3.84 x 10
8
m).
6
m; radius of moon = 1.74 x
3

3.
A 80-kg person rides in an elevator of mass 600 kg.
(5 pts.) If the elevator moves upward with a constant speed of 2.5 m/s, what is the tension in the cable?
(10 pts.) If the elevator accelerates upward at 0.7 m/s
2
, what is the tension in the cable.
(10 pts.) If the elevator accelerates downward at 0.5 m/s
2
, what is the force of the floor on the person?
4

4.
A baseball is thrown with a velocity of 35 m/s at an angle of 45 o
above the horizontal. Neglect air resistance.
(a) (10 pts.) Find the maximum height of the ball.
(b) (10 pts.) Find the total time in the air.
(c) (10 pts.) Find the horizontal distance the ball travels assuming level ground.
5

5.
A 35-kg child on a 5-kg sled starts from rest and slides 30 m down a hill inclined at an angle of 25
0
to the horizontal. The coefficient of kinetic friction is 0.15.
(a) (5 pts.) How much work is done by gravity?
(b) (5 pts.) How much work is done by friction?
(c) (10 pts.) What is the velocity of the child and sled at the bottom of the hill?
(d) (10 pts.) Assuming level terrain and the same coefficient of friction, how far do the child and sled travel before coming to rest?
6

6.
The simple harmonic motion of a 5-kg object is described by the equation: x = (0.40 m) cos(
 t/3)
(a) (5 pts.) What is the object’s position at t = 0.5 s?
(b) (5 pts.) What is the period of the motion in seconds?
(c) (5 pts.) What is the maximum kinetic energy of the motion?
(d) (5 pts.) What is the maximum potential energy?
7

EXTRA CREDIT
MULTIPLE CHOICE
(There are 5 questions. Each one counts as 2 points.)
1.
An ice skater can glide at nearly constant speed in a straight line across the ice because
(a) There is no friction between her and the ice.
(b) Inertia maintains her motion.
(c) The force of the ice pushes her forward.
(d) Gravity does not act her.
Answer:________
2.
A ping-pong ball attached to a string hangs loosely from the rear view mirror of your car. You start up and then travel at a constant speed in a straight line. The ping-pong ball will
(a) Deflect backwards while starting up and then hang straight down when moving at constant speed.
(b) Continually deflect backwards as long as you are moving.
(c) Always hang straight down.
(d) Deflect forward while starting and then hang straight down while traveling at constant speed.
Answer:________
3.
A car accelerates forward starting from rest at a constant rate of 5 mi/hr/sec. After 2 seconds the car travels at a certain speed and it has traveled a certain distance. After 4 seconds
(a) The acceleration of the car will be two times greater.
(b) The car will have traveled twice as far.
(c) Its speed will be twice as great.
(d) Its speed will be four times greater.
Answer:________
4.
A revolving door will be most difficult to rotate if
(a) Most of the mass is concentrated near the center.
(b) The mass is distributed uniformly throughout the door.
(c) Most of the mass is concentrated at the outer edge.
(d) Most of the mass is concentrated along the top and bottom edges.
Answer:________
5.
A gymnast does a somersault and then lands on the mat. When landing she instinctively bends her knees a little bit to
(a) Maximize her momentum in stopping.
(b) Increase her velocity.
(c) Decrease her time in stopping.
(d) Minimize the force of impact.
Answer:________
8

INFORMATION SHEET sin

= opp/hyp cos

= adj/hyp tan

= opp/adj c
2
= a
2
+ b
2 v
 v
0
 at x
 x
0

1
2
( v
 v
0
) t x
 x
0
 v
0 t

1
2 at
2
 r
 r f
 r i v

 r
 t v

 r lim
 t

0
 t f s
  s
N f k
=
 k
N
W = (F-comp.) x (d) = F
 d
 x = x f
- x i v

 x
 t
 x t f f
 x i
 t i v
 a

 lim t

0
 x
 t

 v t
 v t f f
 v i
 t i a

 lim t

0
 v
 t v
2  v
2
0

2 ax
A x

A cos

A y

A sin

A

A 2 x

A 2 y tan
 
A y
A x a a


 v t

 lim t

0
 v
 t v ac
 v ab
 v bc
F net
 
F
 m a v
2 a c
 r
W net
=

KE = ½mv
2
- ½mv
0
2
KE = ½ mv
2
U gravity
= PE gravity
= mgy
F spring
= -kx
U spring
= PE spring
= ½kx
2
E = KE + PE
P


W
 t
 dW dt

F
 v p = m v

= s/r
F

 p
 t
 d p dt
Impulse

F
 t
Impulse =
 p = m v f
- m v i
F net
= 0

P initial
= P final v
1i
– v
2i
= -(v
1f
– v
2f
)
 
 
 t
 
 
 t v t
= r
 a t
= r

L = r
 p; L = I

  
0
  t F c
= ma c
= mv t
2
/r

= r

F
  
0

1
2
(
  
0
) t

= rFsin


= rFsin

= I

; I =

m i
r i
2
  
0
 
0 t

1
2
 t
2
KE = ½ I
 2
 2   2
0

2
 a c
= v t
2
/r = r
 2
F

Gm
1 m
2 r
2
U(r)

;

G m
1 m
2 r
E

KE max

PE max

1
2 kA
2
  d L dt
;
 net
= 0

L init
= L fin x

A cos(
 t
 
) g = 9.8 m/s
2
=32 ft/s
2  
2
 f ; T

1 / f
;
 2  k m
T

2
 m k
;
T

2

L g
9