03 homework ARMSTRONG, DOMINIC Due: Sep 25 2007, 3:00 am Question 1, chap 4, sect 4. part 1 of 1 10 points A ball is thrown and follows the parabolic path shown. Air friction is negligible. Point Q is the highest point on the path. Points P and R are the same height above the ground. Q P R 9.
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Explanation: Since air friction is negligible, the only acceleration on the ball after being thrown is that due to gravity, which acts straight down. Question 2, chap 4, sect 4. part 1 of 1 10 points A cannon fires a 0.784 kg shell with initial velocity vi = 10 m/s in the direction = 56 above the horizontal. The acceleration of gravity is 9.8 m/s2 .
h
Which of the following diagrams best indicates the direction of the acceleration, if any, on the ball at point P ? 1.
m/
s
y
10
56
2.
y
x 3. 4. 5. The shell's trajectory curves downward because of gravity, so at the time t = 0.62 s the shell is below the straight line by some vertical distance. Your task is to calculate this distance h in the absence of air resistance. Correct answer: 1.88356 m (tolerance 1 %). Explanation: In the absence of gravity, the shell would fly along the straight line at constant velocity: x = t vi cos , ^ y = t vi sin . ^ The gravity does not affect the x coordinate of the shell, but it does pull its y coordinate at constant downward acceleration ay = -g,
6. correct 7. The ball is in free-fall and there is no acceleration at any point on its path. 8.
homework 03 ARMSTRONG, DOMINIC Due: Sep 25 2007, 3:00 am hence x = t vi cos , g t2 . 2 1 Thus, x = x but y = y - g t2 , or in other ^ ^ 2 words, the shell deviates from the straight-line path by the vertical distance y = t vi sin - g t2 h = y - y = ^ . 2 Note: This result is completely independent on the initial velocity vi or angle of the shell. It is a simple function of the flight time t and nothing else (besides the constant g = 9.8 m/s2 ). g t2 2 (9.8 m/s2 ) (0.62 s)2 = 2 = 1.88356 m .
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Correct answer: 13.6216 m (tolerance 1 %). Explanation: First, let's find out the flight time of the car. Consider the vertical motion of the car -- it has constant acceleration g. Initially, when the car goes off the cliffs edge, its vertical velocity is v0y = v0 sin = 1.4702 m/s . When the car hits the water, its vertical velocity has increased. To calculate the increase, we use 2 2 vy = v0y + 2 g h = (1.4702 m/s)2 + 2 (9.8 m/s2 ) (44 m) = 864.561 m2 /s2 , which gives us vy = 29.4034 m/s. Consequently, in light of vy = v0y + g t the time of flight is vy - v0y t= = 2.85033 s . g Now consider the horizontal motion of the car. The horizontal velocity is constant, vx v0x = v0 cos = 4.77897 m/s. Therefore, the horizontal displacement of the car during its fall is simply x = vx t = 13.6216 m. Question 4, chap 4, sect 4. part 1 of 2 10 points Consider the motion of a projectile. It is fired at t = 0. The initial velocity vector in rectangular coordinates is v0x and v0y , and in the polar coordinates is v0 and (see figure). A y B
h =
Question 3, chap 4, sect 4. part 1 of 1 10 points For your information: g = 9.8 m/s2 . A car is parked near a cliff overlooking the ocean on an incline that makes an angle = 17.1 with the horizontal. The negligent driver leaves the car in neutral, and the emergency brakes are defective. The car rolls from rest down the incline and has velocity 5 m/s when it reaches the cliff's edge, 44 m above the ocean. When the car hits the water, how far is it from the cliff's base?
v0 h
R
x
homework 03 ARMSTRONG, DOMINIC Due: Sep 25 2007, 3:00 am The height of the trajectory h, is given by v0 1. h = . 2g v0x 2 . 2. h = 2g 3. h = 2 g v0 . 4. h = v0y 2 . correct 2g v0 2 . 5. h = 2g v0y 6. h = . 2g v0x . 2g 1. 2. 3. 4. 5. 6. 7. 8. 9. t2 t1 t2 t1 t2 t1 t2 t1 t2 t1 t2 t1 t2 t1 t2 t1 t2 t1 = = = = = = = = = cos 2 . cos 1 sin 2 . sin2 1 tan 2 . tan 1 sin 2 . correct sin 1 tan 2 . tan2 1 cos 2 . cos2 1 cos 1 . cos 2 tan 1 . tan 2 sin 1 . sin 2
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7. h = 2 g v0 2 . 8. h =
9. h = 2 g v0x 2 . 10. h = 2 g v0y 2 . Explanation: Basic Concepts: Constant acceleration: x - x0 = v0 t + 1 2 at 2 (1) (2)
Explanation: Apply the "vt" relation v = v0 + a t for the vertical motion, one gets 0 = v0y - g trise , (5) (2)
v = v0 + a t .
Solution: Consider the motion in y direction, vy 2 = v0y 2 + 2 a s (3) At the maximum height h, 0 = v0y 2 - 2 g h or v0y 2 . h= 2g Question 5, chap 4, sect 4. part 2 of 2 10 points Consider two cases with the initial angles 1 and 2 where the initial speed v0 is fixed. The ratio of the time intervals for the trip t2 is given by for the two cases t1 (4)
where trise is the time taken to reach the v0y top. So trise = . The time for the trip, g ttrip = 2 trise . 2 v0y2 t2 sin 2 v0 sin 2 g = = . = 2 v0y1 t1 v0 sin 1 sin 1 g Question 6, chap 5, sect 1. part 1 of 1 10 points An elevator is being lifted up an elevator shaft at a constant speed by a steel cable as shown in the figure below. All frictional effects are negligible.
(6)
(4)
homework 03 ARMSTRONG, DOMINIC Due: Sep 25 2007, 3:00 am
4
steel cable
2. No force acts on a body at rest. If at least one force acted on it the body would move. 3. No net force acts on a body at rest. When the net force is zero, the body is in static equilibrium. correct 4. No force acts on a body at rest. All forces are canceled by each other. 5. All are wrong.
Elevator going up at constant speed
In this situation, forces on the elevator are such that 1. the upward force by the cable is greater than the downward force of gravity. 2. the upward force by the cable is smaller than the downward force of gravity. 3. None of these. (The elevator goes up because the cable is being shortened, not because an upward force is exerted on the elevator by the cable.) 4. the upward force by the cable is equal to the downward force of gravity. correct 5. the upward force by the cable is greater than the sum of the downward force of gravity and a downward force due to the air. Explanation: Since the elevator is being lifted at a constant speed, the net force on it is zero, therefore, the upward force by the cable is equal to the downward force of gravity.
Explanation: There may be any number of forces that act to produce a zero net force. When the net force is zero, the body is in static equilibrium. Question 8, chap 5, sect 2. part 1 of 1 10 points An elevator starts from rest with a constant upward acceleration and moves 1 m in the first 1.5 s. A passenger in the elevator is holding a 7.7 kg bundle at the end of a vertical cord. The acceleration of gravity is 9.8 m/s2 . What is the tension in the cord as the elevator accelerates? Correct answer: 82.3044 N (tolerance 1 %). Explanation: T g mg aelevator
Question 7, chap 5, sect 1. part 1 of 1 10 points Is it correct to say that no force acts on a body at rest? 1. No net force acts on a body at rest, because no force acts on the body at all.
Let h be the distance traveled and a the acceleration of the elevator. With the initial velocity being zero, we simplify the following expression and solve for acceleration of the elevator: h = v0 t + = a= 2h . t2 1 2 1 2 at = at 2 2
The equation describing the forces acting on
homework 03 ARMSTRONG, DOMINIC Due: Sep 25 2007, 3:00 am the bundle is Fnet = m a = T - m g T = m (g + a) 2h =m g+ 2 t = (7.7 kg) 9.8 m/s2 + = 82.3044 N . Question 9, chap 5, sect 2. part 1 of 3 10 points You are driving a car down a straight road at a constant 55 miles per hour. Consider the following forces: I) air drag pushing back on the car; II) gravity pushing down on the car; III) the ground pushing up on the car; IV) friction between the wheels and the ground pushing the car forward; and V) friction between the wheels and the ground pushing the car back. Which of the above forces is/are acting on the car? 1. I and II 2. I, III and V 3. II, III and IV 4. I, II and III 5. I, II, III and IV correct Explanation: There are four forces acting on the car. The ground pushing up on the car balances the gravity pushing down in the vertical direction; the friction between the wheels and the ground pushing the car forward balances the air drag pushing back on the car. Question 10, chap 5, sect 2. part 2 of 3 10 points Is there a net or unbalanced force acting on the car? 2 (1 m) (1.5 s)2 1. No; the speed is constant. 2. Cannot be determined.
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3. Yes; there is a net force in the horizontal direction. 4. No; the velocity is constant. correct 5. Yes; there is a net force in the vertical direction. Explanation: According to Newton's law; first a moving object will continue moving in a straight line at a constant speed until acted upon by an unbalanced force. The car's speed is constant and its direction does not change, so, there is no net force acting on the car. Question 11, chap 5, sect 2. part 3 of 3 10 points After a while, the car started to go around a long bend, still maintaining its constant speed of 55 miles per hour. Is there a net or unbalanced force acting on the car? 1. Cannot be determined. 2. Yes; the direction of the net force is not the same as the velocity. correct 3. No; the speed of the car does not change. 4. Yes; the direction of the net force is the same as the velocity. 5. No; the velocity of the car does not change. Explanation: The direction of the net force is the same as the direction of acceleration, which is directed inward (toward the center of the curve). Question 12, chap 5, sect 5. part 1 of 1 10 points
homework 03 ARMSTRONG, DOMINIC Due: Sep 25 2007, 3:00 am If F1 is the magnitude of the force exerted by the Earth on a satellite in orbit about the Earth and F2 is the magnitude of the force exerted by the satellite on the Earth, then which of the following is true? 1. F1 > F2 . 2. F2 > F1 . 3. F1 F2 . 4. F2 F1 . 5. F1 = F2 . correct Explanation: The forces must be equal in magnitude but opposite in direction. Question 13, chap 5, sect 5. part 1 of 3 10 points Consider the following statement made regarding a book at rest on a level table: The two forces exerted on the book are the normal force directed up and the weight of the book directed down. These are equal and opposite to one another. By Newton's third law they are an actionreaction pair, so the normal force is always equal to the weight of the book. Do you agree with the statement? 1. Disagree, because the weight must be slightly greater than the normal force to keep the book in contact with the table. However, the forces are an action-reaction force pair because they are acting on one object. 2. Agree, because a normal force only exerts enough force to keep the object from falling through. So the normal and weight are always equal, even when other forces are present. 3. Agree, according to Newton's third law for every action there is an equal and opposite reaction. Therefore, since gravity pulls the
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book down, to be "equal and opposite" the force of gravity must equal the normal force. 4. Disagree, if there is a force in addition to gravity acting downwards on the book then the normal force must also counter this extra force. correct Explanation: These choices were all taken from student answers to this same question. If we were to put a weight on the book, the normal force would not be simply the weight of the book as the statement implies ( consider the next problem ). Note: This force pair is not a third law action-reaction pair. The force pairs in Newton's third law always act on two different objects! The action-reaction pair is the gravitational force of the Earth-book and an equal and opposite gravitational force of the book-Earth. That is, the Earth exerts a gravitational force on the book and the book exerts a gravitational force on the Earth. Caution: This is a trick question! The explanation introduces something not mentioned in the problem. Question 14, chap 5, sect 5. part 2 of 3 10 points Consider a book on top of a level table while the book is being pressed straight down with a force F by your hand.
F Book
The following figures show several attempts at drawing free-body diagrams for the book. Which figure has the correct directions for each force? Note: The magnitude of the forces are not
homework 03 ARMSTRONG, DOMINIC Due: Sep 25 2007, 3:00 am necessarily drawn to scale. 6. 1. gravitational hand normal normal gravitational hand
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Explanation: The normal force opposes both the gravitational force down and the Fhand down. 2. normal gravitational Question 15, chap 5, sect 5. part 3 of 3 10 points hand What forces change when comparing the free body diagram before the hand was placed on the book to after? 1. gravitational and Fhand correct 2. gravitational, normal, and Fhand gravitational hand 3. None of these 4. normal gravitational 5. gravitational 6. Fhand hand normal 7. gravitational and normal 8. normal and Fhand correct hand normal Explanation: There will be a change in both the normal force and the Fhand . When the hand is removed, the normal force is equal to the weight of the book. Question 16, chap 5, sect 5. part 1 of 1 10 points If a Mack truck and Honda Civic have a head-on collision, upon which vehicle is the impact force greater? Which vehicle experiences the greater acceleration?
3.
normal
4.
5.
gravitational
homework 03 ARMSTRONG, DOMINIC Due: Sep 25 2007, 3:00 am 1. The forces are the same; the accelerations are same. 2. All are wrong. 3. The force on the truck is greater; the accelerations are the same. 4. The forces are the same; the Civic experiences the greater acceleration. correct 5. The forces are the same; the truck experiences the greater acceleration. 35.2422 N 35.2422 N 42 N
8
54
.8
50
Explanation: In accordance with Newton's 3rd law, the force on each will be of the same magnitude. But the effect of the force (acceleration) will be different for each because of the different masses. The more massive truck undergoes less change in motion than the Civic. Question 17, chap 5, sect 5. part 1 of 2 10 points Consider the 42 N weight held by two cables shown below. The left-hand cable is horizontal.
Scale: 10 N Note: The sum of the x- and y-components of T1 , T2 , and Wg are equal to zero. Given : Wg = 42 N and = 50 . Basic Concept: Vertically, we have
50
Fy,net = F1 sin - Wg = 0 Solution:
42 N a) What is the tension in the cable slanted at an angle of 50 ? Correct answer: 54.8271 N (tolerance 1 %).
F1 (sin ) = Wg Wg F1 = sin 42 N = sin 50 = 54.8271 N Question 18, chap 5, sect 5. part 2 of 2 10 points
Explanation: Observe the free-body diagram below.
b) What is the tension in the horizontal cable?
42 N
27
1N
homework 03 ARMSTRONG, DOMINIC Due: Sep 25 2007, 3:00 am Correct answer: 35.2422 N (tolerance 1 %). Explanation: Basic Concept: Horizontally, Fx,net = F1 cos - F2 = 0 Solution: F2 = F1 cos Wg cos = sin (42 N) cos 50 = sin 50 = 35.2422 N Question 19, chap 5, sect 6. part 1 of 1 10 points Given: The initial position of the block is the origin; i.e., x = 0 at t = 0 . Consider down the track to be the positive x-direction. A block with an initial velocity v0 slides up and back down a frictionless incline. x 6. t 4. x t 3. x t
9
x 5. t
x
v0
7.
t
x 8. Which graph best represents a description the position of the block versus time? x 1. t 9. x t t
x 2. t
correct
homework 03 ARMSTRONG, DOMINIC Due: Sep 25 2007, 3:00 am x 10. t Explanation: Given : m = 330 kg , = 38 , and s = 0 .
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Explanation: The block's acceleration a = +g sin > 0 is down the track, hence positive in our coordinate system. It is also constant -- because the forces on the block are constant. The initial velocity of the block is up the track, hence negative in our coordinate system, v0 < 0. Consequently, the block's equation of motion is 1 x(t) = v0 t + g sin t2 , 2 whose plot is a parabola with its horns up. Question 20, chap 5, sect 6. part 1 of 2 10 points Assume: Down the plane is the positive direction and up the plane is the negative direction. A block of mass 330 kg slides on a frictionless plane inclined at 38 with the horizontal under the influence of a restraining force of 1101 N acting parallel to the incline and up the incline. The acceleration of gravity is 9.8 m/s2 .
Consider the free body diagram for the block
os N gc m =
F
m Basic Concepts:
g
si
n
N
W = mg
Fx,net = F cos - W|| = 0 W|| = m g sin = 0 The applied force F acts up the plane and the weight component m g sin acts down the plane. Fnet = m a = m g sin - F m g sin - F a= m (330 kg) (9.8 m/s2 ) sin - 1101 N = 330 kg 2 = 2.69712 m/s a = 2.69712 m/s2 . Question 21, chap 5, sect 6. part 2 of 2 10 points The resulting motion is 1. undetermined.
0 33 F
kg = 0
2. down the plane, since the acceleration is positive. correct 38 3. up the plane, since the acceleration is negative. Explanation: Note: The assumption given in the statement of the problem is that down the plane is the positive direction. From Part 1, the acceleration is 2.69712 m/s2 .
What is the magnitude of the acceleration a of the block? Correct answer: 2.69712 m/s2 (tolerance 1 %).