Chapter 5 - Applying Newton s Laws w./ QuickCheck Questions
|
|
|
- Jerome Newman
- 7 years ago
- Views:
Transcription
1 Chapter 5 - Applying Newton s Laws w./ QuickCheck Questions 2015 Pearson Education, Inc. Anastasia Ierides Department of Physics and Astronomy University of New Mexico September 17, 2015
2 Review of Last Time Force - push/pull; the cause of acceleration; vector; units N or lb; net force Newton s Laws of Motion (3); interactions; equilibrium; dynamics Free body diagrams - identifying forces Different types of forces: weight, tension, normal, friction (static/kinetic), drag, thrust Mass versus weight; apparent weight/weightlessness
3 QuickCheck Question 4.4 A bobsledder pushes her sled across horizontal snow to get it going, then jumps in. After she jumps in, the sled gradually slows to a halt. What forces act on the sled just after she s jumped in? A. Gravity and kinetic friction B. Gravity and a normal force C. Gravity and the force of the push D. Gravity, a normal force, and kinetic friction E. Gravity, a normal force, kinetic friction, and the force of the push
4 QuickCheck Question 4.4 A bobsledder pushes her sled across horizontal snow to get it going, then jumps in. After she jumps in, the sled gradually slows to a halt. What forces act on the sled just after she s jumped in? A. Gravity and kinetic friction B. Gravity and a normal force C. Gravity and the force of the push D. Gravity, a normal force, and kinetic friction E. Gravity, a normal force, kinetic friction, and the force of the push
5 QuickCheck Question 4.5 A cart is pulled to the right with a constant, steady force. How will its acceleration graph look?
6 QuickCheck Question 4.5 A cart is pulled to the right with a constant, steady force. How will its acceleration graph look? Remember: A constant force means a constant acceleration!
7 Free-Body Diagrams
8 Look at the Skier n T fk w
9 QuickCheck Question 4.11 An elevator, lifted by a cable, is moving upward and slowing. Which is the correct free-body diagram? A. B. C. D. E.
10 QuickCheck Question 4.11 An elevator, lifted by a cable, is moving upward and slowing. Which is the correct free-body diagram? T T T T w w w w A. B. C. D. E.
11 QuickCheck Question 4.12 A ball has been tossed straight up. Which is the correct freebody diagram just after the ball has left the hand? Ignore air resistance. A. B. C. D.
12 QuickCheck Question 4.12 A ball has been tossed straight up. Which is the correct freebody diagram just after the ball has left the hand? Ignore air resistance. HINT: This problem is an example of free fall! The only force is that due to gravity! w w w w A. B. C. D.
13 QuickCheck Question 4.13 A ball, hanging from the ceiling by a string, is pulled back and released. Which is the correct free-body diagram just after its release? A. B. C. D. E.
14 QuickCheck Question 4.13 A ball, hanging from the ceiling by a string, is pulled back and released. Which is the correct free-body diagram just after its release? Let s look at the options again - which one has tension and weight? T T T T w w w A. B. C. D. E.
15 QuickCheck Question 4.14 A car is parked on a hill. Which is the correct free-body diagram?
16 QuickCheck Question 4.14 A car is parked on a hill. Which is the correct free-body diagram? Let s look at those options again which one gives you all three? n n n n f f f w w w C.
17 QuickCheck Question 4.14 A car is towed to the right at constant speed. Which is the correct free-body diagram?
18 QuickCheck Question 4.14 A car is towed to the right at constant speed. Which is the correct free-body diagram? D.
19 QuickCheck Question year-old Sarah stands on a skateboard. Her older brother Jack starts pushing her backward and she starts speeding up. The force of Jack on Sarah is A. Greater than the force of Sarah on Jack. B. Equal to the force of Sarah on Jack. C. Less than the force of Sarah on Jack.
20 QuickCheck Question year-old Sarah stands on a skateboard. Her older brother Jack starts pushing her backward and she starts speeding up. The force of Jack on Sarah is A. Greater than the force of Sarah on Jack. B. Equal to the force of Sarah on Jack. C. Less than the force of Sarah on Jack. Remember: Action/reaction pair forces come in pairs, are equal in magnitude and opposite in direction (Newton s third law)
21 QuickCheck Question 4.17 A mosquito runs head-on into a truck. Splat! Which is true during the collision? A. The mosquito exerts more force on the truck than the truck exerts on the mosquito. B. The truck exerts more force on the mosquito than the mosquito exerts on the truck. C. The mosquito exerts the same force on the truck as the truck exerts on the mosquito. D. The truck exerts a force on the mosquito but the mosquito does not exert a force on the truck. E. The mosquito exerts a force on the truck but the truck does not exert a force on the mosquito.
22 QuickCheck Question 4.17 A mosquito runs head-on into a truck. Splat! Which is true during the collision? A. The mosquito exerts more force on the truck than the truck exerts on the mosquito. B. The truck exerts more force on the mosquito than the mosquito exerts on the truck. C. The mosquito exerts the same force on the truck as the truck exerts on the mosquito. D. The truck exerts a force on the mosquito but the mosquito does not exert a force on the truck. E. The mosquito exerts a force on the truck but the truck does not exert a force on the mosquito.
23 Equilibrium in Motion
24 QuickCheck Question 5.1 A ring, seen from above, is pulled on by three forces. The ring is not moving. How big is the force F? A. 20 N B. 10 cosθ N C. 10 sinθ N D. 20 cosθ N E. 20 sinθ N
25 QuickCheck Question 5.1 A ring, seen from above, is pulled on by three forces. The ring is not moving. How big is the force F? A. 20 N B. 10 cosθ N C. 10 sinθ N x-components cancel out leaving only y-components Fy θ θ Fy D. 20 cosθ N E. 20 sinθ N Fy = 2Fy - F = 0 F = 2Fy = 2 (10 N) sin θ = 20 sin θ N
26 Dynamics - non-equilibrium Forces acting on an object determine its acceleration This acceleration can be used in the equations of motion to describe the motion of the object Nonzero acceleration results in dynamic behavior
27 Mass versus Weight Weight is a force that always points straight down, w = mg, where m is the mass It is easy to convert weight to mass and viceversa
28 Apparent Weight Weight is the force you feel due to the Earth s gravitational pull Sensation of weight is due to contact forces supporting you (action/reaction pair) Your apparent weight, w app, in terms of the force you feel is
29 Apparent Weight in an Elevator Forces acting on the man are the upward normal force due to the elevator floor, and his weight n = w + ma = wapp Since w app > w, the man feels heavier
30 Weightlessness An object in free fall, with no contact forces, has zero apparent weight - weightlessness Weightlessness no weight
31 QuickCheck Question 5.4 What are the components of w in the coordinate system shown?
32 QuickCheck Question 5.4 What are the components of w in the coordinate system shown? y x w wy θ wx
33 QuickCheck Question 5.5 A 50-kg student (mg = 490 N) gets in a 1000-kg elevator at rest and stands on a metric bathroom scale. As the elevator accelerates upward, the scale reads A. > 490 N B. 490 N C. < 490 N but not 0 N D. 0 N
34 QuickCheck Question 5.5 A 50-kg student (mg = 490 N) gets in a 1000-kg elevator at rest and stands on a metric bathroom scale. As the elevator accelerates upward, the scale reads A. > 490 N B. 490 N C. < 490 N but not 0 N D. 0 N Remember the apparent weight of the man in the elevator accelerating upward!
35 QuickCheck Question 5.6 A 50-kg student (mg = 490 N) gets in a 1000-kg elevator at rest and stands on a metric bathroom scale. Sadly, the elevator cable breaks. What is the reading on the scale during the few seconds it takes the student to plunge to his doom? A. > 490 N B. 490 N C. < 490 N but not 0 N D. 0 N
36 QuickCheck Question 5.6 A 50-kg student (mg = 490 N) gets in a 1000-kg elevator at rest and stands on a metric bathroom scale. Sadly, the elevator cable breaks. What is the reading on the scale during the few seconds it takes the student to plunge to his doom? A. > 490 N The student feels weightless! B. 490 N C. < 490 N but not 0 N D. 0 N
37 Surface Forces An object on a desk or a person sitting on a chair is subject to an upward (perpendicular or normal to the surface) force due to the desk or chair
38 Surface Forces An object on a desk or a person sitting on a chair is subject to an upward (perpendicular or normal to the surface) force due to the desk or chair This type of force is known as the normal force, denoted by n, and adjusts itself so the object stays on the surface
39 Remember: Towing Car Example A car with a mass of 1500 kg is being towed at a steady speed by a rope held at a 20 angle from the horizontal. A friction force of 320 N opposes the car s motion. What is the tension in the rope?
40 Example 5.4: Tension in towing a car Here we found that the normal force adjusted itself due to the y-component of the tension in the rope n = w - T sinθ
41 Example 5.9: Normal force on a pressed book A 1.2 kg book lies on a table. The book is pressed down from above with a force of 15 N. What is the normal force acting on the book from the table below?
42 Example 5.9: Normal force on a pressed book A 1.2 kg book lies on a table. The book is pressed down from above with a force of 15 N. What is the normal force acting on the book from the table below? PREPARE The book is not moving and is thus in static equilibrium. We need to identify the forces acting on the book and prepare a free-body diagram showing these forces. These steps are illustrated in the figure.
43 Example 5.9: Normal force on a pressed book SOLVE Because the book is in static equilibrium, the net force on it must be zero. The only forces acting are in the y-direction, so Newton s second law is ΣF y = ma y
44 Example 5.9: Normal force on a pressed book SOLVE Because the book is in static equilibrium, the net force on it must be zero. The only forces acting are in the y-direction, so Newton s second law is ΣF y = ma y n w F =
45 Example 5.9: Normal force on a pressed book SOLVE Because the book is in static equilibrium, the net force on it must be zero. The only forces acting are in the y-direction, so Newton s second law is ΣF y = ma y n w F =
46 Example 5.9: Normal force on a pressed book SOLVE Because the book is in static equilibrium, the net force on it must be zero. The only forces acting are in the y-direction, so Newton s second law is ΣF y = ma y n w F = 0
47 Example 5.9: Normal force on a pressed book SOLVE Because the book is in static equilibrium, the net force on it must be zero. The only forces acting are in the y-direction, so Newton s second law is ΣF y = ma y n w F = 0
48 Example 5.9: Normal force on a pressed book SOLVE Because the book is in static equilibrium, the net force on it must be zero. The only forces acting are in the y-direction, so Newton s second law is ΣF y = ma y n w F = 0 n = w + F
49 Example 5.9: Normal force on a pressed book SOLVE Because the book is in static equilibrium, the net force on it must be zero. The only forces acting are in the y-direction, so Newton s second law is ΣF y = ma y n w F = 0 n = w + F We learned during last lecture that the weight force is w = mg, and we know what the force, F, is.
50 Example 5.9: Normal force on a pressed book The weight of the book is thus w = mg = (1.2 kg)(9.8 m/s 2 ) = 12 N
51 Example 5.9: Normal force on a pressed book The weight of the book is thus w = mg = (1.2 kg)(9.8 m/s 2 ) = 12 N With this information, we see that the normal force exerted by the table is n = w + F
52 Example 5.9: Normal force on a pressed book The weight of the book is thus w = mg = (1.2 kg)(9.8 m/s 2 ) = 12 N With this information, we see that the normal force exerted by the table is n = w + F = 12 N + 15 N = 27 N
53 Example 5.9: Normal force on a pressed book ASSESS The magnitude of the normal force is larger than the weight of the book. From the table s perspective, the extra force from the hand pushes the book further into the atomic springs of the table. These springs then push back harder, giving a normal force that is greater than the weight of the book.
54 Normal Forces
55 Normal Forces
56 Normal Forces
57 QuickCheck Question 5.2 The box is sitting on the floor of an elevator. The elevator is accelerating upward. The magnitude of the normal force on the box is n A. n > mg B. n = mg C. n < mg D. n = 0 E. Not enough information to tell
58 QuickCheck Question 5.2 The box is sitting on the floor of an elevator. The elevator is accelerating upward. The magnitude of the normal force on the box is n A. n > mg B. n = mg C. n < mg D. n = 0 E. Not enough information to tell Remember: apparent weight n = w + ma = wapp
59 QuickCheck Question 5.2 The box is sitting on the floor of an elevator. The elevator is accelerating upward. The magnitude of the normal force on the box is upward acceleration requires a net upward force n A. n > mg B. n = mg C. n < mg D. n = 0 E. Not enough information to tell Remember: apparent weight n = w + ma = wapp
60 QuickCheck Question 5.2 The box is sitting on the floor of an elevator. The elevator is accelerating upward. The magnitude of the normal force on the box is upward acceleration requires a net upward force n A. n > mg B. n = mg C. n < mg D. n = 0 E. Not enough information to tell Remember: apparent weight n = w + ma = wapp
61 QuickCheck Question 5.3 A box is being pulled to the right at steady speed by a rope that angles upward. In this situation: A. n > mg B. n = mg C. n < mg D. n = 0 E. Not enough information to judge the size of the normal force
62 QuickCheck Question 5.3 A box is being pulled to the right at steady speed by a rope that angles upward. In this situation: A. n > mg B. n = mg C. n < mg D. n = 0 E. Not enough information to judge the size of the normal force
63 QuickCheck Question 5.3 A box is being pulled to the right at steady speed by a rope that angles upward. In this situation: Remember: the towed car example A. n > mg n = w - T sin θ B. n = mg C. n < mg D. n = 0 E. Not enough information to judge the size of the normal force
64 QuickCheck Question 5.3 A box is being pulled to the right at steady speed by a rope that angles upward. In this situation: Remember: the towed car example A. n > mg n = w - T sin θ B. n = mg C. n < mg D. n = 0 E. Not enough information to judge the size of the normal force
65 Example 5.10: Acceleration of a downhill skier A skier slides down a steep 27 slope. On a slope this steep, friction is much smaller than the other forces at work and can be ignored. What is the skier s acceleration?
66 Example 5.10: Acceleration of a downhill skier A skier slides down a steep 27 slope. On a slope this steep, friction is much smaller than the other forces at work and can be ignored. What is the skier s acceleration? PREPARE We choose a coordinate system tilted so that the x-axis points down the slope. This greatly simplifies the analysis because with this choice a y = 0 (the skier doesn t move in the y-direction).
67 Example 5.10: Acceleration of a downhill skier A skier slides down a steep 27 slope. On a slope this steep, friction is much smaller than the other forces at work and can be ignored. What is the skier s acceleration? PREPARE We choose a coordinate system tilted so that the x-axis points down the slope. This greatly simplifies the analysis because with this choice a y = 0 (the skier doesn t move in the y-direction).
68 Example 5.10: Acceleration of a downhill skier SOLVE We can now use Newton s second law in component form to find the skier s acceleration: ΣF x = w x + n x = ma x ΣF y = w y + n y = ma y
69 Example 5.10: Acceleration of a downhill skier SOLVE Because the normal force, n, points directly in the positive y- direction, n y = n and n x = 0.
70 Example 5.10: Acceleration of a downhill skier SOLVE Because the normal force, n, points directly in the positive y- direction, n y = n and n x = 0. The angle between the weight, w, and the negative y-axis is the same as the slope angle θ. With this information, the components of w are w x = m g sin θ & w y = m g cos θ
71 Example 5.10: Acceleration of a downhill skier SOLVE With these components in hand, Newton s second law becomes ΣF x = mg sin θ = ma x ΣF y = mg cos θ + n = ma y = 0
72 Example 5.10: Acceleration of a downhill skier SOLVE With these components in hand, Newton s second law becomes ΣF x = mg sin θ = ma x ΣF y = mg cos θ + n = ma y = 0 The m cancels out in the first equation, leaving us with a x = g sinθ
73 Example 5.10: Acceleration of a downhill skier SOLVE With these components in hand, Newton s second law becomes ΣF x = mg sin θ = ma x ΣF y = mg cos θ + n = ma y = 0 The m cancels out in the first equation, leaving us with a x = g sinθ This is the expression for acceleration on a frictionless surface that we presented, without proof, in Chapter 3.
74 Example 5.10: Acceleration of a downhill skier SOLVE With these components in hand, Newton s second law becomes ΣF x = mg sin θ = ma x ΣF y = mg cos θ + n = ma y = 0 The m cancels out in the first equation, leaving us with a x = g sinθ This is the expression for acceleration on a frictionless surface that we presented, without proof, in Chapter 3. We can use this to calculate the skier s acceleration: a x = g sinθ = (9.8 m/s 2 ) sin 27 = 4.4 m/s 2
75 Example 5.10: Acceleration of a downhill skier ASSESS Our result shows that when θ = 0, so that the slope is horizontal, the skier s acceleration is zero, as it should be. Further, when θ = 90 (a vertical slope), his acceleration is g, which makes sense because he s in free fall when θ = 90. Notice that the mass canceled out, so we didn t need to know the skier s mass. We first saw the formula for the acceleration in Section 3.4, but now we see the physical reasons behind it.
76 What about Friction? Surfaces are very rough on a microscopic scale
77 What about Friction? Surfaces are very rough on a microscopic scale High points of surfaces become jammed against each other
78 What about Friction? Surfaces are very rough on a microscopic scale High points of surfaces become jammed against each other Amount of contact depends force between objects
79 What Types of Friction? Static - the force a surface exerts on an object to keep it from moving (stuck)
80 What Types of Friction? Static - the force a surface exerts on an object to keep it from moving (stuck) Kinetic - the force a surface exerts on an object in the opposite direction of the object s motion
81 What Types of Friction? Static - the force a surface exerts on an object to keep it from moving (stuck) Kinetic - the force a surface exerts on an object in the opposite direction of the object s motion Rolling - the force a surface exerts on an object rolling across it
82 Static Friction Denoted f s, points in opposite direction of pushing force
83 Static Friction Denoted f s, points in opposite direction of pushing force Object is in static equilibrium, Fpush = fs
84 Static Friction Denoted f s, points in opposite direction of pushing force Object is in static equilibrium, Fpush = fs Maximum magnitude, f s,max = µ s n
85 Static Friction, fs Maximum magnitude, f s,max = µ s n where, µ s, is the coefficient of static friction
86 Static Friction, fs Maximum magnitude, f s,max = µ s n where, µ s, is the coefficient of static friction Adjusts to oppose motion
87 Static Friction, fs Maximum magnitude, f s,max = µ s n where, µ s, is the coefficient of static friction Adjusts to oppose motion Cannot exceed f s,max
88 QuickCheck Question 5.7 A box on a rough surface is pulled by a horizontal rope with tension T. The box is not moving. In this situation: A. f s > T B. f s = T C. f s < T D. f s = µsmg E. f s = 0
89 QuickCheck Question 5.7 A box on a rough surface is pulled by a horizontal rope with tension T. The box is not moving. In this situation: Newton s second law ΣF = ma = 0 A. f s > T B. f s = T C. f s < T D. f s = µsmg E. f s = 0
90 QuickCheck Question 5.7 A box on a rough surface is pulled by a horizontal rope with tension T. The box is not moving. In this situation: Newton s second law ΣF = ma = 0 A. f s > T B. f s = T C. f s < T D. f s = µsmg E. f s = 0
91 Kinetic Friction, fk Nearly constant (doesn t adjust) magnitude f k = µ k n where µ k is called the coefficient of kinetic friction.
92 Kinetic Friction, fk Allows motion in the opposite direction, either at constant speed, or even acceleration ΣF = ma
93 Rolling Friction, fr A wheel rolling on a surface experiences friction on the portion that touches the surface; stationary, not sliding
94 Rolling Friction, fr A wheel rolling on a surface experiences friction on the portion that touches the surface; stationary, not sliding Interaction between a rolling wheel and the road can be quite complicated
95 Rolling Friction, fr A wheel rolling on a surface experiences friction on the portion that touches the surface; stationary, not sliding Interaction between a rolling wheel and the road can be quite complicated Friction force that opposes the motion, one defined by a coefficient of rolling friction µ r : f r = µ r n
96 Coefficients of Friction
97 Working With Friction
98 Working With Friction
99 QuickCheck Question 5.8 A box with a weight of 100 N is at rest. It is then pulled by a 30 N horizontal force. Does the box move? A. Yes B. No C. Not enough information to say
100 QuickCheck Question 5.8 A box with a weight of 100 N is at rest. It is then pulled by a 30 N horizontal force. Does the box move? A. Yes n = w = 100 N B. No C. Not enough information to say
101 QuickCheck Question 5.8 A box with a weight of 100 N is at rest. It is then pulled by a 30 N horizontal force. Does the box move? A. Yes B. No C. Not enough information to say n = w = 100 N f s,max = µ s n = 40 N
102 QuickCheck Question 5.8 A box with a weight of 100 N is at rest. It is then pulled by a 30 N horizontal force. Does the box move? A. Yes B. No C. Not enough information to say n = w = 100 N f s,max = µ s n = 40 N > 30 N
103 QuickCheck Question 5.9 A box is being pulled to the right over a rough surface. T > f k, so the box is speeding up. Suddenly the rope breaks. What happens? The box A. Stops immediately. B. Continues with the speed it had when the rope broke. C. Continues speeding up for a short while, then slows and stops. D. Keeps its speed for a short while, then slows and stops. E. Slows steadily until it stops.
104 QuickCheck Question 5.9 A box is being pulled to the right over a rough surface. T > f k, so the box is speeding up. Suddenly the rope breaks. What happens? The box A. Stops immediately. B. Continues with the speed it had when the rope broke. C. Continues speeding up for a short while, then slows and stops. D. Keeps its speed for a short while, then slows and stops. E. Slows steadily until it stops.
105 What a Drag! Like friction, it is in the opposite direction of the speed, v, of the object
106 What a Drag! Like friction, it is in the opposite direction of the speed, v, of the object Increases as the speed increases; usually ignored when the speed, v, is low
107 What a Drag! Like friction, it is in the opposite direction of the speed, v, of the object Increases as the speed increases; usually ignored when the speed, v, is low Simple model
108 Terminal Speed As an object falls, after some time it feels the Drag force
109 Terminal Speed As an object falls, after some time it feels the Drag force At some point during the fall, the Drag force is equal to the weight
110 Terminal Speed As an object falls, after some time it feels the Drag force At some point during the fall, the Drag force is equal to the weight Speed of object at this point is the terminal speed
111 Catalog of Forces
112 Newton s 3 rd and Interactions How do describe the motion of an object in contact with another?
113 Newton s 3 rd and Interactions How do describe the motion of an object in contact with another? We know about free body diagrams
114 Newton s 3 rd and Interactions How do describe the motion of an object in contact with another? We know about free body diagrams And we know about action/reaction pairs of forces
115 Newton s 3 rd and Interactions
116 QuickCheck Question 5.11 Consider the situation in the figure. Which pair of forces is an action/reaction pair? A. The tension of the string and the friction force acting on A B. The normal force on A due to B and the weight of A C. The normal force on A due to B and the weight of B D. The friction force acting on A and the friction force acting on B
117 QuickCheck Question 5.11 Consider the situation in the figure. Which pair of forces is an action/reaction pair? A. The tension of the string and the friction force acting on A B. The normal force on A due to B and the weight of A C. The normal force on A due to B and the weight of B D. The friction force acting on A and the friction force acting on B
118 Example 5.15: Pushing 2 Blocks The figure below shows a 5.0 kg block A being pushed with a 3.0 N force. In front of this block is a 10 kg block B; the two blocks move together. What force does block A exert on block B?
119 Example 5.15: Pushing 2 Blocks PREPARE The visual overview of the figure lists the known information and identifies F A on B as what we re trying to find. Following the steps of Tactics Box 5.2, we draw separate force identification diagrams and separate free-body diagrams for the two blocks.
120 Example 5.15: Pushing 2 Blocks The force is the contact force that block A exerts on B; it forms an action/reaction pair with the force that block B exerts on A. Notice that force is drawn acting on block B; it is the force of A on B. Force vectors are always drawn on the free-body diagram of the object that experiences the force, not the object exerting the force. Because action/reaction pairs act in opposite directions, force pushes backward on block A and appears on A s free-body diagram.
121 Example 5.15: Pushing 2 Blocks SOLVE We begin by writing Newton s second law in component form for each block. Because the motion is only in the x-direction, we need only the x-component of the second law. For block A, ΣF x = (F H ) x + (F B on A ) x = m A a Ax The force components can be read from the free-body diagram, where we see pointing to the right and pointing to the left. Thus Σ H F B on A = m A a Ax
122 Example 5.15: Pushing 2 Blocks For B, we have ΣF x = (F A on B ) x = F A on B = m B a Bx We have two additional pieces of information: First, Newton s third law tells us that F B on A = F A on B. Second, the boxes are in contact and must have the same acceleration a x ; that is, a Ax = a Bx = a x. With this information, the two x-component equations become F H F A on B = m A a x F A on B = m B a x
123 Example 5.15: Pushing 2 Blocks F H F A on B = m A a x F A on B = m B a x Our goal is to find F A on B, so we need to eliminate the unknown acceleration a x. From the second equation, a x = F A on B /m B. Substituting this into the first equation gives This can be solved for the force of block A on block B, giving
124 Example 5.15: Pushing 2 Blocks ASSESS Force F H accelerates both blocks, a total mass of 15 kg, but force F A on B accelerates only block B, with a mass of 10 kg. Thus it makes sense that F A on B < F H.
125 Interactions With Ropes Consider a box is pulled by a rope with tension, T
126 Interactions With Ropes Consider a box is pulled by a rope with tension, T Make the massless string approximation that m rope = 0
127 Interactions With Ropes Consider a box is pulled by a rope with tension, T Make the massless string approximation that m rope = 0 Newton s second law for the rope is thus ΣF x = F box on rope = F T = m rope a x = 0
128 Interactions With Ropes The tension, T, is the same from one end of the rope to the other
129 Interactions With Ropes The tension, T, is the same from one end of the rope to the other The pulling force does not change across the rope either
130 Interactions With Ropes The tension, T, is the same from one end of the rope to the other The pulling force does not change across the rope either The magnitude of the force pulling the string or rope is equal to the tension in the string
131 QuickCheck Question 5.12 Boxes A and B are being pulled to the right on a frictionless surface; the boxes are speeding up. Box A has a larger mass than Box B. How do the two tension forces compare? A. T 1 > T 2 B. T 1 = T 2 C. T 1 < T 2 D. Not enough information to tell
132 QuickCheck Question 5.12 Boxes A and B are being pulled to the right on a frictionless surface; the boxes are speeding up. Box A has a larger mass than Box B. How do the two tension forces compare? A. T 1 > T 2 B. T 1 = T 2 C. T 1 < T 2 D. Not enough information to tell
133 QuickCheck Question 5.13 Boxes A and B are sliding to the right on a frictionless surface. Hand H is slowing them. Box A has a larger mass than Box B. Considering only the horizontal forces: A. F B on H = F H on B = F A on B = F B on A B. F B on H = F H on B > F A on B = F B on A C. F B on H = F H on B < F A on B = F B on A D. F H on B = F H on A > F A on B
134 QuickCheck Question 5.13 Boxes A and B are sliding to the right on a frictionless surface. Hand H is slowing them. Box A has a larger mass than Box B. Considering only the horizontal forces: A. F B on H = F H on B = F A on B = F B on A B. F B on H = F H on B > F A on B = F B on A C. F B on H = F H on B < F A on B = F B on A D. F H on B = F H on A > F A on B
135 Summary
136 Summary
137 Summary
138 Summary
139 Summary
140 Summary
141 Summary
142 Summary
143 Summary
144 Summary
145 Things that are due Homework #4 Due September 23, 2015 by 11:59 pm Reading Quiz #5 Due September 17, 2015 by 4:59 pm Reading Quiz #6 Due September 22, 2015 by 4:59 pm
146 QUESTIONS?
Chapter 4. Forces and Newton s Laws of Motion. continued
Chapter 4 Forces and Newton s Laws of Motion continued 4.9 Static and Kinetic Frictional Forces When an object is in contact with a surface forces can act on the objects. The component of this force acting
v v ax v a x a v a v = = = Since F = ma, it follows that a = F/m. The mass of the arrow is unchanged, and ( )
Week 3 homework IMPORTANT NOTE ABOUT WEBASSIGN: In the WebAssign versions of these problems, various details have been changed, so that the answers will come out differently. The method to find the solution
Lecture 6. Weight. Tension. Normal Force. Static Friction. Cutnell+Johnson: 4.8-4.12, second half of section 4.7
Lecture 6 Weight Tension Normal Force Static Friction Cutnell+Johnson: 4.8-4.12, second half of section 4.7 In this lecture, I m going to discuss four different kinds of forces: weight, tension, the normal
5. Forces and Motion-I. Force is an interaction that causes the acceleration of a body. A vector quantity.
5. Forces and Motion-I 1 Force is an interaction that causes the acceleration of a body. A vector quantity. Newton's First Law: Consider a body on which no net force acts. If the body is at rest, it will
C B A T 3 T 2 T 1. 1. What is the magnitude of the force T 1? A) 37.5 N B) 75.0 N C) 113 N D) 157 N E) 192 N
Three boxes are connected by massless strings and are resting on a frictionless table. Each box has a mass of 15 kg, and the tension T 1 in the right string is accelerating the boxes to the right at a
Serway_ISM_V1 1 Chapter 4
Serway_ISM_V1 1 Chapter 4 ANSWERS TO MULTIPLE CHOICE QUESTIONS 1. Newton s second law gives the net force acting on the crate as This gives the kinetic friction force as, so choice (a) is correct. 2. As
Physics: Principles and Applications, 6e Giancoli Chapter 4 Dynamics: Newton's Laws of Motion
Physics: Principles and Applications, 6e Giancoli Chapter 4 Dynamics: Newton's Laws of Motion Conceptual Questions 1) Which of Newton's laws best explains why motorists should buckle-up? A) the first law
Physics 11 Assignment KEY Dynamics Chapters 4 & 5
Physics Assignment KEY Dynamics Chapters 4 & 5 ote: for all dynamics problem-solving questions, draw appropriate free body diagrams and use the aforementioned problem-solving method.. Define the following
Chapter 6. Work and Energy
Chapter 6 Work and Energy ENERGY IS THE ABILITY TO DO WORK = TO APPLY A FORCE OVER A DISTANCE= Example: push over a distance, pull over a distance. Mechanical energy comes into 2 forms: Kinetic energy
Conceptual Questions: Forces and Newton s Laws
Conceptual Questions: Forces and Newton s Laws 1. An object can have motion only if a net force acts on it. his statement is a. true b. false 2. And the reason for this (refer to previous question) is
A) N > W B) N = W C) N < W. speed v. Answer: N = W
CTN-12. Consider a person standing in an elevator that is moving upward at constant speed. The magnitude of the upward normal force, N, exerted by the elevator floor on the person's feet is (larger than/same
Newton s Law of Motion
chapter 5 Newton s Law of Motion Static system 1. Hanging two identical masses Context in the textbook: Section 5.3, combination of forces, Example 4. Vertical motion without friction 2. Elevator: Decelerating
LAB 6: GRAVITATIONAL AND PASSIVE FORCES
55 Name Date Partners LAB 6: GRAVITATIONAL AND PASSIVE FORCES And thus Nature will be very conformable to herself and very simple, performing all the great Motions of the heavenly Bodies by the attraction
Chapter 4: Newton s Laws: Explaining Motion
Chapter 4: Newton s Laws: Explaining Motion 1. All except one of the following require the application of a net force. Which one is the exception? A. to change an object from a state of rest to a state
Worksheet #1 Free Body or Force diagrams
Worksheet #1 Free Body or Force diagrams Drawing Free-Body Diagrams Free-body diagrams are diagrams used to show the relative magnitude and direction of all forces acting upon an object in a given situation.
Chapter 5 Using Newton s Laws: Friction, Circular Motion, Drag Forces. Copyright 2009 Pearson Education, Inc.
Chapter 5 Using Newton s Laws: Friction, Circular Motion, Drag Forces Units of Chapter 5 Applications of Newton s Laws Involving Friction Uniform Circular Motion Kinematics Dynamics of Uniform Circular
Objective: Equilibrium Applications of Newton s Laws of Motion I
Type: Single Date: Objective: Equilibrium Applications of Newton s Laws of Motion I Homework: Assignment (1-11) Read (4.1-4.5, 4.8, 4.11); Do PROB # s (46, 47, 52, 58) Ch. 4 AP Physics B Mr. Mirro Equilibrium,
Forces. When an object is pushed or pulled, we say that a force is exerted on it.
Forces When an object is pushed or pulled, we say that a force is exerted on it. Forces can Cause an object to start moving Change the speed of a moving object Cause a moving object to stop moving Change
LAB 6 - GRAVITATIONAL AND PASSIVE FORCES
L06-1 Name Date Partners LAB 6 - GRAVITATIONAL AND PASSIVE FORCES OBJECTIVES And thus Nature will be very conformable to herself and very simple, performing all the great Motions of the heavenly Bodies
VELOCITY, ACCELERATION, FORCE
VELOCITY, ACCELERATION, FORCE velocity Velocity v is a vector, with units of meters per second ( m s ). Velocity indicates the rate of change of the object s position ( r ); i.e., velocity tells you how
Lecture 7 Force and Motion. Practice with Free-body Diagrams and Newton s Laws
Lecture 7 Force and Motion Practice with Free-body Diagrams and Newton s Laws oday we ll just work through as many examples as we can utilizing Newton s Laws and free-body diagrams. Example 1: An eleator
Two-Body System: Two Hanging Masses
Specific Outcome: i. I can apply Newton s laws of motion to solve, algebraically, linear motion problems in horizontal, vertical and inclined planes near the surface of Earth, ignoring air resistance.
TEACHER ANSWER KEY November 12, 2003. Phys - Vectors 11-13-2003
Phys - Vectors 11-13-2003 TEACHER ANSWER KEY November 12, 2003 5 1. A 1.5-kilogram lab cart is accelerated uniformly from rest to a speed of 2.0 meters per second in 0.50 second. What is the magnitude
Physics. Lesson Plan #6 Forces David V. Fansler Beddingfield High School
Physics Lesson Plan #6 Forces David V. Fansler Beddingfield High School Force and Motion Objective Define a force and differentiate between contact forces and long-range forces; Recognize the significance
B Answer: neither of these. Mass A is accelerating, so the net force on A must be non-zero Likewise for mass B.
CTA-1. An Atwood's machine is a pulley with two masses connected by a string as shown. The mass of object A, m A, is twice the mass of object B, m B. The tension T in the string on the left, above mass
Steps to Solving Newtons Laws Problems.
Mathematical Analysis With Newtons Laws similar to projectiles (x y) isolation Steps to Solving Newtons Laws Problems. 1) FBD 2) Axis 3) Components 4) Fnet (x) (y) 5) Subs 1 Visual Samples F 4 1) F 3 F
HW Set II page 1 of 9 PHYSICS 1401 (1) homework solutions
HW Set II page 1 of 9 4-50 When a large star becomes a supernova, its core may be compressed so tightly that it becomes a neutron star, with a radius of about 20 km (about the size of the San Francisco
Recitation Week 4 Chapter 5
Recitation Week 4 Chapter 5 Problem 5.5. A bag of cement whose weight is hangs in equilibrium from three wires shown in igure P5.4. wo of the wires make angles θ = 60.0 and θ = 40.0 with the horizontal.
Resistance in the Mechanical System. Overview
Overview 1. What is resistance? A force that opposes motion 2. In the mechanical system, what are two common forms of resistance? friction and drag 3. What is friction? resistance that is produced when
AP1 Dynamics. Answer: (D) foot applies 200 newton force to nose; nose applies an equal force to the foot. Basic application of Newton s 3rd Law.
1. A mixed martial artist kicks his opponent in the nose with a force of 200 newtons. Identify the action-reaction force pairs in this interchange. (A) foot applies 200 newton force to nose; nose applies
Physics 111: Lecture 4: Chapter 4 - Forces and Newton s Laws of Motion. Physics is about forces and how the world around us reacts to these forces.
Physics 111: Lecture 4: Chapter 4 - Forces and Newton s Laws of Motion Physics is about forces and how the world around us reacts to these forces. Whats a force? Contact and non-contact forces. Whats a
MULTIPLE CHOICE. Choose the one alternative that best completes the statement or answers the question.
MULTIPLE CHOICE. Choose the one alternative that best completes the statement or answers the question. 1) Vector A has length 4 units and directed to the north. Vector B has length 9 units and is directed
When showing forces on diagrams, it is important to show the directions in which they act as well as their magnitudes.
When showing forces on diagrams, it is important to show the directions in which they act as well as their magnitudes. mass M, the force of attraction exerted by the Earth on an object, acts downwards.
Work Energy & Power. September 2000 Number 05. 1. Work If a force acts on a body and causes it to move, then the force is doing work.
PhysicsFactsheet September 2000 Number 05 Work Energy & Power 1. Work If a force acts on a body and causes it to move, then the force is doing work. W = Fs W = work done (J) F = force applied (N) s = distance
Work, Energy and Power Practice Test 1
Name: ate: 1. How much work is required to lift a 2-kilogram mass to a height of 10 meters?. 5 joules. 20 joules. 100 joules. 200 joules 5. ar and car of equal mass travel up a hill. ar moves up the hill
At the skate park on the ramp
At the skate park on the ramp 1 On the ramp When a cart rolls down a ramp, it begins at rest, but starts moving downward upon release covers more distance each second When a cart rolls up a ramp, it rises
Lecture 07: Work and Kinetic Energy. Physics 2210 Fall Semester 2014
Lecture 07: Work and Kinetic Energy Physics 2210 Fall Semester 2014 Announcements Schedule next few weeks: 9/08 Unit 3 9/10 Unit 4 9/15 Unit 5 (guest lecturer) 9/17 Unit 6 (guest lecturer) 9/22 Unit 7,
Chapter 7: Momentum and Impulse
Chapter 7: Momentum and Impulse 1. When a baseball bat hits the ball, the impulse delivered to the ball is increased by A. follow through on the swing. B. rapidly stopping the bat after impact. C. letting
AP Physics Applying Forces
AP Physics Applying Forces This section of your text will be very tedious, very tedious indeed. (The Physics Kahuna is just as sorry as he can be.) It s mostly just a bunch of complicated problems and
Physics 2A, Sec B00: Mechanics -- Winter 2011 Instructor: B. Grinstein Final Exam
Physics 2A, Sec B00: Mechanics -- Winter 2011 Instructor: B. Grinstein Final Exam INSTRUCTIONS: Use a pencil #2 to fill your scantron. Write your code number and bubble it in under "EXAM NUMBER;" an entry
while the force of kinetic friction is fk = µ
19. REASONING AND SOLUION We know that µ s =2.0µ k for a crate in contact with a MAX cement floor. he maximum force of static friction is fs = µ sfn while the force of kinetic friction is fk = µ kfn. As
Solution Derivations for Capa #11
Solution Derivations for Capa #11 1) A horizontal circular platform (M = 128.1 kg, r = 3.11 m) rotates about a frictionless vertical axle. A student (m = 68.3 kg) walks slowly from the rim of the platform
AP Physics - Chapter 8 Practice Test
AP Physics - Chapter 8 Practice Test Multiple Choice Identify the choice that best completes the statement or answers the question. 1. A single conservative force F x = (6.0x 12) N (x is in m) acts on
STATIC AND KINETIC FRICTION
STATIC AND KINETIC FRICTION LAB MECH 3.COMP From Physics with Computers, Vernier Software & Technology, 2000. INTRODUCTION If you try to slide a heavy box resting on the floor, you may find it difficult
WORK DONE BY A CONSTANT FORCE
WORK DONE BY A CONSTANT FORCE The definition of work, W, when a constant force (F) is in the direction of displacement (d) is W = Fd SI unit is the Newton-meter (Nm) = Joule, J If you exert a force of
Force. Force as a Vector Real Forces versus Convenience The System Mass Newton s Second Law. Outline
Force Force as a Vector Real Forces versus Convenience The System Mass Newton s Second Law Outline Force as a Vector Forces are vectors (magnitude and direction) Drawn so the vector s tail originates at
PHY231 Section 2, Form A March 22, 2012. 1. Which one of the following statements concerning kinetic energy is true?
1. Which one of the following statements concerning kinetic energy is true? A) Kinetic energy can be measured in watts. B) Kinetic energy is always equal to the potential energy. C) Kinetic energy is always
KE =? v o. Page 1 of 12
Page 1 of 12 CTEnergy-1. A mass m is at the end of light (massless) rod of length R, the other end of which has a frictionless pivot so the rod can swing in a vertical plane. The rod is initially horizontal
If you put the same book on a tilted surface the normal force will be less. The magnitude of the normal force will equal: N = W cos θ
Experiment 4 ormal and Frictional Forces Preparation Prepare for this week's quiz by reviewing last week's experiment Read this week's experiment and the section in your textbook dealing with normal forces
Chapter 6 Work and Energy
Chapter 6 WORK AND ENERGY PREVIEW Work is the scalar product of the force acting on an object and the displacement through which it acts. When work is done on or by a system, the energy of that system
WORKSHEET: KINETIC AND POTENTIAL ENERGY PROBLEMS
WORKSHEET: KINETIC AND POTENTIAL ENERGY PROBLEMS 1. Stored energy or energy due to position is known as Potential energy. 2. The formula for calculating potential energy is mgh. 3. The three factors that
Fundamental Mechanics: Supplementary Exercises
Phys 131 Fall 2015 Fundamental Mechanics: Supplementary Exercises 1 Motion diagrams: horizontal motion A car moves to the right. For an initial period it slows down and after that it speeds up. Which of
PHYSICS 111 HOMEWORK SOLUTION, week 4, chapter 5, sec 1-7. February 13, 2013
PHYSICS 111 HOMEWORK SOLUTION, week 4, chapter 5, sec 1-7 February 13, 2013 0.1 A 2.00-kg object undergoes an acceleration given by a = (6.00î + 4.00ĵ)m/s 2 a) Find the resultatnt force acting on the object
2 Newton s First Law of Motion Inertia
2 Newton s First Law of Motion Inertia Conceptual Physics Instructor Manual, 11 th Edition SOLUTIONS TO CHAPTER 2 RANKING 1. C, B, A 2. C, A, B, D 3. a. B, A, C, D b. B, A, C, D 4. a. A=B=C (no force)
Physics 201 Homework 8
Physics 201 Homework 8 Feb 27, 2013 1. A ceiling fan is turned on and a net torque of 1.8 N-m is applied to the blades. 8.2 rad/s 2 The blades have a total moment of inertia of 0.22 kg-m 2. What is the
PHY231 Section 1, Form B March 22, 2012
1. A car enters a horizontal, curved roadbed of radius 50 m. The coefficient of static friction between the tires and the roadbed is 0.20. What is the maximum speed with which the car can safely negotiate
The Big Idea. Key Concepts
The Big Idea Acceleration is caused by force. All forces come in pairs because they arise in the interaction of two objects you can t hit without being hit back! The more force applied, the greater the
P211 Midterm 2 Spring 2004 Form D
1. An archer pulls his bow string back 0.4 m by exerting a force that increases uniformly from zero to 230 N. The equivalent spring constant of the bow is: A. 115 N/m B. 575 N/m C. 1150 N/m D. 287.5 N/m
Chapter 11 Equilibrium
11.1 The First Condition of Equilibrium The first condition of equilibrium deals with the forces that cause possible translations of a body. The simplest way to define the translational equilibrium of
F N A) 330 N 0.31 B) 310 N 0.33 C) 250 N 0.27 D) 290 N 0.30 E) 370 N 0.26
Physics 23 Exam 2 Spring 2010 Dr. Alward Page 1 1. A 250-N force is directed horizontally as shown to push a 29-kg box up an inclined plane at a constant speed. Determine the magnitude of the normal force,
B) 286 m C) 325 m D) 367 m Answer: B
Practice Midterm 1 1) When a parachutist jumps from an airplane, he eventually reaches a constant speed, called the terminal velocity. This means that A) the acceleration is equal to g. B) the force of
Name Period WORKSHEET: KINETIC AND POTENTIAL ENERGY PROBLEMS. 1. Stored energy or energy due to position is known as energy.
Name Period Date WORKSHEET: KINETIC AND POTENTIAL ENERGY PROBLEMS 1. Stored energy or energy due to position is known as energy. 2. The formula for calculating potential energy is. 3. The three factors
Review Vocabulary force: a push or a pull. Vocabulary Newton s third law of motion
Standard 7.3.17: Investigate that an unbalanced force, acting on an object, changes its speed or path of motion or both, and know that if the force always acts toward the same center as the object moves,
CHAPTER 6 WORK AND ENERGY
CHAPTER 6 WORK AND ENERGY CONCEPTUAL QUESTIONS. REASONING AND SOLUTION The work done by F in moving the box through a displacement s is W = ( F cos 0 ) s= Fs. The work done by F is W = ( F cos θ). s From
W02D2-2 Table Problem Newton s Laws of Motion: Solution
ASSACHUSETTS INSTITUTE OF TECHNOLOGY Department of Physics Physics 8.01 W0D- Table Problem Newton s Laws of otion: Solution Consider two blocks that are resting one on top of the other. The lower block
NEWTON S LAWS OF MOTION
Name Period Date NEWTON S LAWS OF MOTION If I am anything, which I highly doubt, I have made myself so by hard work. Isaac Newton Goals: 1. Students will use conceptual and mathematical models to predict
1 of 7 9/5/2009 6:12 PM
1 of 7 9/5/2009 6:12 PM Chapter 2 Homework Due: 9:00am on Tuesday, September 8, 2009 Note: To understand how points are awarded, read your instructor's Grading Policy. [Return to Standard Assignment View]
9. The kinetic energy of the moving object is (1) 5 J (3) 15 J (2) 10 J (4) 50 J
1. If the kinetic energy of an object is 16 joules when its speed is 4.0 meters per second, then the mass of the objects is (1) 0.5 kg (3) 8.0 kg (2) 2.0 kg (4) 19.6 kg Base your answers to questions 9
Conservative vs. Non-conservative forces Gravitational Potential Energy. Work done by non-conservative forces and changes in mechanical energy
Next topic Conservative vs. Non-conservative forces Gravitational Potential Energy Mechanical Energy Conservation of Mechanical energy Work done by non-conservative forces and changes in mechanical energy
PHYS 211 FINAL FALL 2004 Form A
1. Two boys with masses of 40 kg and 60 kg are holding onto either end of a 10 m long massless pole which is initially at rest and floating in still water. They pull themselves along the pole toward each
Friction and Gravity. Friction. Section 2. The Causes of Friction
Section 2 Friction and Gravity What happens when you jump on a sled on the side of a snow-covered hill? Without actually doing this, you can predict that the sled will slide down the hill. Now think about
force (mass)(acceleration) or F ma The unbalanced force is called the net force, or resultant of all the forces acting on the system.
4 Forces 4-1 Forces and Acceleration Vocabulary Force: A push or a pull. When an unbalanced force is exerted on an object, the object accelerates in the direction of the force. The acceleration is proportional
Physics 125 Practice Exam #3 Chapters 6-7 Professor Siegel
Physics 125 Practice Exam #3 Chapters 6-7 Professor Siegel Name: Lab Day: 1. A concrete block is pulled 7.0 m across a frictionless surface by means of a rope. The tension in the rope is 40 N; and the
Review Chapters 2, 3, 4, 5
Review Chapters 2, 3, 4, 5 4) The gain in speed each second for a freely-falling object is about A) 0. B) 5 m/s. C) 10 m/s. D) 20 m/s. E) depends on the initial speed 9) Whirl a rock at the end of a string
Problem Set #8 Solutions
MASSACHUSETTS INSTITUTE OF TECHNOLOGY Physics Department 8.01L: Physics I November 7, 2015 Prof. Alan Guth Problem Set #8 Solutions Due by 11:00 am on Friday, November 6 in the bins at the intersection
Chapter 3.8 & 6 Solutions
Chapter 3.8 & 6 Solutions P3.37. Prepare: We are asked to find period, speed and acceleration. Period and frequency are inverses according to Equation 3.26. To find speed we need to know the distance traveled
Reading quiz: In the SI (metric) system, the unit of force is the.. A) galileo B) joule C) newton D) eiffel E) none of these
Reading quiz: In the SI (metric) system, the unit of force is the.. A) galileo B) joule C) newton D) eiffel E) none of these 5-1: An astronaut floating weightlessly in orbit shakes a large iron anvil rapidly
Laboratory Report Scoring and Cover Sheet
Laboratory Report Scoring and Cover Sheet Title of Lab _Newton s Laws Course and Lab Section Number: PHY 1103-100 Date _23 Sept 2014 Principle Investigator _Thomas Edison Co-Investigator _Nikola Tesla
Newton s Laws. Physics 1425 lecture 6. Michael Fowler, UVa.
Newton s Laws Physics 1425 lecture 6 Michael Fowler, UVa. Newton Extended Galileo s Picture of Galileo said: Motion to Include Forces Natural horizontal motion is at constant velocity unless a force acts:
Kinetic Friction. Experiment #13
Kinetic Friction Experiment #13 Joe Solution E01234567 Partner- Jane Answers PHY 221 Lab Instructor- Nathaniel Franklin Wednesday, 11 AM-1 PM Lecture Instructor Dr. Jacobs Abstract The purpose of this
Experiment: Static and Kinetic Friction
PHY 201: General Physics I Lab page 1 of 6 OBJECTIVES Experiment: Static and Kinetic Friction Use a Force Sensor to measure the force of static friction. Determine the relationship between force of static
Name Class Period. F = G m 1 m 2 d 2. G =6.67 x 10-11 Nm 2 /kg 2
Gravitational Forces 13.1 Newton s Law of Universal Gravity Newton discovered that gravity is universal. Everything pulls on everything else in the universe in a way that involves only mass and distance.
AP Physics C Fall Final Web Review
Name: Class: _ Date: _ AP Physics C Fall Final Web Review Multiple Choice Identify the choice that best completes the statement or answers the question. 1. On a position versus time graph, the slope of
Tennessee State University
Tennessee State University Dept. of Physics & Mathematics PHYS 2010 CF SU 2009 Name 30% Time is 2 hours. Cheating will give you an F-grade. Other instructions will be given in the Hall. MULTIPLE CHOICE.
University Physics 226N/231N Old Dominion University. Getting Loopy and Friction
University Physics 226N/231N Old Dominion University Getting Loopy and Friction Dr. Todd Satogata (ODU/Jefferson Lab) satogata@jlab.org http://www.toddsatogata.net/2012-odu Friday, September 28 2012 Happy
LeaPS Workshop March 12, 2010 Morehead Conference Center Morehead, KY
LeaPS Workshop March 12, 2010 Morehead Conference Center Morehead, KY Word Bank: Acceleration, mass, inertia, weight, gravity, work, heat, kinetic energy, potential energy, closed systems, open systems,
5.1 The First Law: The Law of Inertia
The First Law: The Law of Inertia Investigation 5.1 5.1 The First Law: The Law of Inertia How does changing an object s inertia affect its motion? Newton s first law states that objects tend to keep doing
SHORT ANSWER. Write the word or phrase that best completes each statement or answers the question.
Exam Name SHORT ANSWER. Write the word or phrase that best completes each statement or answers the question. 1) A person on a sled coasts down a hill and then goes over a slight rise with speed 2.7 m/s.
Newton s Laws of Motion
Newton s Laws of Motion The Earth revolves around the sun in an elliptical orbit. The moon orbits the Earth in the same way. But what keeps the Earth and the moon in orbit? Why don t they just fly off
FRICTION, WORK, AND THE INCLINED PLANE
FRICTION, WORK, AND THE INCLINED PLANE Objective: To measure the coefficient of static and inetic friction between a bloc and an inclined plane and to examine the relationship between the plane s angle
AP Physics C. Oscillations/SHM Review Packet
AP Physics C Oscillations/SHM Review Packet 1. A 0.5 kg mass on a spring has a displacement as a function of time given by the equation x(t) = 0.8Cos(πt). Find the following: a. The time for one complete
A Determination of g, the Acceleration Due to Gravity, from Newton's Laws of Motion
A Determination of g, the Acceleration Due to Gravity, from Newton's Laws of Motion Objective In the experiment you will determine the cart acceleration, a, and the friction force, f, experimentally for
Work, Energy & Momentum Homework Packet Worksheet 1: This is a lot of work!
Work, Energy & Momentum Homework Packet Worksheet 1: This is a lot of work! 1. A student holds her 1.5-kg psychology textbook out of a second floor classroom window until her arm is tired; then she releases
AP Physics 1 Midterm Exam Review
AP Physics 1 Midterm Exam Review 1. The graph above shows the velocity v as a function of time t for an object moving in a straight line. Which of the following graphs shows the corresponding displacement
Ch 7 Kinetic Energy and Work. Question: 7 Problems: 3, 7, 11, 17, 23, 27, 35, 37, 41, 43
Ch 7 Kinetic Energy and Work Question: 7 Problems: 3, 7, 11, 17, 23, 27, 35, 37, 41, 43 Technical definition of energy a scalar quantity that is associated with that state of one or more objects The state
Exam 2 is at 7 pm tomorrow Conflict is at 5:15 pm in 151 Loomis
* By request, but I m not vouching for these since I didn t write them Exam 2 is at 7 pm tomorrow Conflict is at 5:15 pm in 151 Loomis There are extra office hours today & tomorrow Lots of practice exams
Newton s Second Law. ΣF = m a. (1) In this equation, ΣF is the sum of the forces acting on an object, m is the mass of
Newton s Second Law Objective The Newton s Second Law experiment provides the student a hands on demonstration of forces in motion. A formulated analysis of forces acting on a dynamics cart will be developed
TIME OF COMPLETION DEPARTMENT OF NATURAL SCIENCES. PHYS 1111, Exam 2 Section 1 Version 1 October 30, 2002 Total Weight: 100 points
TIME OF COMPLETION NAME DEPARTMENT OF NATURAL SCIENCES PHYS 1111, Exam 2 Section 1 Version 1 October 30, 2002 Total Weight: 100 points 1. Check your examination for completeness prior to starting. There
4 Gravity: A Force of Attraction
CHAPTER 1 SECTION Matter in Motion 4 Gravity: A Force of Attraction BEFORE YOU READ After you read this section, you should be able to answer these questions: What is gravity? How are weight and mass different?