Forces & Newton s Laws. Teacher Packet

Transcription

1 AP * PHYSICS B Forces & Newton s Laws eacher Packet AP* is a trademark of the College Entrance Examination Board. he College Entrance Examination Board was not involved in the production of this material.

2 Objective o review the student on the concepts, processes and problem solving strategies necessary to successfully answer questions on forces and Newton s laws. Standards Forces and Newton s laws are addressed in the topic outline of the College Board AP* Physics Course Description Guide as described below. I. Newtonian Mechanics B. Newton s laws of motion (including friction and centripetal force) 1. Static equilibrium (first law). Dynamics of a single particle (second law) 3. Systems of two or more bodies (third law) F. Oscillations and Gravitation 4. Newton s law of gravity AP Physics Exam Connections opics relating to forces and Newton s laws are tested every year on the multiple choice and in most years on the free response portion of the exam. he list below identifies free response questions that have been previously asked over forces and Newton s laws. hese questions are available from the College Board and can be downloaded free of charge from AP Central. Free Response Questions 008 Question 008 Form B Question 007 Question Form B Question Question 006 Form B Question Question Form B Question Question Form B Question Question Question AP* is a trademark of the College Entrance Examination Board. he College Entrance Examination Board was not involved in the production of this material.

3 What I Absolutely Have to Know to Survive the AP* Exam Force is any push or pull. It is a vector. Newton s Second Law is the workhorse of the AP Physics B exam. It allows you to write down mathematical relationships that are true. hus, for a single body, if you pick any direction and sum up all the positive and negative forces that act on the body along that line, the sum will equal the product of the body s mass and its acceleration along that line. A Free Body Diagram allows you to identify all of the forces acting on a single body. Neglect one force or add a fictitious force on your FBD and you are in trouble. Newton s 1 st Law: in an inertial frame of reference, an object in a state of constant velocity (including zero velocity) will continue in that state unless impinged upon by a net external force. If ΣF=0, then a=0 and the object is at rest or moving at a constant velocity in a straight line. he converse is true also, if an object is in a state of constant velocity (including zero velocity) then a=0 and ΣF=0. Newton s nd Law: A net force acting on a mass causes that mass to accelerate in the direction of the net force. he acceleration (vector) is directly proportional to the net force (vector) acting on the mass and ΣF inversely proportional to the mass of the object being accelerated. a= or Σ F = ma m Newton s 3 rd Law: For every action force, there exists an equal and opposite reaction force. Let s say you hit a table with your fist. Doing so, applies a force to the table which, if great enough will break the table. Likewise, the table applies a force to your fist which, if great enough, will break your fist. he size and direction of the force you apply must be equal and opposite the force the table applies to you. Hence, the only way forces can be generated are in action/reaction pairs which occur on different objects. If you try to apply 800 Newtons of force to a table that can only provide 600 Newtons of reaction force back on you, you will never succeed. he table will break as soon as you exceed 600 Newtons, which is the maximum force it can apply to you. Key Formulas and Relationships Σ F= Fnet = ma FW = mg Ffsmax μsfn ( static) Ffk = μkfn ( kinetic) Gm1m FG = r kgm ΣF = Sum of the forces is the Net Force Newtons (N) = s a = acceleration m = mass F W = weight g = acceleration due to gravity F fsmax = maximum static frictional force F f k = kinetic frictional force F N = normal force F G = gravitational force r = distance between the centers of two masses AP* is a trademark of the College Entrance Examination Board. he College Entrance Examination Board was not involved in the production of this material.

4 Weight Gravitational F g, W F G Fg = mg Basic Kinds of Forces Always directed toward the Earth s center. Force on a free falling body, if we neglect air friction. mm 1 FG = G r A force of attraction between any two massive objects. When the Earth is one of the two bodies involved, then the force felt by the second body while positioned on the Earth s surface will always be directed toward the Earth s center. A force of support, provided to an object by a surface in which the object is in contact. Always directed perpendicular to and away from the surface providing the support. F N Normal F N, N W In the figure above, a box is supported by a table. he figure shows all the forces acting on the box and is called a Free Body Diagram (FBD). If a box, rests on a level table, then the FN = W = mg. Notice that the normal force sometimes equals the weight but not always. y F N x θ mgcosθ W If the box is placed on an inclined plane, then the FN = mgcosθ, the component of the weight that is equal and opposite the normal force. For the inclined plane above, the normal force and the weight are not equal and not even in the same direction.

5 Friction is produced by the atomic interaction between two bodies as they either slide over one another (kinetic friction) or sit motionless in contact with one another (static friction). Ff smax μ sf N (static) Static Friction opposes the intended direction of relative sliding. he static frictional force will only be as high as it needs to be to keep the system in equilibrium. If successively greater and greater forces are applied, the static frictional force will counter each push with a force of equal and opposite magnitude until the applied force is great enough to shear the bonding between the two surfaces. When you calculate f s or Ff s using the equation above, you are finding the maximum static frictional force, one of an infinite number of possible frictional forces that could be exerted between the two bodies. μs is a proportionality constant called the coefficient of static friction. It is the ratio of the static frictional force between the surfaces divided by the normal force acting on the surface. Friction Ff, f Ffk = μkfn (kinetic) Kinetic Friction or Dynamic Friction or Sliding Friction is always opposite the direction of motion. he statement that kinetic friction is a function of the normal force only (surface area is independent) is true only when dealing with rigid bodies that are sliding relative to each other. When you calculate f k or Ff using the equation above, you are finding the single, constant kinetic frictional force that exists between the two bodies sliding relative to one another. No matter their velocity (assuming heating does not alter the coefficient of kinetic friction) the kinetic frictional force will always be the same. μk is a proportionality constant called the coefficient of kinetic friction. It is the ratio of the kinetic frictional force between the surfaces divided by the normal force acting on the surface WARNING! he two quantities f s and f k may look the same, but they tell us different things. Kinetic fric tion is ty pically less than static friction for the same two surfaces in contact. Note that the normal force sometimes equals the weight but not always. When you draw a free body diagram of forces acting on an object or system of objects, be sure to include the frictional force as opposing the relative motion (or potential for relative motion) of the two surfaces in contact.

6 ension Applied F, F Subscript F is a force that is applied to a body by a rope, string, or cable. F is applied along the line of the string and away from the body in question. Push me, pull you force that does not fall into one of the above categories, for example, a friend shoves you. he magnitude of the force is characterized by an F, with any subscript that makes sense to solve the problem. Later in the year, you will encounter additional forces, like the electric and magnetic forces. Strategy on Force Problems 1. ake one body in the system and draw a Free Body Diagram (FBD) for it.. Choose x and y axes and place them beside your FBD. One axis must be in the direction of the acceleration you are trying to find. If there is no acceleration, then Σ F = If there are forces on the FBD that are not along the x and y directions, find their respective x and y components. 4. Using Newton s nd Law, sum the forces in the x direction and set them equal to ma x. If a second equation is needed, sum the forces in the y direction and set them equal to ma. y 5. Repeat the above process for all the bodies in the system or until you have the same number of equations as unknowns and solve the problem. Effective Problem Solving Strategies Free Body Diagram (FBD) A Free Body Diagram is normally depicted as a box showing all the forces acting on the body. hese forces are depicted as arrows. hey don t have to be drawn to scale, but they should have a length that is appropriate for their magnitude. Also, the force vectors do need to be directionally accurate and labeled. Do not include components of the force vectors on your FBD. When drawing Free Body Diagrams show only the force(s) that act on the body in question and do not show forces that the body applies to other bodies. Also, do not include velocity or acceleration vectors on your free body force diagrams, since you will lose points for extraneous vectors on your FBD.

7 Example 1 A rope supports an empty bucket of mass 3.0 kg. Determine the tension in the rope when the bucket is (a) at rest and (b) the bucket is accelerated upward at.0 m/s. Solution Step 1 Draw a Free Body Diagram for each body in the system. F W Step Choose x and y axes. In this case we could choose the y axes to be in the same dimension as the tension and the weight. Step 3 If there are off axes forces, then find the x and y components. Step 4 Using Newton s nd law, sum the forces in the y dimension and set equal to ma y. (a) When the bucket is at rest, the net force is zero, so that the tension in the rope equals the weight of the bucket. Σ F = ma F y y mg = 0 m F = mg = ( 3.0kg) 10 = 30N s (b) When the bucket accelerates upward, the net force is ma y.

8 Σ F = ma y F mg = ma y ( ) F = mg+ ma = m g+ a m m F = 3.0kg N upward = s s Example In the diagram below, two bodies of different masses (M 1 and M ) are connected by a string which passes over a pulley of negligible mass and friction. What is the magnitude of the acceleration of the system in terms of the given quantities and fundamental constants? M 1 M Solution Step 1: Draw a FBD for each body in the system. here are two forces acting on each of the bodies: weight downward and the tension in the string upward. he tension is distributed throughout the string. he pulley (negligible mass and friction) changes only the direction of motion, not the tension, so the tension is the same on each side of the pulley. Our FBDs should look like this:

9 W 1 W Step Choose x and y axes. In this case we could choose the y axes to be in the same dimension as the tension and the weight for both masses. Step 3 If there are off axes forces, then find the x and y components. Step 4 Using Newton s nd law, for each body sum the forces in the y dimension and set equal to ma y and/or the x dimension. For m 1 the tension is positive and the weight is negative since the acceleration is upward. Σ F = ma y 1 F W = ma y 1 1 For m the tension is negative and the weight is positive since the acceleration is downward. Σ F = m a y F + W = m a y

10 F W = ma 1 1 F + W = m a ( ) W W = m + m a a = = = ( ) ( ) ( 1) ( ) W W m g m g m m g m + m m + m m + m Example 3 A block of mass m rests on a horizontal table. A string is tied to the block, passed over a pulley, and another block of mass M is hung on the other end of the string, as shown in the figure below. he coefficient of kinetic friction between block m and the table is μ k. Find the magnitude of the acceleration of the system in terms of the given quantities and fundamental constants. μ k m M

11 Solution Step 1: Draw a FBD for each body. Note that according to the FBD, the vertical acceleration of the block on the table is zero, since the normal force is directed upward and the weight force is directed downward. here is a horizontal acceleration for the block since the tension is greater than the frictional force. F N f k W 1 W Step Choose x and y axes. In this case we could choose the y axes to be in the same dimension as the tension and the weight for both masses. We could choose the x axis to be in the same dimension as the frictional force and the tension pulling the mass to the right. Step 3 If there are off axes forces, then find the x and y components. Step 4 Using Newton s nd law, for each body sum the forces in the y dimension and set equal to ma y and/or the x dimension. For block m the net force vertically is 0 since the block is accelerating to the right and not upward nor downward. Summing the forces in the y dimension we find that the normal force equals the weight of the block. Σ Fy = may F W = 0 N 1 F = W = mg N 1 Summing the forces in the horizontal dimension, the direction of the acceleration for block m, we find that

12 Σ F = ma x f = ma k x For mass two, summing the forces in the vertical dimension, the vertical acceleration for block M, we find that Σ F = M a y + W = Ma y Adding the two equations together we determine the acceleration of the system f = ma k + W = Ma k ( ) W f = m+ M a a W f Mg μ mg k = = a = ( m+ M) ( m+ M) ( M μkm) g ( m+ M) Since f = μ F = μ mg k k N k k

13 Example 4 hree blocks of mass m 1, m, and m 3 are connected by a string passing over a pulley attached to a plane inclined at an angle θ as shown below. m m 1 m 3` θ θ he coefficient of kinetic friction between blocks m, m 1 and the table is μ k. Assuming that m 3 is large enough to descend and cause the system to accelerate, determine the acceleration of the system in terms of the given quantities and fundamental constants. Solution Step 1: Draw a FBD for each body. When drawing a FBD, be sure the orientation of the box representing the body is the same as the actual orientation of the body in the problem. Note the weight force is always directed straight downward and the normal force perpendicular to the surface.

14 F N 1 1 F N1 f W f W 1 W 3 Step Choose x and y axes. Place axes next to FBD. One axis must be in the direction of the acceleration you are trying to determine. Step 3 If there are off axes forces, then find the x and y components. here is one force that is off axis, the weight force. he weight of each block must be broken into two vectors parallel and perpendicular to the inclined plane. m 1 x1 1 y1 1 the parallel and perpendicular components of the weight force are w = F = m gsinθ w = F = m gcosθ m the parallel and perpendicular components of the weight force are w = F = m gsinθ w x = F = m gcosθ y Step 4 Using Newton s nd law, for each body, sum the forces in the y dimension and set equal to ma y. Sum the forces in the x dimension and set equal to ma x.

15 m 1 vertical motion perpendicular to the incline plane Σ F = 0 F F N N y m gcosθ = 0 1 = m gcosθ 1 horizontal motion parallel to the incline plane Σ F = ma x 1 f F = ma 1 k 1 μ m gcosθ m gsinθ = ma 1 k m vertical motion perpendicular to the incline plane Σ F = 0 F F N N y m gcosθ = 0 = m gcosθ horizontal motion parallel to the incline plane Σ F = ma x f F = m a 1 k μ mgcosθ mgsinθ = ma 1 k m 3 vertical motion Σ F = ma y 3 + W = m a m g = m a 3 3 Looking at the motion along the axis of acceleration, up the incline plane

16 μ m gcosθ m gsinθ = ma 1 k μ mgcosθ mgcosθ = ma 1 k k ( ) ( ) ( μ cosθ sinθ)( ) μ gcosθ m + m gsinθ m + m = ma+ m a g + g m + m = ma+ m a k m g = m a 3 3 = m g m a 3 3 ( μk cosθ sinθ)( ) ( μk cosθ sinθ)( ) ( ( μk cosθ + sinθ)( + ) ( m + m + m ) mg ma g + g m + m = ma+ ma m g g + g m + m = m + m + m mg g g m m a = ) a

17 Free Response Question 1 (15 pts) wo blocks, both of mass 0.5 kg, are connected to each other by a thin string which is passed over a pulley as shown in the diagram. Block 1 sits on a rough horizontal part where the coefficient of static friction is 0.6 and the coefficient of kinetic friction is 0.4. Block sits on a frictionless incline which forms an angle θ with the horizontal. 1 θ A. On the diagram below, draw and label vectors to represent the forces acting on block. (3 points max) N 1 point for each correctly drawn and labeled vector (tension, normal, weight) 1 point deducted for each extraneous vector with a minimum score being 0 W

18 B. Determine the maximum angle θ the incline can make without the blocks sliding down. (6 points max) 1 point for a correct statement that When block is not sliding, block 1 is not when the blocks are not sliding the sliding and the ΣF=0, so the maximum ΣF=0 angle means maximal frictional force. Block Σ F = 0 F = 0 = mgsinθ Block 1 Σ F = 0 f = 0 hus s N s max = f = μ F μ F s max s N = mgsinθ μ mg = mg sinθ s sinθ = μ = 0.6 θ = ( ) 1 sin 0.6 θ = 37 s 1 point for a correct application of NSL to block such that the tension equals the parallel component of the weight 1 point for a correct application of NSL to block 1 such that the tension equals the maximum frictional force 1 point for a correct statement that the maximum angle occurs when the frictional force is maximum on block 1 using 0.6 for the coefficient of friction 1 point for a correct statement that the maximum frictional force on block 1 equals the parallel component of the weight on block 1 point for the correct answer including correct units and reasonable number of significant digits

19 C. Assume θ is 30º, find the magnitude of the acceleration of the system (6 points max) Block 1 Σ F = ma f = ma k μ mg = ma k Block Σ F = ma + F = ma + mgsinθ = ma μ mg = ma k + mgsinθ = ma mg sinθ μ mg = ma k gsinθ μk g = a m m 10 ( sin 30 ) ( 0.4) 10 s a s = m a = 0.5 s 1 point for any indication that there are two forces acting on Block 1 parallel to the surface 1 point for a correct application of NSL to block 1 that includes the two forces (tension and friction) and a non zero acceleration 1 point for a correct statement that the frictional force is kinetic and using 0.4 for the coefficient of friction 1 point for any indication that there are two forces acting on Block parallel to the surface 1 point for a correct application of NSL to block that includes the two forces (tension and parallel component of the weight) and a non zero acceleration 1 point for the correct answer including correct units and reasonable number of significant digits

20 Question (10 pts) A block of mass m rests on an air table (no friction), and is pulled with a force probe, producing the Force vs. Acceleration graph shown below. A. Determine the mass of the block. Newton s nd law states that ratio is the slope. Fnet m =. his a 1 point for using the slope of the F vs. a graph to determine the mass or the correct use of Newton s second law ΔF 9N 3N m = slope = = = 1.5 kg Δa 6m/s m/s 1 point for the correct answer including correct units and reasonable number of significant digits he block is now placed on a rough horizontal board having a coefficient of static friction μ s = 0., and a coefficient of kinetic (sliding) friction μ k = 0.1.

21 B. What is the minimum value of the force F which will cause the block to just begin to move? he block will just begin to move when the force F overcomes the maximum static frictional force: Fmin = fs = μsfn = μsmg Fmin = kg 10 m/s = 3 N ( )( )( ) 1 point for the correct equation for the maximum static frictional force 1 point for the correct answer or an answer consistent with part (A), including correct units and reasonable number of significant digits he block rests on the rough horizontal board. One end of the board is slowly lifted until the block just begins to slide down the board. At the instant the block begins to slide, the angle of inclination for the board is θ. C. Determine the relationship between the angle θ and the coefficient of static friction, μ s. (6 points max) 1 point for a correct statement of net force parallel to the plane is zero or FBD indicating the same At the instant the block is just about to move, the maximum frictional force directed up the incline is equal and opposite to the parallel component of the weight down the incline, and the normal force is equal and opposite to the perpendicular component of the weight. Σ Fx = 0 fs max F = 0 f = mgsinθ s max Σ Fy = 0 FN F = 0 F = mgcosθ N 1 point for a correct statement of net force perpendicular to the plane is zero or FBD indicating the same 1 point for a correct statement that the maximum static frictional force equals the parallel component of the weight or FBD indicating the same 1 point for a correct statement that the normal force equals the perpendicular component of the weight or FBD indicating the same hus μ f mg sinθ s max s = = = FN mgcosθ tanθ 1 point for a correct statement that μ s is the ratio between the parallel and perpendicular components of the weight 1 point for the correct relationship between θ and μ s

22 Questions 1 Multiple Choice m m A block of mass m is suspended by a string, the other end of which is passed over a pulley of negligible mass and friction and tied to a block of mass m. 1. he acceleration of the system is a) 1 3 g b) 1 g c) 3 g d) g e) 4 3 g ΣF = ma Find net force on object mg mg he forces acting on each block are shown above. From this, we write two net force equations, one for each block (the smaller block is block 1) and apply ΣF = ma. F = mg = ma 1 F = mg = ma hese equations can be added together, eliminating. ma = mg + ma = mg ma + ma = mg mg Simplifying and solving for a gives g a = 3 A

23 . he tension in the string is a) 1 3 mg b) 1 mg c) 3 mg d) mg e) 4 3 mg ΣF = ma Find net force on object Use either net force equation from above and substitute for a to solve for ma = mg m g 3 = mg mg = + mg 3 4mg = 3 E F θ m 3. An object of mass m is pushed to the right by a force F that forms an angle θ with the horizontal as shown. he object moves with constant velocity. he coefficient of kinetic friction between the object and the surface is a) F cosθ b) F sinθ c) cos θ sinθ F cosθ d) mg F cosθ e) Fsinθ + mg

24 Resolve for vector into perpendicular and parallel components ΣF = ma F f = μf N Object moving with constant velocity has zero net force. Resolve the force vector F into components and draw a FBD. We know the net force in both directions is zero, since the velocity is constant. F N F x Analyzing the forces in the vertical direction, we can write F = mg + F = mg + Fsinθ and F = F = Fcosθ N y We now apply the equation for kinetic friction: F = μf f F y N mg Substituting from our other two equations: F cosθ = μ( mg + Fsinθ) And solving for μ gives F cosθ μ = mg + F sinθ ( ) f x E 4. A 75 kg man and a 5 kg girl stand facing each other on a frictionless sheet of ice. he girl exerts a force of 30 N on the man, causing him to accelerate to the north. he magnitude and direction of the reaction force on the girl is Magnitude Direction a) 10 N north b) 10 N south c) 30 N north d) 30 N south e) 90 N south Reaction forces are equal and opposite he force exerted by the girl on the man is 30 N to the north, therefore the reaction force must be equal (30 N) and in the opposite direction (south) D

25 Questions kg kg F A block of mass 1 kg is sitting on top of a block of mass kg. he 1 kg block is tied to a thin thread which extends horizontally to a wall where the other end is attached. he kg block is pulled by an external horizontal force F with constant velocity to the right. he coefficient of kinetic friction between the two blocks is.5 and the coefficient of kinetic friction between the kg block and the surface is As the bottom block is pulled to the right, the tension in the thread is most nearly a). N b) 1 N c) 5 N d) 8 N e) 15 N F f = μf N Object moving with constant velocity has zero net force. We know the net force on the top block is zero (it remains at rest). Drawing the forces on the top block: F N F f,k1 m 1 g he force of kinetic friction opposes the relative motion of the two blocks. Since the bottom block moves to the right, friction is attempting to pull the top block to the right so the two don t move relative to each other. From the above diagram, we can see that = F f,k1. Substituting and solving for gives (1 designates the top block, the bottom) = F = μ F = μ m g = = 5N f 1, k k,1 N,1 k,1 1 (.5)( 1kg)( 10 m ) s C

26 6. he force F required to pull the bottom block with constant velocity is most nearly a) 15 N b) 1 N c) 4 N d) 9 N e) 34 N F f = μf N Object moving with constant velocity has zero net force. he net force on the bottom block is zero as well (it moves with constant speed). he forces on the bottom block are: F N F f,k F F f,k1 m 1 g m g In the y-direction: F = mg+ mg N 1 In the x-direction: F = F + F f, k1 f, k F f,k1 is already known to be 5 N, so we find F f,k and substitute to find F. ( ) m ( ) ( g) F = 5N + μ F = 5N + μ m g + m k, N k, 1 F = 5 N kg + 1kg s F = 5N + 4N F = 9N D 7. An object moves with constant velocity while three forces act on it. Which of the following must be true? I. the three forces have equal magnitude II. the vector sum of the three forces is zero III. the forces must be perpendicular to the direction the object is traveling a) I only b) II only c) I and II d) I and III e) II and III

27 Object moving with constant velocity has zero net force. Since the object is moving with zero acceleration, it must experience zero net force: the vector sum of the three forces must equal zero. No other requirement is needed; the forces can be of any magnitude and direction so long as their vector sum is zero. B 8. A 4 kg mass is hung vertically from an ideal spring, which causes it to stretch 8 cm. If the 4 kg mass is replaced by a mass of 8 kg, how far will the spring stretch? a) cm b) 4 cm c) 8 cm d) 16 cm e) 64 cm Hooke s Law: F s = -kx Hooke s Law shows that the stretch of a spring is directly proportional to the force it exerts: hence double the force on a spring and the distance it is stretched will double as well. hus, the 8 kg mass with stretch the spring 16 cm. D 9. As a 40 N ball falls through Earth s atmosphere, it experiences a drag force, which is given by the expression FD = bv, where b is known to be 5 kg/s. he terminal velocity of the ball close to Earth s surface is a) 5 m/s b) 8 m/s c) 5 m/s d) 40 m/s e) 45 m/s Object moving with constant velocity has zero net force. erminal velocity occurs when F D = F g he force of drag increases with the velocity, reducing the acceleration until the force of drag equals the weight. At this point we can write F = F D bv v g = 40N 40N = = 8 m kg s 5 s B

28 Questions m θ An object of mass m is at rest on an inclined plane as shown above. he inclined plane forms an angle θ with the horizontal. he coefficient of static friction between the object and the plane is μ s. 10. Which of the following diagrams correctly shows the forces acting on the object? a) b) c) d) e) Resolve forces on an incline Direction of forces D correctly shows the forces. Weight always pulls an object down, a normal force is always perpendicular to the surface, and friction is always parallel to the surface. In this case, friction points up the ramp because it opposes the motion of the object sliding down the ramp. D 11. he force of static friction acting on the object is a) μ mg s b) μ mg s cos θ c) μ mg s sin θ d) mg cosθ e) mg sinθ

29 Resolve forces on an incline Static friction is variable In this case, the relationship between static friction and the coefficient of static friction cannot be used because it is an inequality: 0 F μ F f, s s N. We must determine the force of friction by force analysis. We know that the net force is zero (it is at rest), so parallel to the ramp we have: F = F f, s g, x Resolving the weight vector and substituting: θ F g,y = F g cosθ E F g F g,x = F g sinθ F = mgsinθ f, s 1. hree forces act on an object that is moving at constant velocity. wo of the forces are shown on the diagram above. Which of the following correctly shows the third force vector? a) b) c) d) e) Object moving with constant velocity has zero net force. In order for the object to move with constant velocity, the vector sum must be zero. Adding the vector shown in answer choice A would accomplish this. A

30 13. A 0 N bucket is at the bottom of a 9 m deep well and is initially at rest. he bucket is then pulled to the top of the well by a string that exerts a constant tension force, such that it reaches the top in 3.0 s. During this interval, the tension in the string is a) 4 N b) 6 N c) 0 N d) 4 N e) 40 N Newton s Second Law & Kinematics Use kinematics to find the acceleration of the bucket: 1 x = x + v t + at x = at ( m) ( 3s) x 9 a = = = m t s he forces on the bucket are: D And the net force equation is F = mg Substituting Newton s Second Law and solving for : ma = mg m ( ) = ma + mg = kg s + 0N = 4N

31 14. Which of the following graphs shows an object whose net force is zero? I. x II. x III. v t t t a) I only b) II only c) I and II d) I and III e) II and III Interpreting Motion Graphs Object moving with constant velocity has zero net force. Zero net force means the velocity must be constant. Graphs I and II show objects moving with constant velocity: graph III shows an object with decreasing velocity. hus, the answer is C C Q P R 15. he diagram above shows the path of a projectile as it moves from left to right without friction or drag. At which point along the projectile s path is the acceleration of the projectile the greatest? a) P b) Q c) R d) P and R e) the acceleration is the same at all points A projectile is free falling due to gravity he diagram shows a projectile: we know that the only force acting on a projectile is gravity, which is constant. hus, the acceleration is the same at all points. he projectile moves horizontally due to its initial velocity and inertia. E