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Transcript

Chapter 4 Forces and Newton’s Laws of Motion 4.1 The Concepts of Force and Mass 1) A force is a push or a pull. Contact forces arise from physical contact . Action-at-a-distance forces do not require contact and include gravity and electrical forces. 4.1 The Concepts of Force and Mass Arrows are used to represent forces. The length of the arrow is proportional to the magnitude of the force. 15 N 5N 4.1 The Concepts of Force and Mass 2) Mass is a measure of the amount of “stuff” contained in an object. SI Unit of Mass: kilogram (kg) Galileo vs. Aristotle In our experience, objects must be pushed in order to keep moving. So a force would be needed to have a constant velocity. This is what Aristotle claimed in his in his series of books entitled "Physics", written 2400 years ago. But 400 years ago, another scientist and astronomer, Galileo, proposed the following thought experiment which revealed another perspective. Thought Experiment Imagine two perfectly smooth ramps connected together by a perfectly smooth surface. If a ball is let go at the top of the one ramp, what will happen? Thought Experiment Imagine two perfectly smooth ramps connected together by a perfectly smooth surface. If a ball is let go at the top of the one ramp, what will happen? Thought Experiment Imagine two perfectly smooth ramps connected together by a perfectly smooth surface. If a ball is let go at the top of the one ramp, what will happen? Thought Experiment Imagine two perfectly smooth ramps connected together by a perfectly smooth surface. If a ball is let go at the top of the one ramp, what will happen? Thought Experiment If a ball rolls down one ramp, it keeps rolling up the other side until it reaches the same height. Thought Experiment Now repeat that experiment, but make the second ramp less steep. What Will Happen? Thought Experiment Now repeat that experiment, but make the second ramp less steep. What Will Happen? Thought Experiment Now repeat that experiment, but make the second ramp less steep. What Will Happen? Thought Experiment Now repeat that experiment, but make the second ramp less steep. What Will Happen? Thought Experiment It will still keep rolling until it reaches the same height, but it has to roll farther! Thought Experiment Finally, make the ramp flat. Now what will happen? Thought Experiment Finally, make the ramp flat. Now what will happen? Thought Experiment Finally, make the ramp flat. Now what will happen? Thought Experiment Finally, make the ramp flat. Now what will happen? Thought Experiment Finally, make the ramp flat. Now what will happen? Thought Experiment Finally, make the ramp flat. Now what will happen? Thought Experiment It will keep rolling forever, no external force is necessary. Galileo vs. Aristotle It's not that Aristotle was wrong. In everyday life, objects do need to keep being pushed in order to keep moving. Push a book across the table. When you stop pushing, it stops moving. Aristotle is right in terms of what we see around us every day. Force and Motion It's just that Galileo, and later Newton, imagined a world where friction could be eliminated. Friction represents an external force acting on the object, just as your push is an external force. Fapplied Ffriction In the absence of all external forces, an object's velocity remains constant. Two equal and opposite forces have the same effect, they cancel to create zero net force. 4.2 Newton's 1st Law of Motion Sir Isaac Newton Galileo's observations were more fully formed in 1687 by the 'father of physics, ' Sir Isaac Newton, who called this observation "The First Law of Motion". 4.2 Newton’s First Law of Motion An object at rest remains at rest, and an object in motion remains in motion, unless acted on by a net external force. In other words, an object maintains its velocity (both speed and direction) unless acted upon by a nonzero net force. Having zero velocity, being at rest, is not special, it is just one possible velocity…a velocity which is no more special than any other. 4.2 Newton’s First Law of Motion 3) The net force on an object is the vector sum of all forces acting on that object. The SI unit of force is the Newton (N). Individual Forces 4N 10 N Net Force 6N Net Force: Mathematically, the net force is written as Fnet or F where the Greek letter sigma denotes the vector sum. 1 In the absence of an external force, a moving object will c A stop immediately. c B slow down and eventually come to a stop. c C go faster and faster. c D move with constant velocity. Ans: D 2 When the rocket engines on a spacecraft are suddenly turned off while traveling in empty space, the starship will c A stop immediately. c B slowly slow down, and then stop. c C go faster and faster. c D move with constant velocity. Ans: D 3 A rocket moves through empty space in a straight line with constant speed. It is far from the gravitational effect of any star or planet. Under these conditions, the force that must be applied to the rocket in order to sustain its motion is c A equal to its weight. c B equal to its mass. c C dependent on how fast it is moving. c D zero. Ans: D 4 You are standing in a moving bus, facing forward, and you suddenly fall forward. You can infer from this that the bus‘s c A velocity decreased. c B velocity increased. c C speed remained the same, but it's turning to the right. c D speed remained the same, but it's turning to the left. Ans: A 5 You are standing in a moving bus, facing forward, and you suddenly fall forward as the bus comes to an immediate stop. What force caused you to fall forward? c A gravity c B normal force due to your contact with the floor of the bus c C force due to friction between you and the floor of the bus c D There is not a force leading to your fall. Ans: D 4.2 Newton’s First Law of Motion 4) Inertia is the natural tendency of an object to remain at rest or in motion at a constant speed along a straight line (constant velocity). Inertia is the resistance of any physical object to any change in its state of motion, including changes to its speed and direction. The mass of an object is a quantitative measure of inertia. 4.2 Newton’s First Law of Motion 5) All Newton's laws are only valid in inertial reference frames An inertial reference frame is one which is not accelerating or rotating. It is a space in which every body remains in a state of rest unless acted on by an external unbalanced force. Newton’s First Law is also called Law of inertia 6) Relative Motion • Relative motion – motion depends on your frame of reference • Are you moving if you are sitting 100% still in your chair?? – YES! The earth is moving and you are on the earth, so you are moving relative to the other planets, the sun, etc. – NO! Relative to the floor, you are not moving – It all depends on your frame of reference – http://www.phy.ntnu.edu.tw/ntnujava/index.php?topic= 140 Check your understanding http://www.physicsclassroom.com/getattachment/curriculum/ne wtlaws/newtl2.pdf http://www.physicsclassroom.com/class/newtlaws/Lesson1/Inertia-and-Mass 4.3 Newton’s Second Law of Motion 4.3 Newton’s Second Law of Motion When a net external force acts on an object of mass m, the acceleration that results is directly proportional to the net force and has a magnitude that is inversely proportional to the mass. The direction of the acceleration is the same as the direction of the net force. F a m F ma 4.3 Newton’s Second Law of Motion SI Unit for Force m kg m kg 2 2 s s Fnet ma 4.3 Newton’s Second Law of Motion 4.4 Free body diagram A free-body-diagram is a diagram that represents the object and the forces that act on it. 4.3 Newton’s Second Law of Motion The net force in this case is: 275 N + 395 N – 560 N = +110 N and is directed along the + x axis of the coordinate system. 4.3 Newton’s Second Law of Motion If the mass of the car is 1850 kg then, by Newton’s second law, the acceleration is F 110 N a 0.059 m s m 1850 kg 2 Rules for free body diagram: 1. Identify the object--represent the object by a box or a circle 2. draw the force arrow from the object outward in the direction that the force is acting. 3. label each force arrow according to its type. 4. Choose coordinate direction. https://phet.colorado.edu/en/simulation/forces-and-motion-basics https://phet.colorado.edu/en/simulation/forces-1d https://phet.colorado.edu/en/simulation/forces-and-motion 6 A 3.5 kg object experiences an acceleration of 0.5 2 m/s . What net force does the object feel? 7 A 12 N net force acts on a 36 kg object? How much does it accelerate? 8 How much net force is required to accelerate a 0.5 kg toy car, initially at rest to a velocity of 2.4 m/s in 6 s? 9 A net force F accelerates a mass m with an acceleration a. If the same net force is applied to mass 2m, then the acceleration will be c A 4a c B 2a c C a/2 c D a/4 10 A constant net force acts on an object. The object moves with: c A constant acceleration c B constant speed c C constant velocity c D increasing acceleration 11 A net force F acts on a mass m and produces an acceleration a. What acceleration results if a net force 2F acts on mass 4m? c A a/2 c B 8a c C 4a c D 2a 12 The acceleration of an object is inversely proportional to: c A the net force acting on it. c c c B its position. C D its velocity. its mass. Newton’s Second Law of Motion ΣF Example: A 5.0 kg object is being acted on by a 20N force to the right (F1), and a 30N force, also to the right (F2). What is the net force on the object? F2 F1 The second force, F2, acts to the right also, with a greater magnitude of 30N. This is drawn slightly larger than F1. Newton’s Second Law of Motion ΣF Example: A 5.0 kg object is being acted on by a 20N force to the right (F1), and a 30N force, also to the right (F2). What is the net force on the object? To add vectors, move the second vector so it starts where the first one ends. F1 F2 The sum is a vector which starts where the first vector started, and ends where the last one ends. Newton’s Second Law of Motion ΣF Example: A 5.0 kg object is being acted on by a 20N force to the right (F1), and a 30N force, also to the right (F2). What is the net force on the object? These free body diagrams are critically important to our work. Once done, the problem can be translated into an algebra problem. ΣF F1 F2 Newton’s Second Law of Motion For example: A 5.0 kg object is being acted on by a 20N force to the right (F1), and a 30N force, also to the right (F2). What is the net force on the object? ΣF F1 F2 First we will define "to the right" as positive. Then we can interpret our diagram to read: ΣF = F1 + F2 ΣF = 20N + 30N ΣF = 50N to the right we get the direction from our diagram and from our positive answer, which we defined as meaning "to the right" 13 Two forces act on an object. One force is 40N to the west and the other force is 40N to the east. What is the net force acting on the object? 14 Two forces act on an object. One force is 8.0 N to the north and the other force is 6.0N to the south. What is the net force acting on the object? c A 14 N to the north c B 14 N to the south c C 2 N to the north c D 2 N to the south Newton’s Second Law of Motion For example: A 5.0 kg object is being acted on by a 20N force to the right (F1), and a 30N force, also to the right (F2). We found the net force on the object to be 50N to the right. Now let's find its acceleration. ΣF F1 F2 ΣF = ma Answer a = ΣF /m a = 50N / 5.0 kg 2 2 a = 10 m/s to the right s to the right 4.3 Newton’s Second Law of Motion Force is a vector, so ΣF = ma is true along each coordinate axis. F1 F3 a = 1 m/s 2 F2 That means we can add up all the forces in the vertical direction and those will equal "ma" in the vertical direction. F1 F1 +(-F2) = m*a F2 F1 - F2 = 0 And then can do the same thing in the horizontal direction. F3 = m*a F3 2 a = 1 m/s 2 F3 = (2kg)(1 m/s ) F3 = 2 N 15 A force F1 = 50N acts to the right on a 5.0 kg object. Another force, F2 = 30N, acts to the left. Find the acceleration of the object: 16 A force F1 = 350N pushes upward on 20.0 kg object. Another force, F2 = 450N pulls downward. Find the acceleration of the object: 17 An object accelerates downward at a rate of 4.9 2 m/s . If the downward force on the object is 500N and the upward force is 250N, what is the mass of the object? 4.5 Newton’s Third Law of Motion Whenever one body exerts a force on a second body, the second body exerts an oppositely directed force of equal magnitude on the first body. For every action, there is an equal, opposite reaction. This is another way to state Newton's 3rd Law. It is important to remember that the forces (or actions) are always applied to two different objects. About action and reaction force: • All forces come in action-reaction pairs • Action and reaction forces are equal in size • Action and reaction forces are opposite in direction • They are same type of force • They acted on different objects • will never cancel out each other Newton’s Third Law of Motion A key to the correct application of the third law is that the forces are exerted on different objects. Make sure you don’t use them as if they were acting on the same object. Then they would add to zero! Force on hands Force on floor 4.5 Newton’s Third Law of Motion Suppose that the magnitude of the force is 36 N. If the mass of the spacecraft is 11,000 kg and the mass of the astronaut is 92 kg, what are the accelerations? 4.5 Newton’s Third Law of Motion F P. On the astronaut F P. On the spacecraft Acceleration of spacecraft: P 36 N as 0.0033 m s 2 ms 11,000 kg Acceleration of astronaut: P 36 N 2 aA 0.39 m s mA 92 kg Newton’s Third Law of Motion Rocket propulsion can also be explained using Newton’s third law. Hot gases from combustion spew out of the tail of the rocket at high speeds. The reaction force is what propels the rocket. Note that the rocket does not need anything to “push” against. Newton’s Third Law of Motion Helpful notation: the first subscript is the object that the force is being exerted on; the second is the source. Subscripts help keep your ideas and equations clear. FGP = -FPG Horizontal force exerted on the ground by person's foot FGP Horizontal force exerted on the person's foot by ground FPG 27 When you sit on a chair, the net external force on you is c A zero. c B dependant on your weight. c C up. c D down. 28 An object of mass m sits on a flat table. The Earth pulls on this object with force mg, which we will call the action force. What is the reaction force? c A The table pushing up on the object with force mg c B The object pushing down on the table with force mg c C The table pushing down on the floor with force mg c D The object pulling upward on the Earth with force mg 2 9 A 20-ton truck collides with a 1500-lb car and causes a lot of damage to the car. Since a lot of damage is done on the car c A the force on the truck is greater then the force on the car c B the force on the truck is equal to the force on the car c C the force on the truck is smaller than the force on the car c D the truck did not slow down during the collision 30 As you are sitting in a chair, you feel the chair pushing up on you. The reaction force in this situation is: c A The chair pushing down on the ground c B Gravity pulling down on you c C You pushing down on the chair c D The ground pushing up on the chair 31 A student is doing push-ups in gym class. A reaction pair of forces is best described as: c A The student pushes down on the ground The ground pushes up on the student c B Gravity is pulling the student down The ground is pushing the student up c C Gravity is pulling the student down The student's arms push the student up c D The student's hands push down on the ground The students arms push the student up 33 If you blow up a balloon, and then release it, the balloon will fly away. This is an illustration of: (Note: there may be more than one answer. Be prepared to explain WHY!) c A Newton's first law c B Newton's second law c C Newton's third law c D Galileo's law of inertia 4.5 Types of Forces: An Overview In nature there are two general types of forces, fundamental and nonfundamental. Fundamental Forces 1. Gravitational force 2. Strong Nuclear force 3. Electroweak force 4.6 Types of Forces: An Overview Examples of nonfundamental forces: Friction force tension in a rope normal or support forces 4.6 Gravitational force, Weight, apparent weight and Normal Force 1) The Gravitational Force Newton’s Law of Universal Gravitation Every particle in the universe exerts an attractive force on every other particle. A particle is a piece of matter, small enough in size to be regarded as a mathematical point. The force that each exerts on the other is directed along the line joining the particles. 1) The Gravitational Force For two particles that have masses m1 and m2 and are separated by a distance, then there will be gravitational force acting on each mass by the other, 4.6 The Gravitational Force 2) Weight – the Force of Gravity Weight is the force exerted on an object by gravity. Close to the surface of Earth, where the gravitational acceleration is nearly constant, weight can be calculated with: Fg = mg 2 Near the surface of Earth, g is 9.8 m/s downwards. 4.6.2 weight The weight of an object on or above the earth is the gravitational force that the earth exerts on the object. The weight always acts downwards, toward the center of the earth. On or above another astronomical body, the weight is the gravitational force exerted on the object by that body. SI Unit of Weight: newton (N) 18 Determine the Force of Gravity (weight) of a 6.0 kg bowling ball. 19 Determine the weight of a small car with a mass of 900 kg. 20 Using a spring scale, you find that the weight of a friction block in the lab is around 24 N. What is the mass of the block, in kilograms? 21 A 120 lb woman has a mass of about 54.5 kg. What is her weight? 22 What is the weight of a 25 kg object located near the surface of Earth? 23 Which of the following properties of an object is likely to change on another planet? c A Mass c B Weight c C Color c D Volume (size and shape) 24 The acceleration due to gravity is lower on the Moon than on Earth. Which of the following is true about the mass and weight of an astronaut on the Moon's surface, compared to Earth? c A Mass is less, weight is same c B Mass is same, weight is less c C Both mass and weight are less c D Both mass and weight are the same 3) Normal Force An object at rest must have no net force on it. If it is sitting on a table, the force of gravity is still there; what other force is there? FG Normal Force An object at rest must have no net force on it. If it is sitting on a table, the force of gravity is still there; what other force is there? The force exerted perpendicular to a surface is called the normal force. It is exactly as large as needed to balance the force from the object (if the required force gets too big, something breaks!) The words "normal" and "perpendicular" are synonyms. FN FG 25 A 14 N brick is sitting on a table. What is the normal force supplied by the table? c A 14 N upwards c B 28 N upwards c C 14 N downwards c D 28 N downwards 26 What normal force is supplied by a desk to a 2.0 kg box sitting on it? 3) The Normal Force FN 11 N 15 N 0 FN 26 N FN 11 N 15 N 0 FN 4 N 4) Apparent Weight The apparent weight of an object is the reading of the scale. It is equal to the normal force the man exerts on the scale. The Normal Force F y FN mg ma FN mg ma apparent weight true weight 4.7 Friction Static and Kinetic Frictional Forces Friction - A Resistive Force There are many different types of forces that occur in nature, but perhaps none is more familiar to us than the force of friction (Ffr). Friction is a resistive force that opposes the motion of an object. What does sandpaper have to do with friction? Friction is resistance to the relative or tendency of relative motion between the object and the contact surface. Friction - A Resistive Force Friction is the reason objects stop rolling or sliding along a surface. It is the reason it is difficult to start pushing a heavy box along the floor. There are many different types of friction: Friction between solid objects and air is often called air resistance. Friction between two fluids is called viscosity, and so on. Static and Kinetic Frictional Forces When an object is in contact with a surface there is a force acting on that object. Frictional force is the component of this force that is parallel to the surface. 1) Static Frictional Forces When the two surfaces are not sliding across one another the friction is called static friction. Static Frictional Forces The magnitude of the static frictional force can have any value from zero up to a maximum value. fs f f 0 s MAX s MAX s s FN is called the coefficient of static friction. 4.9 Static and Kinetic Frictional Forces Note that the magnitude of the frictional force does not depend on the contact area of the surfaces. 2) Kinetic Frictional Forces While Static friction opposes the impending relative motion between two objects. Kinetic friction opposes the relative sliding motion motions that actually does occur. f k k FN 0 k is called the coefficient of kinetic friction. 4.7 Static and Kinetic Frictional Forces 4.7 Static and Kinetic Frictional Forces The sled comes to a halt because the kinetic frictional force opposes its motion and causes the sled to slow down. 4.7 Static and Kinetic Frictional Forces Suppose the coefficient of kinetic friction is 0.05 and the total mass is 40kg. What is the kinetic frictional force? f k k FN k mg 0.0540kg 9.80 m s 20N 2 Example: The force of friction between an object and the surface upon which it is sliding is 8 N. The mass of the object is 20kg. How much is the object’s weight? How much is the normal force? What is the coefficient of kinetic friction? Example: The minimum horizontal force to move an object on a surface is 400N. The force of friction between an object and the surface upon which it is sliding is 360N. The mass of the object is 95kg. What is the coefficient of kinetic friction? What is the coefficient of static friction?. 4.8 The Tension Force Cables and ropes transmit forces through tension. 4.8 The Tension Force A massless rope will transmit tension undiminished from one end to the other. If the rope passes around a massless, frictionless pulley, the tension will be transmitted to the other end of the rope undiminished. Tension Force The tension in a rope is the same everywhere in the rope. If two masses hang down from either side of a cable, for instance, the tension in both sides must be the same. T1 = T2 20 kg 50 kg 4.9 Air resistance force--Fair • special type of frictional force • oppose the motion of an object • It is most noticeable for objects that travel at high speeds • Usually is speed related. 4.10 Spring force--Fspring • force exerted by a compressed or stretched spring upon any object that is attached to it • to its rest or equilibrium position • Hook’s Law-magnitude of the force is directly proportional to the amount of stretch or compression of the spring Practice of free body diagram and problem solving 1. Select a “object”("system“) to which you intend to apply Newton's Second Law. 2. Draw free body diagram, identify all forces 3. Pick a coordinate system. Find out forces relations and list them out 4. Find what you need to solve and which what relations needed 5. Write Newton's Second Law, and equations 6. Solve equations 4.11 Equilibrium Application of Newton’s Laws of Motion Definition of Equilibrium An object is in equilibrium when it has zero acceleration. F F net 0 4.12 Force at Angles-- When force vector has an angle to coordinate direction: Assume that the chain is exerting a 60 N force upon Fido at an angle of 40 degrees above the horizontal http://www.physicsclassroom.com/class/vectors/Lesson-3/Resolution-of-Forces 4.12 Vector nature of force -- When force vector has an angle to coordinate direction: Consider the tow truck at the right. If the tensional force in the cable is 1000 N and if the cable makes a 60-degree angle with the horizontal, then what is the vertical component of force that lifts the car off the ground? The vertical component can be found if a triangle is constructed with the 1000 N diagonal force as the hypotenuse. The vertical component is the length of the side opposite the hypotenuse. Thus, sin (60o) = (Fvert) / (1000 N) Solving for Fvert will give the answer 866 N. 4.12 Vector nature of force -- When force vector has an angle to coordinate direction: Example: A 10.0 kg box is pulled along a horizontal surface by a rope that makes a 30.0o angle with the horizontal. The tension in the rope is 30.0 N. There is 10.0 N of friction acting on the box. Find: a) the acceleration of the box b) The normal force of the ground 4.12 Vector nature of force -- When force vector has an angle to coordinate direction: variables: m=10.0kg FA= 30.0N, angle θ= 30.0o Ff = 10.0N a) a=? b) FN=? FN FN Ff Ff 4.12 Vector nature of force -- When force vector has an angle to coordinate direction: FN Ff 4.12 connected objects eg. Two blocks, with masses m1 = 0.4kg and m2 = 0.6kg, are connected by a string and lie on a frictionless tabletop. A force F = 3.5 N is applied to block m2. a. Draw a free-body diagram for each block showing all applied forces to scale. Next to each diagram show the direction of the acceleration of that object. b. Find the acceleration of each object. c. Find the tension force in the string between two objects. m1 = 0.4kg, m2 = 0.6kg, frictionless tabletop. Fapp = 3.5 N on m2. Key: both blocks have the same acceleration (motion) a. free-body diagram for each block. FN2 FN1 Ften Fg1 Fapp Ften Fg2 b. Find the acceleration of each object. Two blocks, with masses m1 = 400 g and m2 = 600 g, are connected by a string and lie on a frictionless tabletop. A force F = 3.5 N is applied to block m2. c. Find the tension force in the string between two objects. 15 kg 12 kg # 79. A 12 kg load hangs from one end of a rope that passes over a small frictionless pulley. A 15 kg counterweight is suspended from the other end of the rope. The system is released from rest. a. Draw a free-body diagram for each object showing all applied forces in relative scale. Next to each diagram show the direction of the acceleration of that object. b. Find the acceleration of each mass. c. What is the tension force in the rope? d. What distance does the 12 kg load move in the first 3 s? e. What is the velocity of 15 kg mass at the end of 5 s? A 12 kg load hangs from one end of a rope that passes over a small frictionless pulley. A 15 kg counterweight is suspended from the other end of the rope. The system is released from rest. a. Draw a free-body diagram for each object showing all applied forces in relative scale. Next to each diagram show the direction of the acceleration of that object. A 12 kg load hangs from one end of a rope that passes over a small frictionless pulley. A 15 kg counterweight is suspended from the other end of the rope. The system is released from rest. b. Find the acceleration of each mass. A 12 kg load hangs from one end of a rope that passes over a small frictionless pulley. A 15 kg counterweight is suspended from the other end of the rope. The system is released from rest. c. What is the tension force in the rope? A 12 kg load hangs from one end of a rope that passes over a small frictionless pulley. A 15 kg counterweight is suspended from the other end of the rope. The system is released from rest. d. What distance does the 12 kg load move in the first 3 s? A 12 kg load hangs from one end of a rope that passes over a small frictionless pulley. A 15 kg counterweight is suspended from the other end of the rope. The system is released from rest. e. What is the velocity of 15 kg mass at the end of 5 s? 500 g 300 g #81. A 500 g block lies on a horizontal tabletop. The coefficient of kinetic friction between the block and the surface is 0.25. The block is connected by a massless string to the second block with a mass of 300 g. The string passes over a light frictionless pulley as shown above. The system is released from rest. a. Draw clearly labeled free-body diagrams for each of the 500 g and the 300g masses. Include all forces and draw them to relative scale. Draw the expected direction of acceleration next to each free-body diagram. b. Use Newton’s Second Law to write an equation for the 500 g mass. c. Use Newton’s Second Law to write an equation for the 300 g mass. d. Find the acceleration of the system by simultaneously solving the system of two equations. e. What is the tension force in the string? 500 g 300 g A 500 g block lies on a horizontal tabletop. The coefficient of kinetic friction between the block and the surface is 0.25. The block is connected by a massless string to the second block with a mass of 300 g. The string passes over a light frictionless pulley as shown above. The system is released from rest. a. Draw clearly labeled free-body diagrams for each of the 500 g and the 300g masses. Include all forces and draw them to relative scale. Draw the expected direction of acceleration next to each free-body diagram. 500 g 300 g A 500 g block lies on a horizontal tabletop. The coefficient of kinetic friction between the block and the surface is 0.25. The block is connected by a massless string to the second block with a mass of 300 g. The string passes over a light frictionless pulley as shown above. The system is released from rest. b. Use Newton’s Second Law to write an equation for the 500 g mass. 500 g 300 g A 500 g block lies on a horizontal tabletop. The coefficient of kinetic friction between the block and the surface is 0.25. The block is connected by a massless string to the second block with a mass of 300 g. The string passes over a light frictionless pulley as shown above. The system is released from rest. c. Use Newton’s Second Law to write an equation for the 300 g mass. 500 g 300 g A 500 g block lies on a horizontal tabletop. The coefficient of kinetic friction between the block and the surface is 0.25. The block is connected by a massless string to the second block with a mass of 300 g. The string passes over a light frictionless pulley as shown above. The system is released from rest. d. Find the acceleration of the system by simultaneously solving the system of two equations. 500 g 300 g A 500 g block lies on a horizontal tabletop. The coefficient of kinetic friction between the block and the surface is 0.25. The block is connected by a massless string to the second block with a mass of 300 g. The string passes over a light frictionless pulley as shown above. The system is released from rest. e. What is the tension force in the string? ** An elevator (mass 4850 kg) is to be designed so that the maximum acceleration is 0.0680g. What are the maximum and minimum forces the motor should exert on the supporting cable? Two boxes are connected by a cord. A person pulls horizontally on box A with force F = 40.0 N. The boxes have masses of 10 kg and 12 kg. Ignore friction between the boxes and the tabletop. a) Show the free-body diagram of the box B. b) Show the free-body diagram of the box A. c) Find the acceleration of the system. d) Find the tension in the cord. mB = 12 kg mA = 10 kg FT = 40 N Two boxes are connected by a cord. A person pulls horizontally on box A with force F = 40.0 N. The boxes have masses of 10 kg and 12 kg. Ignore friction between the boxes and the tabletop. a) Show the free-body diagram of the box B. Two boxes are connected by a cord. A person pulls horizontally on box A with force F = 40.0 N. The boxes have masses of 10 kg and 12 kg. Ignore friction between the boxes and the tabletop. b) Show the free-body diagram of the box A. Two boxes are connected by a cord. A person pulls horizontally on box A with force F = 40.0 N. The boxes have masses of 10 kg and 12 kg. Ignore friction between the boxes and the tabletop. c) Find the acceleration of the system. Given: mB = 12 kg mA = 10 kg Fapp = 40 N 2 g = 9.8 m/s a=? Two boxes are connected by a cord. A person pulls horizontally on box A with force F = 40.0 N. The boxes have masses of 10 kg and 12 kg. Ignore friction between the boxes and the tabletop. d) Find the tension in the cord. Given: mB = 12 kg mA = 10 kg Fapp = 40 N 2 g = 9.8 m/s 2 a = 1.82 m/s Two masses are suspended over a pulley by a cable, as shown on the diagram. Let the mass of the elevator be 1150 kg and the counterweight 1000 kg. a E Elevator car a) Show the free-body diagram of the elevator b) Show the free-body diagram of the counterweight aC mE = 1150 kg c) Calculate the acceleration of the system Counterweight mC = 1000 kg d) Calculate the tension in the cable Two masses are suspended over a pulley by a cable, as shown on the diagram. Let the mass of the elevator be 1150 kg and the counterweight 1000 kg. a) Show the free-body diagram of the elevator a E Elevator car a mE = 1150 kg C Counterweight mC = 1000 kg Two masses are suspended over a pulley by a cable, as shown on the diagram. Let the mass of the elevator be 1150 kg and the counterweight 1000 kg. b) Show the free-body diagram of the counterweight aE Elevator car mE = 1150 kg aC Counterweight mC = 1000 kg Two masses are suspended over a pulley by a cable, as shown on the diagram. Let the mass of the elevator be 1150 kg and the counterweight 1000 kg. c) Calculate the acceleration of the system aE Elevator car mE = 1150 kg aC Counterweight mC = 1000 kg Two masses are suspended over a pulley by a cable, as shown on the diagram. Let the mass of the elevator be 1150 kg and the counterweight 1000 kg. d) Calculate the tension in the cable aE Elevator car mE = 1150 kg aC Counterweight mC = 1000 kg Two boxes are connected by a cord running over a pulley. The coefficient of kinetic friction between box A and the table is 0.2 5.0 kg A a) Show the free-body diagrams of box A and box B B 2.0 kg b) Find the acceleration of the system of two boxes c) Find the tension in the cord Two boxes are connected by a cord running over a pulley. The coefficient of kinetic friction between box A and the table is 0.2 a) Show the free-body diagrams of box A and box B 5.0 kg A B 2.0 kg Two boxes are connected by a cord running over a pulley. The coefficient of kinetic friction between box A and the table is 0.2 b) Find the acceleration of the system of two boxes Given: mA = 5 kg mB = 2 kg 2 g = 9.8 m/s μ = .2 a=? 5.0 kg A B 2.0 kg Two boxes are connected by a cord running over a pulley. The coefficient of kinetic friction between box A and the table is 0.2 c) Find the tension in the cord Given: mA = 5 kg mB = 2 kg g = 29.8 m/s μ = .2 a = 1.4 2 m/s T=?