Chapter 4 Homework Packet

Transcription

1 Chapter 4 Homework Packet Conceptual Questions 1) Which of Newton's laws best explains why motorists should buckle-up? A) the first law B) the second law C) the third law D) the law of gravitation Inertia is the tendency of an object in motion to continue in that motion and resist a change of motion. The mass of the object is a measure of this tendency the greater the mass of an object the greater the tendency to resist a change in motion. In fact, for linear motion (motion in a straight line) the mass is stated to be the same thing as the inertia. Newton s statement of this tendency of an object to resist a change of motion is his first law. If a motorist is traveling in a car at some velocity, if the car stopped suddenly, the motorist would have a tendency, because of his mass (his inertia), to continue moving in the same direction. The only thing that would prevent the motorist from continuing in his motion is a restraint that is, a seat belt. This example is best described by Newton s First Law of Motion. Answer A (1 point/1) 2) When you sit on a chair, the resultant force on you is A) zero. B) up. C) down. D) depending on your weight. The resultant force is a synonym for the net force or the sum of forces. Remember, there are only two options. You are either accelerating, in which case there is a non-zero, or unbalanced net force; or you are not accelerating, in which case there is a net force of zero. If you are sitting in a chair you are not accelerating therefore, the net or resultant force must be balanced, or zero. Answer A (1 point/2) 3) In the absence of an external force, a moving object will A) stop immediately. B) slow down and eventually come to a stop. C) go faster and faster. D) move with constant velocity. In the absence of an external force is another way of saying no net force or a net force of zero. According to Newton s first law, the law of inertia, if there is no net force on an object, because of its inertia it will continue to travel in the direction and at the velocity it is already traveling it will resist a change in motion. So, in the absence of an external force, a moving object will move with a constant velocity. Answer D (1 point/3) 4) When the rocket engines on the starship NO-PAIN-NO-GAIN are suddenly turned off, while traveling in empty space, the starship will A) stop immediately. B) slowly slow down, and then stop. C) go faster and faster. D) move with constant speed. According to Newton s first law, the law of inertia, if there is no net force on an object, because of its inertia it will continue to travel in the direction and at the velocity it is already traveling it will resist a change in motion. So, in the absence of an external force, a moving

2 object will move with a constant velocity. Even though the rockets are turned off, after that point in time there are no net forces acting on the spaceship so it will continue traveling at constant speed. Answer D (1 point/4) 5) 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 A) equal to its weight. B) equal to its mass. C) dependent on how fast it is moving. D) zero. According to Newton s first law, the law of inertia, if there is no net force on an object, because of its inertia it will continue to travel in the direction and at the velocity it is already traveling it will resist a change in motion. So, in the absence of an external force, a moving object will move with a constant velocity. Newton s first law tells us that to continue in the constant motion, then, NO force must be applied to the rocket in order to sustain its motion. Answer D (1 point/5) 6) You are standing in a moving bus, facing forward, and you suddenly fall forward. You can imply from this that the bus's A) velocity decreased. B) velocity increased. C) speed remained the same, but it's turning to the right. D) speed remained the same, but it's turning to the left. According to Newton s first law of motion, because you have mass you have inertia, and because you are in motion, even if the bus decreases its velocity, you will still want to continue in the same motion. So, if the bus decreases its motion, if you continue in the same motion you were in, it will seem as if you are falling forward, or as if a force pushed you, when all that is really happening is that you are just continuing in your same motion while the bus is lessening its motion and moving out from under you. If the bus increased its velocity it would seem as if you were falling backward. If the bus turned to the right, it would seem like you were falling to the left, and if the bus turned to the left, it would seem like the bus was turning to the right. Answer A (1 point/6) 7) 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? A) gravity B) normal force due to your contact with the floor of the bus C) force due to friction between you and the floor of the bus D) There is not a force leading to your fall. According to Newton s first law of motion, because you have mass you have inertia, and because you are in motion, even if the bus comes to an immediate stop, you will still want to continue in the same motion. So, if the bus stops, if you continue in the same motion you were in, it will seem as if you are falling forward, or as if a force pushed you, when all that is really happening is that you are just continuing in your same motion while the bus is lessening its motion and moving out from under you. There is no force pushing you. Answer D (1 point/7)

3 8) A constant net force acts on an object. Describe the motion of the object. A) constant acceleration B) constant speed C) constant velocity D) increasing acceleration Read this question carefully a constant net force means that there is a non-zero net force acting on the object it s just that the net force is constant. Remember there are only two possibilities either an object is accelerating or it is not accelerating. If the object is not accelerating, then there can be no net force on the object. If it is accelerating, then there must be a net force acting on the object. We are told that there is a constant net force acting on the object therefore, there must be a constant acceleration. Answer A (1 point/8) 9) The acceleration of an object is inversely proportional to A) the net force acting on it. B) its position. C) its velocity. D) its mass. F = ma, therefore, a =F/m this is saying that acceleration must be inversely proportional to the mass if the mass increases, this will cause a decrease in acceleration. If the mass increases this will cause a decreasing acceleration. Answer D (1 point/9) 10) 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 A) 4a. B) 2a. C) a/2. D) a/4. As we want to see what the affect of doubling the mass is on the acceleration, write out F=ma and isolate a, highlighting in some way that you want to know what happens when you double mass. I would write it out like this. a = F 2 m a 2 I have indicated by drawing a box around the 2 in front of the m that the change we are making is to double the mass. Then I write after that, the affect this has on the acceleration. This basically shows, if I double the mass, I halve the acceleration. You can waste large quantities of time trying to reason something out like this in your head if you don t have a visual cue in front of you. I highly recommend this approach. Answer C (1 point/10) 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? A) a/2 B) 8a C) 4a D) 2a As we want to see what the affect of doubling the F is on the acceleration, write out F = ma and isolate a, highlighting in some way that you want to know what happens when you double the F I would write it out like this.

4 a = 2 F m 2 a I have indicated by drawing a box around the 2 in front of the F that the change we are making is to double F. Then I write after that, the affect this has on the acceleration. This basically shows, if I double the F, I double the acceleration. You can waste large quantities of time trying to reason something out like this in your head if you don t have a visual cue in front of you. I highly recommend this approach. Answer D (1 point/11) 12) If you blow up a balloon, and then release it, the balloon will fly away. This is an illustration of A) Newton's first law. B) Newton's second law. C) Newton's third law. D) Galileo's law of inertia. The force of the air flowing out of the balloon is an action force. Because the air leaving the balloon is in the balloon, the action force of the air flowing out creates a corresponding reaction force on the balloon. The process of an action force creating a reaction force is the basis of Newton s third law. Answer C (1 point/12) 13) Two cars collide head-on. At every moment during the collision, the magnitude of the force the first car exerts on the second is exactly equal to the magnitude of the force the second car exerts on the first. This is an example of A) Newton's first law. B) Newton's second law. C) Newton's third law. D) Newton's law of gravitation. At every moment during the collision the force the first car applied on the second car would be considered an action force while the second must exert an equal and opposite reaction force on the first car. Action-reaction forces are the core idea of Newton s third law. Answer C (1 point/13) 14) If you exert a force F on an object, the force which the object exerts on you will A) depend on whether or not the object is moving. B) depend on whether or not you are moving. C) depend on the relative masses of you and the object. D) always be F. According to Newton s third law of motion, if you apply a force to an object, the object will apply an equal and opposite force on you therefore, the force the object applies to you will always be F. Answer D (1 point/14) 15) Action-reaction forces A) sometimes act on the same object. B) always act on the same object. C) may be at right angles. D) always act on different objects. According to Newton s third law of motion, one object applies an action force on a second object, and the second object applies an equal and opposite reaction force on the first object. So, action-reaction forces are always going to act on different objects. Answer D (1 point/15) 16) Action-reaction forces are A) equal in magnitude and point in the same direction.

5 B) equal in magnitude but point in opposite directions. C) unequal in magnitude but point in the same direction. D) unequal in magnitude and point in opposite directions. According to Newton s third law of motion, one object applies an action force on a second object, and the second object applies an equal and opposite reaction force on the first object. Therefore, action-reaction forces must necessarily by equal in magnitude and point in the opposite directions. Answer A (1 point/16) 17) 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 A) the force on the truck is greater then the force on the car. B) the force on the truck is equal to the force on the car. C) the force on the truck is smaller than the force on the car. D) the truck did not slow down during the collision. The amount of damage done on the car is irrelevant. The truck applied an action force on the car, and according to Newton s third law, the car must have applied an equal and opposite reaction force on the truck. Even thought the force on both was the same, the reason more damage was done on the car is that the car is much smaller and that amount of force would do more damage on the car than that same force would have done on the much larger truck. Answer C (1 point/17) 18) 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? A) The table pushing up on the object with force mg. B) The object pushing down on the table with force mg. C) The table pushing down on the floor with force mg. D) The object pulling upward on the Earth with force mg. Action reaction forces must be between the same two objects. Therefore, as the action force is between the object and the earth, the reaction force must be between the object and the earth. Therefore, the reaction force is the object pulling up with a gravity force on the earth. It s just that the force mg will have a much greater effect on the smaller object than on the much larger earth. Also, while the surface of the table will certainly be pushing up on the object, and the object will be pushing downward on the table, this is a different interaction than between the object and the earth. Answer D (1 point/18) 19) A child's toy is suspended from the ceiling by means of a string. The Earth pulls downward on the toy with its weight force of 8.0 N. If this is the "action force," what is the "reaction force"? A) The string pulling upward on the toy with an 8.0-N force. B) The ceiling pulling upward on the string with an 8.0-N force. C) The string pulling downward on the ceiling with an 8.0-N force. D) The toy pulling upward on the Earth with an 8.0-N force. Action reaction forces must be between the same two objects. Therefore, as the action force is between the earth and the toy, the reaction force must be between the earth and the toy. Therefore, the reaction force is the toy pulling up with a force of 8.0N on the earth. It s just that this force will have a much greater effect on the smaller toy than on the much larger earth. Also, while the toy will certainly be exerting a force on the string, and the string will be exerting a force of tension on the toy, this is a different interaction than between the toy and the earth. Answer D (1 point/19) 20) A golf club hits a golf ball with a force of 2400 N. The golf ball hits the club with a force

6 A) slightly less than 2400 N. B) exactly 2400 N. C) slightly more than 2400 N. D) close to 0 N. According to Newton s third law, if the golf club hits the ball with a force of 2400 N, which would be an action force, the golf ball would apply an equal and opposite reaction force of 2400 N on the club. Do not get wrapped up in thinking that this should somehow stop the club, or that the force of the ball on the club should somehow be less (we will explain this in greater detail when we talk about momentum). For now, just know that if the club smacks the ball with a force of 2400 N, the ball smacks the club with a force of 2400 N. Answer B (1 point/20) 21) Your bat hits the ball pitched to you with a 1500-N instantaneous force. The ball hits the bat with an instantaneous force, whose magnitude is A) somewhat less than 1500 N. B) somewhat greater than 1500 N. C) exactly equal to 1500 N. D) essentially zero. According to Newton s third law, if the bat hits the ball with a force of 1500 N, which would be an action force, the ball would apply an equal and opposite reaction force of 1500 N on the bat. Do not get wrapped up in thinking that this should somehow stop the bat, or that the force of the ball on the bat should somehow be less (we will explain this in greater detail when we talk about momentum). For now, just know that if the bat smacks the ball with a force of 1500 N, the ball smacks the bat with a force of 1500 N. Answer C (1 point/21) 22) State Newton's three laws. Newton's first law: Every object continues in its state of rest, or of uniform velocity in a straight line, as long as no net force acts on it. Newton's second law: The acceleration of an object is directly proportional to the net force acting on it, and is inversely proportional to its mass. The direction of the acceleration is in the direction of the net force acting on the object. Newton's third law: Whenever one object exerts a force on a second object, the second exerts an equal force in the opposite direction on the first. (1 point/22) 23) Mass and weight A) both measure the same thing. B) are exactly equal. C) are two different quantities. D) are both measured in kilograms. Please make sure you thoroughly understand that weight and mass are not the same thing. Mass is the amount of matter contained in an object. This amount of matter will be the same no matter where the mass is on the earth, on the moon or in outer space. It is this property of an object that is, the mass or its inertia that resists acceleration. The weight of an object is the effect of a force due to gravity on the mass. While the mass of an object on earth will be the same as that in outer space, the weight of the object on earth will be 9.8 times the mass, while in outer space, the weight will be 0. It does not matter what the weight is because of the object s mass (inertia), it will resist acceleration to the same the same degree anywhere. If

7 you push me with a force of 10N in outer space, I will accelerate to the same degree as if you push me horizontally with a force of 10 N while a hover a meter above the surface of the earth, despite the fact that I would have weight in the second situation. Answer C (1 point/23) 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? A) Mass is less, weight is same. B) Mass is same, weight is less. C) Both mass and weight are less. D) Both mass and weight are the same. Please make sure you thoroughly understand that weight and mass are not the same thing. Mass is the amount of matter contained in an object. This amount of matter will be the same no matter where the mass is on the earth, on the moon or in outer space. It is this property of an object that is, the mass or its inertia that resists acceleration. The weight of an object is the effect of a force due to gravity on the mass. If there is no gravity there is no weight. If gravity is less on the moon, then weight will be less on the moon. Therefore, in this case, the mass of the astronaut would be the same wherever it is, while the weight of the astronaut will be less on the moon. Answer B (1 point/24) 25) An example of a force which acts at a distance is A) tension. B) weight. C) static friction. D) kinetic friction. Some forces require contact between objects or surfaces in order to transmit the force. Other forces do not require contact for example, forces due to various fields, like gravitation, electric or magnetic fields. In this problem, the exerting of tension requires contact between a cord and an object; the exerting of static or kinetic friction requires contact between surfaces. On the other hand, weight is the application of force due to a gravitational field, and the object does not have to be in contact with the gravitational object. Answer B (1 point/25) 26) Who has a greater weight to mass ratio, a person weighing 400 N or a person weighing 600 N? A) the person weighing 400 N B) the person weighing 600 N C) Neither; their ratios are the same. D) The question can't be answered; not enough information is given. As F = ma, a = F, and weight is a force, a weight to mass ratio is just another way of m saying acceleration. You are aware that for weight, the acceleration we are referring to is the acceleration due to gravity. Because you know that all things near the surface of the earth are accelerated at the same rate by gravity, asking which person has the greater weight to mass ratio is the same thing as asking, which person is accelerated more by gravity? Without performing any calculations, you know the answer is, neither, their ratios are the same. However, you could certainly find the masses of each person by dividing their weights by g, and then comparing their weight to mass ratios which in both cases would turn out to be 9.8:1, or just 9.8. Answer C (1 point/26) 27) A stone is thrown straight up. At the top of its path, the net force acting on it is A) greater than its weight.

8 B) greater than zero, but less than its weight. C) instantaneously equal to zero. D) equal to its weight. Once the stone leaves the thrower s hand, the only force acting on the stone will be the force due to gravity. Therefore, the net force will also be just the force due to gravity. It doesn t matter at what part of the flight the stone is in, the net force on it will be just the force due to gravity and that will first cause it to decelerate as it is moving upward, and then accelerate in a negative direction as it is coming back down. In any case, as F = ma, the force will equal mg which is the same thing as its weight. Answer D (1 point/27) 28) A 20-N weight and a 5.0-N weight are dropped simultaneously from the same height. Ignore air resistance. Compare their accelerations. A) The 20 N weight accelerates faster because it is heavier. B) The 20 N weight accelerates faster because it has more inertia. C) The 5.0 N weight accelerates faster because it has a smaller mass. D) They both accelerate at the same rate because they have the same weight to mass ratio. As F = ma, a = F, and weight is a force, a weight to mass ratio is just another way of m saying acceleration. You are aware that for weight, the acceleration we are referring to is the acceleration due to gravity. Because you know that all things near the surface of the earth are accelerated at the same rate by gravity, even without performing any calculations, you know that both accelerate at the same rate, so any of the answers, if they say one weight will accelerate faster than the others, will be incorrect. This only leaves one possibility, D. Make sure you read the whole answer though. Because we just said that the weight to mass ratio is like saying acceleration, this statement is basically saying, they both accelerate at the same rate because the have the same acceleration. If you are not convinced, you can find masses for each object by dividing their weights (the force) by g, and then dividing the force by the mass (the weight to mass ratio) and see that you get 9.8 m/s 2 for both. Answer D (1 point/28) 29) A brick and a feather fall to the earth at their respective terminal velocities. Which object experiences the greater force of air friction? A) the feather B) the brick C) Neither, both experience the same amount of air friction. D) It cannot be determined because there is not enough information given. If any object is at a terminal velocity, this means it is no longer accelerating. At that point, the force of air friction upward must equal to the force due to gravity in the opposite direction. Because the force due to gravity on the brick (its weight) will be greater than the force due to gravity (its weight) on the feather, the force of friction on the brick must also be greater than the force of air friction on the feather. Please understand the ACCELERATION due to gravity is the same on both. However, as the force due to gravity (that is, the weight) is mg, the force due to gravity on the brick is much greater than that on the feather. Answer B (1 point/29) 30) An object of mass m is hanging by a string from the ceiling of an elevator. The elevator is moving up at constant speed. What is the tension in the string? A) less than mg B) exactly mg C) greater than mg

9 D) cannot be determined without knowing the speed As the motion of the object is in the vertical direction, the forces controlling its motion (whether it accelerates or travels at a constant velocity) are also in the vertical direction. As the force due to gravity will always be the same no matter what the motion of the elevator is, whether or not the object is accelerating will depend on the force due to tension. If the elevator is at constant velocity, as the object will also be at a constant velocity, the tension in the string must also be equal to the force due to gravity. Answer B (1 point/30) 31) An object of mass m is hanging by a string from the ceiling of an elevator. The elevator is moving upward, but slowing down. What is the tension in the string? A) less than mg B) exactly mg C) greater than mg D) zero As the motion of the object is in the vertical direction, the forces controlling its motion (whether it accelerates or travels at a constant velocity) are also in the vertical direction. As the force due to gravity will always be the same no matter what the motion of the elevator is, whether or not the object is accelerating will depend on the force due to tension. If the elevator is at constant velocity, as the object will also be at a constant velocity, the tension in the string must also be equal to the force due to gravity. If the elevator is accelerating upward, the object must also be accelerating. Therefore, the tension would have to be greater than the acceleration due to gravity. However, in this case, the elevator is slowing down so the object must also be undergoing a negative acceleration. Therefore, the force due to gravity must be greater than the force due to tension or tension is less than mg. Answer A (1 point/31) 32) The force that keeps you from sliding on an icy sidewalk is A) weight. B) kinetic friction. C) static friction. D) normal force. The force that keeps you from sliding would be that force that acknowledges that your foot is not moving or slipping as you push off on it against the icy sidewalk this would be the force of static friction, as static means not moving. That is, the force of static friction keeps your foot from moving as you push against the surface of the earth with it. Although the normal force contributes to the static friction, it is not the static friction. Therefore, as weight is the opposite of normal force, it also is not the static friction force. Finally, kinetic friction is the force of friction present as the foot is moving against the ice. If this is the case, the foot is slipping. Answer C (1 point/32) 33) A horizontal force accelerates a box from rest across a horizontal surface (friction is present) at a constant rate. The experiment is repeated, and all conditions remain the same with the exception that the horizontal force is doubled. What happens to the box's acceleration? A) It increases to more than double its original value. B) It increases to exactly double its original value. C) It increases to less than double its original value. D) It increases somewhat. When considering this problem, you have to keep in mind that the force that causes an acceleration is net force, not the applied force. If the sum of forces is 0 N, there will not be an acceleration, even if you have a significant positive applied force it is the net force that

10 equals mass times acceleration. Therefore, if you double the applied force, you will not be doubling the net force. To see your way through a problem like this it would probably be best that you not spend a lot of time trying to reason through it but simply create a quick illustration. If you had an acceleration of a 1 kg box caused by an F A of 2 N and F f of -1 N, the acceleration would be 1 N. However, if you doubled the applied force (F A is now 4 N), the net force is now 3N and the acceleration would be 3 m/s 2. So, the box increases to more than double its original value. Answer A (1 point/33) First Experiment : F net = F f + F A = ma F net = 1N + 2N = (1kg) a a = 1 m / s 2 Second Experiment : F net = F f + F A = ma F net = 1N + 4N = (1kg) a a = 3 m / s 2 34) A packing crate slides down an inclined ramp at constant velocity. Thus we can deduce that A) a frictional force is acting on it. B) a net downward force is acting on it. C) it may be accelerating. D) it is not acted on by appreciable gravitational force. Remember that forces in the parallel direction control the motion of an object down an incline. Although we may add in other forces in the parallel direction, usually we will only consider the force due to gravity in the parallel (x) direction, and a friction force. These would usually be the main forces in a net force equation that determined whether or not there would be an acceleration. Because there will always be a parallel force due to gravity down an incline, if there is no acceleration, there must be an opposing force of friction. Therefore, if there is no acceleration, we can deduce that there must be a force of friction. For answer B, if there was a net downward force, saying there is a net force means that it must be accelerating. For answer C, we have already stated that it is at constant velocity so it cannot be accelerating. For answer D, if it was not acted on by any appreciable force, the force of friction would cause the object to be at rest. Answer A. (1 point/34) 35) A block of mass M slides down a frictionless plane inclined at an angle θ with the horizontal. The normal reaction force exerted by the plane on the block is A) Mg. B) Mg sin θ. C) Mg cos θ. D) zero, since the plane is frictionless. Remember that in your diagram of forces associated with an incline, we have stated that forces are as follows: The perpendicular component (the y component) of the force due to gravity will be mgcos θ, and so the normal force would be - mgcos θ. We normally let the negative sign of g provide the direction of the force so as we need to make the normal force a force in a positive direction, we always add in the negative sign to make it positive. However, not every one does this. It is common practice when we are talking about g as just an absolute value to make it positive. Therefore, on many multiple choice questions you will have to read into the question that it is just using the absolute value of g. That is the case in this question. If we treat g as positive here, the correct answer is Mgcos θ. Answer C (1 point/35)

11 36) A block of mass M slides down a frictionless plane inclined at an angle θ with the horizontal. The normal reaction force exerted by the plane on the block is directed A) parallel to the plane in the same direction as the movement of the block. B) parallel to the plane in the opposite direction as the movement of the block C) perpendicular to the plane. D) toward the center of the Earth. The definition of the normal force is that force of a surface supporting an object, directed perpendicular to the surface. Therefore, the normal force on an incline is always directed perpendicular to the plane of the incline. Answer C (1 point/36) 37) A block of mass M slides down a frictionless plane inclined at an angle θ with the horizontal. The gravitational force is directed A) parallel to the plane in the same direction as the movement of the block. B) parallel to the plane in the opposite direction as the movement of the block C) perpendicular to the plane. D) toward the center of the Earth. Remember, while a component of the force due to gravity (the parallel or x component) is directed down the incline of the plane, the total gravitational force is ALWAYS directed toward the center of the earth. Answer D (1 point/37) 38) Two toy cars (16 kg and 2.0 kg) are released simultaneously on an inclined plane that makes an angle of 30 with the horizontal. Make a statement which best describes their acceleration after being released. A) The 16-kg car accelerates 8 times faster than the 2.0-kg car. B) The 2.0-kg car accelerates 8 times faster than the 16-kg car. C) Both cars accelerate at the same rate. D) none of the above Remember, the two forces affecting the acceleration down the incline would be the force due to gravity in the parallel direction, (which is -mgsinθ ) and friction, which is µ mgcosθ. A net force equation to solve for a would look like: F! net = mg sinθ + µmg cosθ = ma. You can see that mass cancels out of both sides of the equation meaning that a = g sinθ + µg cosθ. Regardless of what the mass is, the acceleration will be the same this is a good thing to simply remember for multiple choice questions. Don t have to reason this out, just know it. Answer C (1 point/38) 39) If you push a 4.0-kg mass with the same force that you push a 10-kg mass from rest, A) the 10-kg mass accelerates 2.5 times faster than the 4.0-kg mass. B) the 4.0-kg mass accelerates 2.5 times faster than the 10-kg mass. C) both masses accelerate at the same rate. D) None of the above is true. Wanting to see what happens if you push a 4.0 kg mass instead of a 10 kg mass is the same thing as asking what would happen if we multiplied the mass by 4/10. As we want to see what the affect of this is on the acceleration, I would isolate a in the F = ma formula, highlighting in some way that you want to know what happens when you multiply the mass by.4 I would write it out like this. a = F.4 m 2.5 a I have indicated by drawing a box around the.4 in front of the m that the change we are making is to multiply m by.4. Then I write after that, the affect this has on the acceleration. This basically shows, if I multiply m by.4, I multiply acceleration by 2.5. Therefore, the 4.0

12 kg mass accelerates 2.5 times faster. You can waste large quantities of time trying to reason something out like this in your head if you don t have a visual cue in front of you. I highly recommend this approach. Answer B (1 point/39) Quantitative Problems 1) An object sits on a frictionless surface. A 16-N force is applied to the object, and it accelerates at 2.0 m/s2. What is the mass of the object? You probably don t need a free body diagram and a net force equation to visualize this but if you do it would look like this. There is only one force so the sum of forces in the x direction in the net force equation is just the applied force. Isolate for m, plug in values and solve. All I would need to see is what is in the box. (3 points/3) 2) A sports car of mass 1000 kg can accelerate from rest to 27 m/s in 7.0 s. What is the average forward force on the car? You know that F net = ma. You have been given the mass but will need to calculate the acceleration. You should be able to do so without a motion diagram. You can immediately see that you have initial and final velocities, and you also have time. Therefore, you can use v = v 0 + at isolated for a. If you need to draw a quick motion diagram to collect data for the motion part of the problem it would look like the diagram below. Once you find acceleration, plug it back into the F=ma formula. I would only need to see what is in the box. (6 points/9) 3) Starting from rest, a 4.0-kg body reaches a speed of 8.0 m/s in 2.0 s. What is the net force acting on the body? You know that F net = ma. You have been given the mass but will need to calculate the acceleration. You should be able to do so without a motion diagram. You can immediately see that you have initial and final velocities, and you also have time. Therefore, you can use v = v 0 + at isolated for a. If you need to draw a quick motion diagram to collect data for the motion part of the problem it would look like the diagram below. Once you find acceleration, plug it back into the F=ma formula. I would only need to see what is in the box. (6 points/15) 4) An antitank weapon fires a 3.00-kg rocket which acquires a speed of 50.0 m/s after traveling 90.0 cm down a launching tube. Assuming the rocket was accelerated uniformly, what is the average force acted on it?

13 You know that F net = ma. You have been given the mass but will need to calculate the acceleration. You should be able to do so without a motion diagram. You can immediately see that you have initial and final velocities, and you also have displacement. What you do not have is time, so you need to use v 2 = v a(x x 0 ) isolated for a. If you need to draw a quick motion diagram to collect data for the motion part of the problem it would look like the diagram below. Once you find acceleration, plug it back into the F=ma formula. I would only need to see what is in the box. Make sure you convert cm to m. I would also carry an extra digit in the acceleration calculation to help avoid rounding error. (6 points/ 21) 5) What is the mass of a person who weighs 110 lb? Remember that weight is a force. Therefore, as F = ma, the weight of a person equals mass times acceleration, with the acceleration being the acceleration due to gravity. Therefore, isolate for mass, plug in the values and solve (recognize, we cannot have negative mass so we will have to use an absolute value for acceleration due to gravity)(also, you will need to convert pounds to kg). (3 points/24) m = F a = (110lb) ( 4.45N 1lb ) 9.8 m / s 2 = 50 kg 6) What is the mass of an object that weighs 250 N on the surface of the Earth where the acceleration due to gravity is 9.80 m/s2? Remember that weight is a force. Therefore, as F = ma, the weight of a person equals mass times acceleration, with the acceleration being the acceleration due to gravity. Therefore, isolate for mass, plug in the values and solve (recognize, we cannot have negative mass so we will have to use an absolute value for acceleration due to gravity). (3 points/27) m = F a = 250N = 25.5 kg m / s 7) An object has a mass of 60 kg on the Earth. What is the mass of the object on the surface of the Moon where the acceleration due to gravity is only 1/6 of that on the Earth? This is a trick question no calculations are involved. The mass of an object is the same no matter where the mass is. If the object has a mass of 60 kg on earth, it will still have a mass of 60 kg on the moon. However, its weight will be 1/6 that of what it is on the moon. (1 point/28) 8) Object A weighs 40 N on Earth, and object B weighs 40 N on the Moon. The Moon's gravity is one sixth of Earth's. Compare the masses of the objects. As the force (weight) of each object is 40N, and as F = ma, and as the acceleration due to gravity on the moon is 1/6 the mass on the earth, the approach to this problem is to create an equation for the force of each object and set them equal to each other to see how the mass of B compares to the mass of A.

14 1 For object A we know F = m A g. For object B we know that F = m B ( 6 )g Therefore, we can 1 say that m A g = m B ( 6 )g or m A is 1 m or m 6 B B is 6 times m A. (3 points/31) 9) Sue and Sean are having a tug-of-war by pulling on opposite ends of a 5.0-kg rope. Sue pulls with a 15-N force. What is Sean's force if the rope accelerates toward Sue at 2.0 m/s2? In this problem we are told that a 5.0 kg rope is accelerating. We know that in order for the rope to accelerate there needs to be a non-zero net force acting on it. This is a simple enough problem that we could probably just write out a net force equation to solve for Sean s force but I will include a free body diagram anyway. Starting with a free body dot to represent the rope that is accelerating we are given that there is a force of 15 N applied by Sue. We draw this in as a force to the right and label it F Sue =15N. We know that Sean is applying an opposing force but that the arrow must be shorter because the acceleration is toward Sue. Label this F Sean =? Since we know that these two forces are related to each other through the net force equation we should write out the net force equation to see if we can create an equation that will solve for F Sean. The net force equation reveals that we can solve for F Sean by subtracting F Sue from ma. We have all of the information we need so we plug this in and find that Sean is applying a force to the rope of 5N in the direction opposite that of Sue. (4 points/35) F x net = F Sue + F Sean = ma F Sean = ma F Sue = (5.0kg)(2.0m / s 2 ) 15N = 5N 10) A stack of books rests on a level frictionless surface. A force F acts on the stack, and it accelerates at 3.0 m/s2. A 1.0 kg book is then added to the stack. The same force is applied, and now the stack accelerates at 2.0 m/s2. What was the mass of the original stack? Any time we have two different scenarios, and one of the variables is the same for both scenarios (in this case, force is stated to be the same) we can create an equation for both scenarios for the variable that is the same, and then set these equations equal to each other to help us solve the desired variable. For the first situation we are told that force F accelerates a stack of books 3.0 m/s 2. We are not told what the mass is but we can just state that it is m, and create the equation F = (3.0m / s 2 )(m) Then we are told that the same force is used to push the stack, but that 1.0 kg has been added to the stack, and it now accelerates at 2.0 m/s 2. We can now create an equation that says F = (2.0m / s 2 )(m +1.0), indicating that, whatever the mass is, when we add 1.0 kg to it, the acceleration lessens to 2.0 m/s 2. Then, because the force was equal for both accelerations, we can set the two equations equal to each other and solve for m. We find that the original mass was 2.0 kg. Notice it will help if we remove the units from the equation until we finish solving. (4 points/39) F = (3.0m / s 2 )(m) F = (2.0m / s 2 )(m +1.0) 3m = 2(m + 1) 3m = 2m + 2 3m 2m = 2 m = 2.0 kg

15 ELEVATOR with SCALE PROBLEMS For any scale in an elevator problem, you need to remember that we only need to consider the forces in the y direction, because these are the forces that will determine the whether or not the person accelerates. The two forces will always be the force due to gravity in the downward direction, and for a particular problem, this force will not change it will always be mg. The force in the opposite direction will be the normal force of the scale upward on the person standing on the scale. You are aware that what the scale actually measures as weight is the value of the normal force on the person. This force can change, and therefore, what the scale reads can change (even though the force due to gravity does not), depending on what the elevator is doing. If the elevator is moving with constant velocity, in either direction, the person standing on the scale will not be accelerating in either direction. As the only forces we can consider in determining whether or not the person is accelerating are the forces on the person, as the person would not be accelerating, the force due to gravity must equal the normal force of the scale on the person. Therefore, if there is no acceleration in either direction, the normal force will equal the force due to gravity and the scale will show that actual value of the force due to gravity, or the weight. If the elevator is accelerating in an upward direction, or decelerating in a downward direction, in order to cause this acceleration the normal force of the scale on the person must be greater than the force due to gravity. Therefore, under these circumstances, even though the force due to gravity (the actual weight) has not changed, because the normal force on the person has increased, the scale will show an increased weight and the person will actually feel heavier. If the elevator is decelerating in an upward direction, or accelerating in a downward direction, in order to cause this acceleration, the normal force of the scale on the person must be less than the force due to gravity. Therefore, under these circumstances, even though the force due to gravity (the actual weight) has not changed, because the normal force on the person has decreased, the scale will show a decreased weight and the person will actually feel lighter. To solve problems related to these concepts simply draw free body diagrams showing the relationship of the normal force of the scale on the person (F scale ) in relationship to the force due to gravity (mg) and then create net force equations to isolate and solve for desired values. 12) A person of weight 480 N stands on a scale in an elevator. What will the scale be reading when the elevator is accelerating downward at 4.00 m/s2? See the above section on elevator problems. We are already given that the force due to gravity is 480 N so we won t have to calculate this using mg. We should also immediately know this is in the negative direction. We are also told that the elevator is accelerating downward at 4.00 m/s 2, which means that the person is also accelerating downward at 4.00 m/s 2 (-4.00 m/s 2 ). In these kinds of problems I always draw a free body diagram to make sure I am visualizing the problem correctly. In this case the gravity vector should be longer than the F scale vector so you immediately know that the scale will read less than the 480 N of the individual. To find F scale create a net force equation and isolate F scale. You will also need to solve the given weight for mass by dividing the weight by g. Then plug in the values and solve for F scale. As we predicted, the reading of the scale is less than the actual weight. (4 points/43)

16 13) A person on a scale rides in an elevator. If the mass of the person is 60.0 kg and the elevator accelerates downward with an acceleration of 4.90 m/s2, what is the reading on the scale? See the above section on elevator problems. We are told that the elevator is accelerating downward with an acceleration of 4.90 m/s 2 (meaning m/s 2 ). In these kinds of problems I always draw a free body diagram to make sure I am visualizing the problem correctly. In this case the gravity vector should be longer than the F scale vector so you immediately know that the scale will read less than the weight (mg) of the individual. To find F scale create a net force equation and isolate F scale, plug in the values and solve. (4 points/47) 14) A person on a scale rides in an elevator. If the mass of the person is 60.0 kg and the elevator accelerates upward with an acceleration of 4.90 m/s2, what is the reading on the scale? See the above section on elevator problems. We are told that the elevator is accelerating upward with an acceleration of 4.90 m/s 2. In these kinds of problems I always draw a free body diagram to make sure I am visualizing the problem correctly. In this case the gravity vector should be shorter than the F scale vector so you immediately know that the scale will read more than the weight (mg) of the individual. To find F scale create a net force equation and isolate F scale, plug in the values and solve. (4 points/51) 15) Two horizontal forces act on a 5.0-kg mass. One force has a magnitude of 8.0 N and is directed due north. The second force toward the east has a magnitude of 6.0 N. What is the acceleration of the mass? Don t forget that we can add force vectors in vector addition problems the same way we can add any other vectors. We may need to use vector resolution (that process where, if the vectors are not added together at right angles, we find the x and y components of both vectors we are adding to create a combined x vector and a combined y vector to find the resultant vector and the direction of the resultant). However, in this case, the force vectors are being added at right angles to each other so we can just use the Pythagorean theorem and inverse tangent. First draw a vector diagram to show the relationship between the two vectors. As you will want the direction of the force vector you will have to decide which axis you want the angle of the vector to be with reference to. It does not matter I have drawn my diagram so the vector is with reference to the vertical axis. We have found the resultant force but we are not through yet. We need to know the acceleration. We have not been given information about any other forces acting on the object, including friction so, we can assume that there are no other forces. The resultant force vector we just identified is the only force acting on the object. Therefore, it is the net force on the object as well, and we know F net = ma. Isolating a, we get a = F net m. Plug in the values and solve, making sure you include the direction of the acceleration in your final answer. (9 points/60)

17 16) An object of mass 6000 kg rests on the flatbed of a truck. It is held in place by metal brackets that can exert a maximum horizontal force of 9000 N. When the truck is traveling 15 m/s, what is the minimum stopping distance if the load is not to slide forward into the cab? You are familiar by now with the concept that some problems will have both a force and a kinematic component. When you have both elements you will want to start solving the problem from the perspective of what the question is asking for. In this question you are asked to find a the minimum stopping distance, which is a displacement, so start the problem from a kinematic perspective, and this will lead you through the problem. I would draw a motion diagram to collect information to determine what I needed to ultimately get from the force problem, although, you might be able to see right away that as you are being asked to find a displacement, and you have not been given a time, you will probably have to use v 2 = v a(x x 0 ) isolated for x, to solve for this displacement. And you will need the force part of the problem to solve for acceleration. If you cannot reason this out right away, again, draw the motion diagram and let the information you collect tell you that you need to use this formula to find x, and complete the isolation so you can see that the missing piece of information is acceleration. Once you determine that you need the force part of this problem to solve for acceleration, create a free body diagram so you can visualize the forces involved. Creating the free body diagram will help you to realize that there is actually only one force involved in the acceleration of the object. THERE IS NO FORWARD APPLIED FORCE ON THE OBJECT. Once the truck begins to decelerate any forces acting to push the object in a forward direction are no longer acting (we are ignoring friction in this problem). The only force on the object is the force in the negative direction of the metal brackets, a force of 9000 N. This is the only force we need to be concerned about in terms of causing the acceleration (deceleration) of the object. Once you have visualized this, you can then write down a net force equation, for which the sum of forces will only be this one force (the force of the metal brackets), relating the sum of forces to ma, and solve for a. Then, once you have a, you can plug this back into the motion formula to solve for the displacement. (6 points/66) 17) An object of mass 6000 kg rests on the flatbed of a truck. It is held in place by metal brackets that can exert a maximum horizontal force of 9000 N. When the truck is traveling 15 m/s, what is the minimum stopping time if the load is not to slide forward into the cab? On the AP test, this would be a second part to question 16. This is a kinematic problem. You already have the initial and final velocities. In question 16 you found the acceleration and minimum displacement for the truck to stop. You would have three formulas available to you to find time (the three involving acceleration NOT the one involving a constant velocity). You should use the easiest one, v = v 0 + at, isolated for time. Just plug the values in and solve. (3 points/69)

18 v = v 0 + at t = v a = 15m / s 1.5 m / s 2 = 10s MULTIPLE MOVING MASS PROBLEMS When we have problems containing the motion of two (or more) objects, we need to analyze the motion of each object separately, remember that ONLY FORCES ACTING ON A GIVEN OBJECT ACCOUNT FOR THE MOTION OF THAT OBJECT. Additionally, in this type of problem, it will be that acceleration for each object, and tensions in the cord connecting the object, will be equal (T will be equal but in the opposite direction). This means that we will either be able to create equations for tension from the net force equations of both objects and set them equal to each other to solve for acceleration; or, we will be able to create equations for acceleration from the net force equations of both objects and set them equal to each other to solve for tension. Remember this as a basic problem solving strategy that we have for many types of problems. Once we find either the acceleration or tension, we can plug these values back into the original equations we created to find the other variables. Direction of motion one thing to always remember about these problems is that the direction of motion of the connected objects will be considered to always be the positive direction even if the direction has one of the objects traveling downward. Therefore, the acceleration due to gravity (g) must be considered to have a positive value for the object traveling in the positive direction. If the other object is traveling upward, the acceleration due to gravity for it would be in the negative direction. If the other object is on a flat surface, there is not an acceleration due to gravity vector involved in the direction of motion but it will be involved in determining the force due to friction, so g in this case would also be negative. Additionally, tension of the connecting cable on the mass moving downward must be considered to be in a negative direction. FIGURE ) In the Atwood machine shown in Fig. 4-1, if M = 0.60 kg and m = 0.40 kg, what is the magnitude of the acceleration of the system? (Ignore friction and the mass of the pulley.) (See MULTIPLE MOVING MASS PROBLEMS above). In this problem you have two masses moving, with the heavier mass moving downward while the lighter mass will be moving upward. The positive direction of motion will be counterclockwise (leftward) around the pulley. Therefore, acceleration due to gravity for the heavier mass will be in the positive direction while acceleration due to gravity for the lighter mass will be in the negative direction. We will need to analyze the motion of each mass separately, each with its own free body diagram and net force equation. Because we have been asked to find acceleration, we will need to create equations from each net force equation for tension (T). This will allow us to set these equations equal to each other and solve for acceleration. For the heavier mass the force due to gravity vector is in the positive direction while the tension vector is in the negative direction, and the gravity vector must be longer otherwise the mass would not be accelerating in the downward direction. (see below) Then, create a net force equation specifically for that mass (F Mnet), relating it to m M a, and then solving for T (see below) Make sure you subscript appropriately to be able to keep things straight later on. I would subscript g in this case just to remind you that it needs to be positive. In the net force

19 equation, T has a negative sign make sure in isolating it you move it to the opposite side of the equation so it is positive. For the lighter mass the force due to gravity vector is in the negative direction while the tension vector is in the positive direction, and the tension vector must be longer otherwise the mass would not be accelerating in the upward direction. (see below) Then, create a net force equation specifically for that mass (F mnet), relating it to m m a, and then solving for T (see below) Make sure you subscript appropriately to be able to keep things straight later on. I would subscript g in this case just to remind you that it needs to be negative. Then, set the two tension equations equal to each other and solve for acceleration, making sure to keep negative signs straight. I would need to see all of this, including the free body diagrams. (13 points/82) 19) In the Atwood machine shown in Fig. 4-1, if M = 0.60 kg and m = 0.40 kg, what is the tension in the string? (Ignore friction and the mass of the pulley.) Recognize that on the AP test, question 19 would be the second part of question 18, and you would be able to use the answer you got in the previous question to help in this question. Here, you are asked for the tension in the string. In the previous problem you created two equations for tension and both should work. For our purposes, we would want to check both to see if they gave us the same answer. However, on the AP test, you would pick one or the other. Because of rounding error, you will likely get slightly different answers if you use both. That s okay for us but again, just use one or the other on the test. For this question, just write out both formulas you created and plug in the 1.96 m/s 2. What s in the box is all I would need to see for this. (6 points/88) T = m M g M m M a = (.6)(9.8) (.6)(1.96) = 4.7N T = m m a m m g m = (.4)(1.96) (.4)( 9.8) = 4.7N 20) A student pulls a box of books on a smooth horizontal floor with a force of 100 N in a direction of 37 above the horizontal. If the mass of the box and the books is 40.0 kg, what is the acceleration of the box? As with all force problems the first thing we need to do is draw a free body diagram. As we have not been specifically asked to draw a free body diagram we can draw ours with components on it and we will need to see these visually in this problem. We have been told that the box is on a smooth floor, so one thing we won t have to worry about is friction, so this won t go in our free body diagram. Our free body diagram should include force due to gravity (mg), force applied (F A ) at a 37 degree angle, x and y components of F, (F Ax and F Ay ) and normal force (F N ), although realize as there is no force of friction, we will not need F N or F Ay. We want to find the acceleration and we know that way to find this is through the net force equation. The motion is in the horizontal direction so the only force that will be affecting the acceleration of is the x component of the applied force. So, write out the net force equation only including this force in the sum of forces, and relate it to ma. Then isolate a, plug in the given values and solve. (4 points/92)

20 21) A student pulls a box of books on a smooth horizontal floor with a force of 100 N in a direction of 37.0 above the horizontal. If the mass of the box and the books is 40.0 kg, what is the normal force on the box? As with all force problems the first thing we need to do is draw a free body diagram. As we have not been specifically asked to draw a free body diagram we can draw ours with components on it and we will need to see these visually in this problem. We have been told that the box is on a smooth floor, so one thing we won t have to worry about is friction, so this won t go in our free body diagram. Our free body diagram should include force due to gravity (mg), force applied (F A ) at a 37 degree angle, x and y components of F, (F Ax and F Ay ) and normal force (F N ). You will recognize this problem as being similar to the previous problem, and although it is often the case you can use information from previous sections of problems to help, in this case, nothing from question 20 will really be helpful so we will treat 21 as a separate problem. In this problem, we have been asked to find the normal force. As the normal force is in the y direction of forces we can use a net force equation in the y direction to help us find the normal force. After completing the free body diagram we should see that mg, normal force and the y component of the applied force are all in the y direction. As there is no acceleration in the y direction, the sum of forces for this net force equation should sum to 0 N. Write out the net force equation, isolate F N and see where this leads us. Actually, once we isolate the normal force all we really have to do is plug in values and solve. (Note look at the equation for the normal force although we created it form the net force equation, recognize that all it is really telling us to do is subtract the value of the y component of the applied force from the force due to gravity). FIGURE ) A traffic light of weight 100 N is supported by two ropes as shown in Fig What are the tensions in the ropes? As with all force problems the first thing we need to do is draw a free body diagram. As we have not been specifically asked to draw a free body diagram we can draw ours with components on it and we will need to see these visually in this problem. The free body of the free body diagram in this case would actually be the point at which the three cables join. Because the lengths of the cables on either side are the same length and at the same angle, the tension in each cable would be the same. We have also been given the force due to gravity (mg = 100N).

21 After we see the free body diagram we should see that in order to find T, we will have to identify some other formula with T in it, for which we can isolate T and solve. The only two candidates would either be the x component of T, which is Tcosθ, or the y component of T, which is Tsinθ. If we can equate something else with either of these, we can solve for T. First, lets look at forces in the x direction. Notice that the only two forces in the x direction of the individual x components of the force due to tension, each in the opposite direction of the other. This will not help us because we have no way of finding the value of the x component of T (yet). Next, lets look at the y component of the tension forces. Notice that there are two of these, and the sum of these two y components of T is balanced by the force due to gravity. We know that these must be equal because there is no acceleration in the y direction. Therefore, we can write a net force equation for forces in the y direction, and equate this with 0N (because ma will equal 0). Also, after isolating T, make sure you recognize that mg is -100N. 23) In Fig. 4-2, if the tensions in the ropes are 50 N, what is the mass of the traffic light? As with all force problems the first thing we need to do is draw a free body diagram. As we have not been specifically asked to draw a free body diagram we can draw ours with components on it and we will need to see these visually in this problem. The free body of the free body diagram in this case would actually be the point at which the three cables join. Because the lengths of the cables on either side are the same length and at the same angle, the tension in each cable would be the same. We have been given that the tension in each rope is 50 N. After seeing the free body diagram we should be able to see that as what we want to find, mass, is part of a force in the y direction, we should be able to create a net force equation for forces in the y direction and solve for mass. As the forces in the y direction must be balanced (there is no acceleration in the y direction), the sum of forces must equal 0. The forces in the y direction include the force due to gravity and the two y components of the tension. Once m is isolated, we have the other values, plug them in and solve. FIGURE ) Two boxes of masses m and 2m are in contact with each other on a frictionless surface. (See Fig. 4-3.) What is the acceleration of the more massive box? A) F/m B) F/(2m) C) F/(3m) D) F/(4m)