Homework 4, Newton's laws 23-24-solutions

levesque (aml6939) – Homework 4, Newton’s laws 23-24 – tejeda – (WilkesDPHY1 1) 1 This print-out should have 24 questions. Multiple-choice questions may continue on the next column or page – find all choices before answering. You will have approximately two weeks to complete this homework assignment. Collab- oration on homework is encouraged. How- ever, you must submit your own responses, and you may be required to show your in- dependent work to your HS Instructor. You are expected to access Quest daily to progress through this homework assignment. 001(part1of2)10.0points A book is at rest on an incline as shown below. A hand, in contact with the top of the book, produces a constant force F hand vertically downward. F hand Book The following figures show several attempts at drawing free-body diagrams for the book. Which figure has the correct directions for each force? The magnitudes of the forces are not necessarily drawn to scale. 1. F g F hand F N F f 2. F g F hand F N F f 3. F g F f F N F hand 4. F N F f F g F hand 5. F g F N F f F hand 6. F N F hand F f F g 7. F g F f F hand F N

levesque (aml6939) – Homework 4, Newton’s laws 23-24 – tejeda – (WilkesDPHY1 1) 2 8. F g F hand F f F N correct Explanation: The normal force points perpendicular to the surface of the inclined plane. The weight force points down. The F hand also points down. The friction force keeps the book from sliding and consequently points up the incline. 002(part2of2)10.0points For the normal force exerted on the book by the wedge in the diagram, which force(s) complete(s) the force pair for Newton’s third law (action-reaction)? 1. the component of gravity pointing parallel to the surface of the incline 2. the pull of the earth on the book 3. the component of F hand pointing perpen- dicular to the surface of the incline 4. the normal force exerted on the wedge by the book correct 5. the pull of the book on the earth 6. the sum of the component of gravity per- pendicular to the surface of the incline and the component of F hand perpendicular to the surface of the incline 7. the component of gravity pointing per- pendicular to the surface of the incline Explanation: The force that completes the third law pair with the normal force of the wedge on the book is the normal force of the book on the wedge. The “the sum of the component of gravity perpendicular to the surface of the incline and the component of F hand perpendicular to the surface of the incline” is equal and opposite to the normal force, but it is not a action-reactionpair ; e.g. , two forces equal and opposite to each other. 003(part1of2)10.0points A balloon is waiting to take off. As seen in the figure below, the balloon’s basket sits on a platform which rests on the ground. The balloon is pulling up on the basket, but not hard enough to lift it off the platform. platform basket ground balloon What is the free-body diagram for the plat- form? Some parallel vectors are offset hori- zontally for clarity. 1. F ground on plat F grav on plat F bask on plat correct 2. F bal on bask F grav on bask F grav on plat 3. F bal on plat F grav on plat 4. F bal on plat F plat on ground F grav on plat 5. F plat on bask F plat on ground F grav on plat Explanation:

levesque (aml6939) – Homework 4, Newton’s laws 23-24 – tejeda – (WilkesDPHY1 1) 4 F net - 9 . 27 N 1 . 13 N θ F net = radicalBig ( F x ) 2 + ( F y ) 2 = radicalBig (1 . 13 N) 2 + ( - 9 . 27 N) 2 = 9 . 33862 N . 008(part2of2)10.0points What is the direction of the net external force on the apple (measured from the downward vertical, so that the angle to the right of downward is positive)? Answer in units of ◦ . Correct answer: 6 . 94999 ◦ . Explanation: tan θ = F x F y θ = tan − 1 parenleftbigg F x F y parenrightbigg = tan − 1 parenleftbigg 1 . 13 N - 9 . 27 N parenrightbigg = - 6 . 94999 ◦ which is 6 . 94999 ◦ to the right of downward. 009(part1of2)10.0points Consider the collision between a car and a light truck whose weights are equal ( M = m ). m M v v If they are moving at the same speed when they collide, compare the size (or magnitude) of the forces between the car and the truck. Friction is so small that it can be ignored. 1. The truck exerts the same amount of force on the car as the car exerts on the truck. correct 2. The truck exerts a greater amount of force on the car than the car exerts on the truck. 3. Neither exerts a force on the other; the car gets smashed simply because it is in the way of the truck. 4. None of the answers correctly describes the situation. 5. Not enough information is given to pick one of the answers. 6. The car exerts a greater amount of force on the truck than the truck exerts on the car. Explanation: By Newton’s third law, action and reaction are of the same magnitude and in the opposite direction. 010(part2of2)10.0points Compare the forces if the light truck is stand- ing still when the car hits it. 1. The truck exerts the same amount of force on the car as the car exerts on the truck. correct 2. The truck exerts a greater amount of force on the car than the car exerts on the truck. 3. Neither exerts a force on the other; the car gets smashed simply because it is in the way of the truck. 4. Not enough information is given to pick one of the answers. 5. None of the answers correctly describes the situation. 6. The car exerts a greater amount of force on the truck than the truck exerts on the car.

levesque (aml6939) – Homework 4, Newton’s laws 23-24 – tejeda – (WilkesDPHY1 1) 5 Explanation: The same reason as Part 1. 011 10.0points Consider a horse pulling a buggy. Is the following statement true? The weight of the horse and normal force exerted by the ground on the horse constitute an interaction pair that are always equal and opposite according to Newton’s third law. 1. no correct 2. yes Explanation: The normal force is a repulsive contact force between the ground and the horse. The weight is the gravitational force exerted by the Earth on the horse. These two forces are of different origin and do not constitute an interaction pair. 012 10.0points A book rests on the shelf of a bookcase. The reaction force to the force of gravity acting on the book is 1. The force exerted by the book on the earth. correct 2. The force of the shelf holding the book up. 3. None of these. 4. The weight of the book. 5. The frictional force between book and shelf. Explanation: The gravity is exerted by the earth on the book, so its reaction is exerted by the book on the earth. 013 10.0points When you drop a 0 . 37 kg apple, Earth exerts a force on it that accelerates it at 9 . 8 m / s 2 to- ward the earth’s surface. According to New- ton’s third law, the apple must exert an equal but opposite force on Earth. If the mass of the earth 5 . 98 × 10 24 kg, what is the magnitude of the earth’s acceleration toward the apple? Answer in units of m / s 2 . Correct answer: 6 . 06355 × 10 − 25 m / s 2 . Explanation: Let : m = 0 . 37 kg , g = 9 . 8 m / s 2 , and m e = 5 . 98 × 10 24 kg . Applying Newton’s Third Law of Motion, F a = F e m a a a = m e a e a e = m a a a m e = m g m e = (0 . 37 kg) (9 . 8 m / s 2 ) 5 . 98 × 10 24 kg = 6 . 06355 × 10 − 25 m / s 2 . 014 10.0points Consider the following system, where F = 150 N, m = 1 kg, and M = 2 kg M m F What is the magnitude of the force with which one block acts on the other? 1. None of these 2. 50 N correct 3. 100 N 4. 150 N 5. 300 N 6. 75 N

levesque (aml6939) – Homework 4, Newton’s laws 23-24 – tejeda – (WilkesDPHY1 1) 7 T = m 1 ( g + a ) = (2 . 5 kg) (9 . 8 m / s 2 + 7 . 8 m / s 2 ) = 44 N . 017(part1of2)10.0points An elevator accelerates upward at 1 . 2 m / s 2 . The acceleration of gravity is 9 . 8 m / s 2 . What is the upward force exerted by the floor of the elevator on a(n) 52 kg passenger? Answer in units of N. Correct answer: 572 N. Explanation: N a mg When the elevator is accelerating upward, F net = m a = N - m g N = m a + m g. 018(part2of2)10.0points If the same elevator accelerates downwards with an acceleration of 1 . 2 m / s 2 , what is the upward force exerted by the elevator floor on the passenger? Answer in units of N. Correct answer: 447 . 2 N. Explanation: N a mg When the elevator is accelerating down- ward, F net = m a = m g - N 2 N 2 = m g - m a. 019(part1of3)10.0points You are driving at the speed of 25 . 7 m / s (57 . 5016 mph) when suddenly the car in front of you (previously traveling at the same speed) brakes and begins to slow down with the largest deceleration possible without skid- ding. Considering an average human reaction, you press your brakes 0 . 484 s later. You also brake and decelerate as rapidly as possible without skidding. Assume that the coefficient of static friction is 0 . 742 between both cars’ wheels and the road. The acceleration of gravity is 9 . 8 m / s 2 . Calculate the acceleration of the car in front of you when it brakes. Answer in units of m / s 2 . Correct answer: - 7 . 2716 m / s 2 . Explanation: The only force on the car horizontally when braking is the force of friction, F f . Thus F net = ma = - F f = - μ N = - μmg since the braking force is a deceleration. Thus ma = - μmg a = - μg = (0 . 742)(9 . 8 m / s 2 ) = - 7 . 2716 m / s 2 . 020(part2of3)10.0points Calculate the braking distance for the car in front of you. Answer in units of m. Correct answer: 45 . 4157 m. Explanation: For constant acceleration a , v 2 - v 2 0 = 2 a ( x - x 0 ) . The final velocity v f = 0, and for d br = x - x 0 , - v 2 0 = 2 a d br d br = - v 2 0 2 a = (25 . 7 m / s) 2 2( - 7 . 2716 m / s 2 ) = 45 . 4157 m 021(part3of3)10.0points Find the minimum safe distance at which you can follow the car in front of you and avoid hitting it (in the case of emergency braking described here).

levesque (aml6939) – Homework 4, Newton’s laws 23-24 – tejeda – (WilkesDPHY1 1) 8 Answer in units of m. Correct answer: 12 . 4388 m. Explanation: The car in front of you moves a total dis- tance of d br , which also is the distance you travel while braking. You also traveled a dis- tance d react = v 0 ∆ t because of your reaction time ∆ t . Thus the minimum safe distance is the distance you travel during your reaction time: d safe = v 0 ∆ t = (25 . 7 m / s)(0 . 484 s) = 12 . 4388 m 022(part1of3)10.0points A block is at rest on the incline shown in the figure. The coefficients of static and ki- netic friction are μ s = 0 . 81 and μ k = 0 . 69, respectively. The acceleration of gravity is 9 . 8 m / s 2 . 23 kg μ 36 ◦ What is the frictional force acting on the 23 kg mass? Answer in units of N. Correct answer: 132 . 487 N. Explanation: F f N m g 36 ◦ The forces acting on the block are shown in the figure. Since the block is at rest, the magnitude of the friction force should be equal to the component of the weight on the plane of the incline F f = M g sin θ = (23 kg) (9 . 8 m / s 2 ) sin 36 ◦ = 132 . 487 N . 023(part2of3)10.0points What is the largest angle which the incline can have so that the mass does not slide down the incline? Answer in units of ◦ . Correct answer: 39 . 0075 ◦ . Explanation: The largest possible value the static friction force can have is F f,max = μ s N , where the normal force is N = Mg cos θ . Thus, since F f = M g sin θ , M g sin θ m = μ s M g cos θ m tan θ m = μ s θ m = tan − 1 ( μ s ) = tan − 1 (0 . 81) = 39 . 0075 ◦ . 024(part3of3)10.0points What is the acceleration of the block down the incline if the angle of the incline is 44 ◦ ? Answer in units of m / s 2 . Correct answer: 1 . 94348 m / s 2 . Explanation: When θ exceeds the value found in part 2, the block starts moving and the friction force is the kinetic friction F k = μ k N = μ k M g cos θ. Newton’s equation for the block then becomes M a = M g sin θ - F f = M g sin θ - μ k M g cos θ and a = g [sin θ - μ k cos θ ] = (9 . 8 m / s 2 ) [sin 44 ◦ - (0 . 69) cos 44 ◦ ] = 1 . 94348 m / s 2 .