SCIENCE 103Chapter 5 (Forces and Motion)-solutions WELL REVISED, ALL ANSWERS CORRECT, GRADED A+

bharti (vpb246) – Chapter 5 (Forces and Motion) – stanescu – (26505) 1 This print-out should have 79 questions. Multiple-choice questions may continue on the next column or page – find all choices before answering. ...

bharti (vpb246) – Chapter 5 (Forces and Motion) – stanescu – (26505) 1 This print-out should have 79 questions. Multiple-choice questions may continue on the next column or page – find all choices before answering. 001 10.0 points Start a ball rolling down a bowling alley and you’ll find it moves slightly slower with time. Does this violate Newton’s law of inertia? Defend your answer. 1. No; air resistance and friction act upon the ball. correct 2. Yes; the air resistance cancels the friction and the total force on the ball is zero. 3. Yes; no force acts upon it. 4. No; the law of inertia can also be applied to moving objects. 5. None of these Explanation: If there were no force acting on the ball it would continue to move without slowing down. Air resistance along with slight friction with the lane slows the ball down. This doesn’t violate the law of inertia because external forces act on the ball. 002 10.0 points If an object is not accelerating, how many forces act on it? 1. 0 2. 3 3. Unable to determine correct 4. 2 5. 1 Explanation: If it isn’t accelerating, no net force acts on it; you do not know how many forces are involved. 003 10.0 points A ball falls straight down through the air under the influence of gravity. There is a retarding force F~ on the ball with magnitude given by F = b v, where v is the speed of the ball and b is a positive constant. What is the magnitude of the acceleration of the ball at any time? 1. k~ak = b v m 2. k~ak = g + b v m 3. k~ak = g − b 4. k~ak = g − b v m correct 5. k~ak = g b Explanation: There are two forces acting (in opposite directions) on the ball; the gravity m g and the retarding force F = b v: m a = m g − b v a = g − b v m . 004 (part 1 of 2) 10.0 points Three objects can only move along a straight, level path. The graphs below show the position d of each of the objects plotted as a function of time t. d t I d t II d t III The magnitude of the velocity k~vk of the object increases in which of the cases? 1. I only 2. III only correct 3. I, II, and III 4. I and II onlybharti (vpb246) – Chapter 5 (Forces and Motion) – stanescu – (26505) 2 5. II and III only 6. II only 7. I and III only Explanation: Case I: The object moves at constant speed. Case II: The object remains at rest. Case III: The speed of the object increases with time; i.e., constant acceleration. Thus, the magnitude of the velocity of the object increases only in case III. 005 (part 2 of 2) 10.0 points The sum of the forces XFi on the object is zero in which of the cases? 1. III only 2. II and III only 3. I and II only correct 4. I only 5. II only 6. I, II, and III 7. I and III only Explanation: When the sum of the forces on the object is zero, the acceleration of the object is zero by Newton’s second law. In cases I and II, the velocity of the object doesn’t change with time, so the sum of the forces on the object is zero. In case III, the object is accelerating, so the sum of the forces on the object is not zero. 006 (part 1 of 2) 10.0 points The position of a toy locomotive moving on a straight track along the x-axis is given by the equation x = b t3 − c t2 + d t , where b = 5 m/s3, c = 60 m/s2, and d = 245 m/s, x is in meters and t is in seconds. Find the time t when the net force on the locomotive is equal to zero. Correct answer: 4 s. Explanation: From Newton’s second law of motion the force is directly proportional to the acceleration, which means zero acceleration will give a zero force. The velocity is v = d x dt = 3 b t2 − 2 c t + d , and the acceleration is a = d2 x dt2 = d v dt = 6 b t − 2 c = 0 , so t = c 3 b = (60 m/s2) 3 (5 m/s3) = 4 s . 007 (part 2 of 2) 10.0 points At the time when the net force on the locomotive is equal to zero, the velocity of the locomotive is 1. positive. correct 2. negative. 3. zero. Explanation: Since v = 5 m/s, the velocity is negative, see Part 1. 008 10.0 points When you jump vertically off the ground, what is your acceleration when you reach your highest point? Up is positive. 1. All are wrong. 2. g 2 3. g 4. − g 3bharti (vpb246) – Chapter 5 (Forces and Motion) – stanescu – (26505) 3 5. g 3 6. −g correct 7. − g 2 8. 0 m/s2 Explanation: At the top of your jump your acceleration is still −g . Let the equation for acceleration via Newton’s second law guide your thinking: a = F m = −m g m = −g . Gravity does not cease to act at any point of your jump. The acceleration of gravity is directed towards the center of the Earth. 009 (part 1 of 3) 10.0 points A boat moves through the water with two forces acting on it. One is a 2332 N forward push by the motor on the propeller, and the other is an 2103 N resistive force due to the water around the bow. What is the acceleration of the 1035 kg boat? Correct answer: 0.221256 m/s2. Explanation: Given : F1 = 2332 N , F2 = 2103 N , and m = 1035 kg . 2103 N 2332 N The forces are unbalanced, so the net force is Fnet = m a = F1 − F2 and a = F1 − F2 m = 2332 N − 2103 N 1035 kg = 0.221256 m/s2 forward. 010 (part 2 of 3) 10.0 points If it starts from rest, how far will it move in 17.9 s? Correct answer: 35.4463 m. Explanation: Given : ∆t = 17.9 s Under acceleration, ∆x = vi t + 1 2 a (∆t)2 = 1 2 a (∆t)2 since vi = 0 m/s. Thus ∆x = 1 2 a (∆t)