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Unformatted text preview: pels (kcp389) Homework 3 hill (666) 1 This printout should have 40 questions. Multiplechoice questions may continue on the next column or page find all choices before answering. 001 (part 1 of 2) 10.0 points Sue and Jenny kick a soccer ball at exactly the same time. Sues foot exerts a force of 44 N to the north. Jennys foot exerts a force of 99 . 7 N to the east. a) What is the magnitude of the resultant force on the ball? Correct answer: 108 . 977 N. Explanation: f j F f s The magnitude of the resultant can be cal culated using the Pythagorean Theorem: F = radicalBig f 2 s + f 2 j = radicalBig (44 N) 2 + (99 . 7 N) 2 = 108 . 977 N 002 (part 2 of 2) 10.0 points b) What is the direction of the resultant force (measured from East)? Correct answer: 23 . 813 . Explanation: Jennys force is directed east, so f s is the side opposite and f j is the side adjacent, and tan = f s f j = arctan parenleftbigg f s f j parenrightbigg = arctan parenleftbigg 44 N 99 . 7 N parenrightbigg = 23 . 813 003 (part 1 of 2) 10.0 points A 17 . 3 kg mass attached to a spring scale rests on a smooth, horizontal surface. The spring scale, attached to the front end of a boxcar, reads T = 31 N when the car is in motion. m If the spring scale reads zero when the car is at rest, determine the acceleration of the car, when it is in motion as indicated above. Correct answer: 1 . 79191 m / s 2 . Explanation: Basic Concept: Fictitious forces Inertial/noninertial frames Solution: In the laboratory frame (which is an inertial frame): The block is accelerating with the car, with an acceleration of a . This requires an external force which must equal ma . Since the tension T is the ONLY available, external force we know that ma = T, so that a = T m = 31 N 17 . 3 kg = 1 . 79191 m / s 2 Alternative derivation in the acceler ated frame of the car: The acceleration of the mass in the refer ence frame of the car is a = 0; on the other hand, the forces acting on the mass are the tension T and the fictitious force F fic = ma , which acts in the direction opposite to the cars motion. Thus, T F fic = T ma = ma = 0 and we get a = T m = 1 . 79191 m / s 2 pels (kcp389) Homework 3 hill (666) 2 004 (part 2 of 2) 10.0 points What would be the reading on the scale if the boxcar were moving at a constant velocity? 1. less than T , but greater than 0 N 2. greater than T 3. 0 N correct 4. T 5. There is not enough information given to tell which is correct. Explanation: The reading will now be zero. The easy way to see this is to let the acceleration be zero, and use this in the equation in part 1. A more sophisticated way to see this is by recognizing that the boxcar is now an inertial frame since it is not accelerating. The block is moving at the same velocity as the boxcar, so the block is at rest in the boxcars frame of reference. The net force on the object at rest in an inertial frame is zero. (Newtons 1st law) 005...
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This note was uploaded on 10/19/2008 for the course PHY 2048 taught by Professor Guzman during the Spring '08 term at FAU.
 Spring '08
 Guzman
 Physics, Work

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