PHY302K-Sai-HW3-solutions - Oslund (hno56) Homework 3 sai...

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Unformatted text preview: Oslund (hno56) Homework 3 sai (58015) 1 This print-out should have 21 questions. Multiple-choice questions may continue on the next column or page find all choices before answering. 001 10.0 points A 58 kg boy and a 37 kg girl use an elastic rope while engaged in a tug-of-war on a frictionless icy surface. If the acceleration of the girl toward the boy is 3 m / s 2 , determine the magnitude of the acceleration of the boy toward the girl. Correct answer: 1 . 91379 m / s 2 . Explanation: According to Newtons Third Law of Mo- tion the force the boy exerts on the girl is the same as the force the girl exerts on the boy, so F g = F b m g a g = m b a b a b = m g a g m b 002 10.0 points A ball initially moves horizontally with ve- locity v i , as shown. It is then struck by a stick. After leaving the stick, the ball moves vertically with a velocity v f , which has the same magnitude as v i . v i v f Which of the following vectors best repre- sents the direction of the average force that the stick exerts on the ball? 1. 2. 3. None of these graphs is correct. 4. 5. 6. 7. correct 8. 9. Explanation: The v f velocity is the resultant of v i and v s , so it is the diagonal of a parallelogram whose sides are v i and v s . v i v f v s 003 10.0 points An elevator starts from rest with a constant upward acceleration and moves 1 m in the first 1 . 7 s. A passenger in the elevator is holding a 5 . 6 kg bundle at the end of a vertical cord. The acceleration of gravity is 9 . 8 m / s 2 . What is the tension in the cord as the ele- vator accelerates? Correct answer: 58 . 7554 N. Explanation: T mg a elevator g Let h be the distance traveled and a the acceleration of the elevator. With the initial velocity being zero, we simplify the following expression and solve for acceleration of the Oslund (hno56) Homework 3 sai (58015) 2 elevator: h = v t + 1 2 a t 2 = 1 2 a t 2 = a = 2 h t 2 . The equation describing the forces acting on the bundle is F net = ma = T- mg T = m ( g + a ) = m g + 2 h t 2 = (5 . 6 kg) 9 . 8 m / s 2 + 2 (1 m) (1 . 7 s) 2 = 58 . 7554 N . 004 (part 1 of 2) 10.0 points A 24 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 = 27 . 9 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 . 1625 m / s 2 . Explanation: Basic Concept: Fictitious forces Inertial/non-inertial 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 = 27 . 9 N 24 kg = 1 . 1625 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...
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This note was uploaded on 02/22/2012 for the course PHY 302K taught by Professor Kaplunovsky during the Spring '08 term at University of Texas at Austin.

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PHY302K-Sai-HW3-solutions - Oslund (hno56) Homework 3 sai...

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