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Unformatted text preview: rodriguez (fr4537) – Homework 6 Aplications of Newton’s Laws – catala – (20104) 1 This printout should have 31 questions. Multiplechoice questions may continue on the next column or page – find all choices before answering. 001 (part 1 of 3) 10.0 points A 16 kg block with a pulley attached slides along a frictionless surface. It is connected by a massless string to a 5 . 3 kg block via the arrangement shown. The acceleration of gravity is 9 . 81 m / s 2 . 16 kg 5 . 3 kg Find the horizontal distance the 16 kg block moves when the 5 . 3 kg block descends a dis tance of 9 . 2 cm . The pulleys are massless and frictionless. Correct answer: 4 . 6 cm. Explanation: Let : m 1 = 16 kg , m 2 = 5 . 3 kg , Δ x 2 = 9 . 2 cm , and g = 9 . 81 m / s 2 . Choose a coordinate system in which the positive x direction is to the right and consider the forces acting on each body. m 1 m 2 a 1 2 T N m 1 g a 2 = 2 a 1 T m 2 g Because of the pulley, the mass m 1 moves half the distance as the suspended mass: Δ x 1 = 1 2 Δ x 2 = 1 2 (9 . 2 cm) = 4 . 6 cm . 002 (part 2 of 3) 10.0 points Find the acceleration of the 16 kg block. Correct answer: 2 . 79532 m / s 2 . Explanation: The pulley causes the same relationship on the accelerations: a 1 = 1 2 a 2 = a . Applying summationdisplay F 1 = m 1 a 1 to the m 1 block, 2 T = m 1 a . (1) Applying summationdisplay F 2 = m 2 a 2 to the m 2 block, m 2 g T = m 2 (2 a ) . (2) Double eq (2) and add to eq (1): 2 m 2 g = m 1 a + 4 m 2 a a = 2 m 2 g m 1 + 4 m 2 (4) = 2 (5 . 3 kg) ( 9 . 81 m / s 2 ) 16 kg + 4 (5 . 3 kg) = 2 . 79532 m / s 2 . 003 (part 3 of 3) 10.0 points Find the tension in the connecting string. Correct answer: 22 . 3626 N. Explanation: Using Eq (1), T = m 1 a 2 = (16 kg) ( 2 . 79532 m / s 2 ) 2 = 22 . 3626 N . 004 (part 1 of 4) 10.0 points Two 125 kg boxes are dragged along a fric tionless surface with a constant acceleration of 1 . 22 m / s 2 , as shown in the figure. rodriguez (fr4537) – Homework 6 Aplications of Newton’s Laws – catala – (20104) 2 Each rope has a mass of 1 . 03 kg. 125 kg 125 kg 1 . 03 kg 1 . 03 kg F Find the force F . Correct answer: 307 . 513 N. Explanation: Let : m 1 = 125 kg , m 2 = 1 . 03 kg , m 3 = 125 kg , m 4 = 1 . 03 kg , and a = 1 . 22 m / s 2 . The freebody diagrams for the boxes and the ropes are below. F 12 m 1 F 21 F 23 m 2 F 32 F 34 m 3 F 43 F m 4 Because the vertical forces have no bearing on the problem they have not been included. vector F 34 = tension force exerted by m 4 on m 3 , vector F 43 = tension force exerted by m 3 on m 4 , vector F 23 = tension force exerted by m 3 on m 2 , vector F 32 = tension force exerted by m 2 on m 3 , vector F 12 = tension force exerted by m 2 on m 1 , vector F 21 = tension force exerted by m 1 on m 2 , The equal and opposite pairs of forces are vector F 21 = vector F 12 , vector F 32 = vector F 23 , vector F 43 = vector F 34 ....
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This note was uploaded on 04/15/2011 for the course PHY 2048 taught by Professor Unkown during the Spring '06 term at Miami Dade College, Miami.
 Spring '06
 UNKOWN
 Acceleration

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