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hw 16 - homework 16 HETTIGER CHRISTOF Due 4:00 am Question...

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homework 16 – HETTIGER, CHRISTOF – Due: Feb 26 2008, 4:00 am 1 Question 1, chap 8, sect 1. part 1 of 1 10 points A 4 . 7 kg mass is attached to a light cord that passes over a massless, frictionless pulley. The other end of the cord is attached to a 3 . 3 kg mass. The acceleration of gravity is 9 . 8 m / s 2 . 6 . 4 m ω 4 . 7 kg 3 . 3 kg Use conservation of energy to determine the final speed of the first mass after it has fallen (starting from rest) 6 . 4 m. Correct answer: 4 . 6853 m / s (tolerance ± 1 %). Explanation: Let : m 1 = 4 . 7 kg , m 2 = 3 . 3 kg , and = 6 . 4 m . Consider the free body diagrams 4 . 7 kg 3 . 3 kg T T m 1 g m 2 g a a Let the figure represent the initial config- uration of the pulley system (before m 1 falls down). From the conservation of energy K i + U i = K f + U f 0 + m 1 g ℓ = m 2 g ℓ + 1 2 m 1 v 2 + 1 2 m 2 v 2 ( m 1 - m 2 ) g ℓ = 1 2 ( m 1 + m 2 ) v 2 Therefore v = radicalBigg ( m 1 - m 2 ) ( m 1 + m 2 ) 2 g ℓ = bracketleftbigg 4 . 7 kg - 3 . 3 kg 4 . 7 kg + 3 . 3 kg × 2 (9 . 8 m / s 2 )(6 . 4 m) bracketrightbigg 1 / 2 = 4 . 6853 m / s . Question 2, chap 8, sect 1. part 1 of 3 10 points A block starts at rest and slides down a fric- tionless track (as shown in the figure below). It leaves the track horizontally, flies through the air, and subsequently strikes the ground. The acceleration of gravity is 9 . 81 m / s 2 . 506 g 4 . 2 m 2 m x 9 . 81 m / s 2 v What is the speed v of the ball when it leaves the track? Correct answer: 6 . 56993 m / s (tolerance ± 1 %). Explanation: Let : x = 4 . 19524 m , g = 9 . 81 m / s 2 , m = 506 g , h 1 = - 2 . 2 m , h 2 = - 2 m , h = h 1 + h 2 , and v x = v .
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homework 16 – HETTIGER, CHRISTOF – Due: Feb 26 2008, 4:00 am 2 m h h 1 h 2 4 . 2 m g v Basic Concepts: Conservation of Me- chanical Energy U i = U f + K f + W . (1) since v i = 0 m/s. K = 1 2 m v 2 (2) U g = m g h (3) W = μ m g ℓ . (4) Choosing the point where the block leaves the track as the origin of the coordinate system, Δ x = v x Δ t (5) h 2 = - 1 2 g Δ t 2 (6) since a x i = 0 m/s 2 and v y i = 0 m/s. Solution: 1 2 m v 2 x = - m g h 1 v 2 x = - 2 g ( h - h 2 ) = - 2 g h 1 v x = radicalbig - 2 g h 1 (4) = radicalBig - 2 (9 . 81 m / s 2 ) ( - 2 . 2 m) = 6 . 56993 m / s . Question 3, chap 8, sect 1.
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