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Solution23 - fullwood(tf5349 – Homework 5 Forces – Catala –(49033 1 This print-out should have 20 questions Multiple-choice questions may

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Unformatted text preview: fullwood (tf5349) – Homework 5 Forces – Catala – (49033) 1 This print-out should have 20 questions. Multiple-choice questions may continue on the next column or page – find all choices before answering. 001 (part 1 of 2) 10.0 points A man is pulling on a rope with a force of 51 N directed at an angle of 65 ◦ to the horizontal. What is the x-component of this force? Correct answer: 21 . 5535 N. Explanation: Let : F = 51 N and θ = 65 ◦ . 5 1 N Scale: 10 N 65 ◦ F x = F cos θ = (51 N) cos65 ◦ = 21 . 5535 N . 002 (part 2 of 2) 10.0 points What is the y-component of this force? Correct answer: 46 . 2217 N. Explanation: F y = F sin θ = (51 N) sin65 ◦ = 46 . 2217 N . 003 (part 1 of 4) 10.0 points A 2.70 kg block starts from rest at the top of a 30.0 ◦ incline and accelerates uniformly down the incline, moving 2.36 m in 2.39 s. The acceleration of gravity is 9 . 81 m / s 2 . a) What is the magnitude of the accelera- tion of the block? Correct answer: 0 . 826316 m / s 2 . Explanation: 2 . 7 k g F k a F n F g , y F g , x F g 30 ◦ Note: Figure is not drawn to scale. Basic Concept: Δ x = 1 2 a (Δ t ) 2 since v i = 0 m/s. Given: Δ x = 2 . 36 m Δ t = 2 . 39 s Solution: a = 2Δ x (Δ t ) 2 = 2(2 . 36 m) (2 . 39 s) 2 = 0 . 826316 m / s 2 004 (part 2 of 4) 10.0 points b) What is the coefficient of kinetic friction between the block and the incline? Correct answer: 0 . 480088. fullwood (tf5349) – Homework 5 Forces – Catala – (49033) 2 Explanation: Basic Concepts: F g,x = mg sin θ F g,y = mg cos θ F x,net = ma x = F g,x- F k F y,net = F n- F g,y = 0 F k = μ k F n Given: m = 2 . 70 kg θ = 30 . ◦ g = 9 . 81 m / s 2 Solution: F g,x = (2 . 7 kg) ( 9 . 81 m / s 2 ) sin 30 ◦ = 13 . 2435 N F g,y = (2 . 7 kg) ( 9 . 81 m / s 2 ) cos 30 ◦ = 22 . 9384 N ma x = F g,x- μ k F g,y μ k = F g,x- ma x F g,y = 13 . 2435 N- (2 . 7 kg)(0 . 826316 m / s 2 ) 22 . 9384 N = 0 . 480088 005 (part 3 of 4) 10.0 points c) What is the magnitude of the frictional force acting on the block? Correct answer: 11 . 0124 N. Explanation: Basic Concept: F k = μ k F n Solution: F k = (0 . 480088)(22 . 9384 N) = 11 . 0124 N 006 (part 4 of 4) 10.0 points d) What is the speed of the block after it slides the distance of 2.36? Correct answer: 1 . 9749 m / s. Explanation: Basic Concept: v 2 f = 2 a x Δ x since v i = 0 m/s. Solution: v f = radicalbig 2 a x Δ x = radicalBig 2 (0 . 826316 m / s 2 )(2 . 36 m) = 1 . 9749 m / s 007 10.0 points A 5.7 kg bucket of water is raised from a well by a rope....
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This note was uploaded on 06/10/2010 for the course PHY 100 taught by Professor Pro.kiroshima during the Spring '10 term at Adelphi.

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Solution23 - fullwood(tf5349 – Homework 5 Forces – Catala –(49033 1 This print-out should have 20 questions Multiple-choice questions may

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