Online Homework Answers 7 (Nov. 2, 2006)

Online Homework Answers 7 (Nov. 2, 2006) - Hood Charles...

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Hood, Charles – Homework 7 – Due: Nov 2 2006, 11:00 pm – Inst: Jan Opyrchal 1 This print-out should have 9 questions. Multiple-choice questions may continue on the next column or page – find all choices before answering. The due time is Central time. 001 (part 1 of 1) 10 points Given: g = 9 . 8 m / s 2 . Your teacher placed a 8 kg block at the po- sition marked with a “ + ” (horizontally, 0 . 5 m from the origin) on a large incline outlined on the graph below and let it slide, starting from rest. The kinetic coefficient of friction for the block on the incline is 0 . 84 . 0 1 2 3 4 5 6 7 8 9 10 0 1 2 3 4 5 6 7 8 9 10 Vertical Height, y (m) Horizontal Distance, x (m) Figure: Drawn to scale. Calculate the horizontal distance from the bottom edge (right-hand edge) of the incline to where the block hits the floor ( i.e. , y = 0). Your answer must be within ± 2%. Correct answer: 2 . 06774 N. Explanation: Given : μ = 0 . 84 , x b = 4 . 5 m , y b = 4 . 5 m , x t = 0 m , y t = 9 . 5 m , θ = arctan y t - y b x b - x t , = arctan 9 . 5 m - 4 . 5 m 4 . 5 m - 0 m , = 48 . 0128 , and x = 0 . 5 m . The graph below shows the trajectory of the block while it is on the incline and after it slides off the incline where it finally hits the floor. 0 1 2 3 4 5 6 7 8 9 10 0 1 2 3 4 5 6 7 8 9 10 Vertical Height, y (m) Horizontal Distance, x (m) θ The free-body diagram when the block is on the incline is shown below. m g m g sin θ μ m g cos θ m g cos θ N Solution: y = y t - y t - y b x b = 9 . 5 m - 9 . 5 m - 4 . 5 m 4 . 5 m = 8 . 94444 m . a = g (sin θ - μ cos θ ) = (9 . 8 m / s 2 ) h sin 48 . 0128 - (0 . 84) cos 48 . 0128 i = 1 . 77736 m / s 2 .
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Hood, Charles – Homework 7 – Due: Nov 2 2006, 11:00 pm – Inst: Jan Opyrchal 2 x d = x b - x = 4 . 5 m - 0 . 5 m = 4 m , y d = y - y b = 8 . 94444 m - 4 . 5 m = 4 . 44444 m , d = q x 2 d + y 2 d = q (4 m) 2 + (4 . 44444 m) 2 = 5 . 97939 m . v b = 2 a d = h 2 (1 . 77736 m / s 2 ) × (5 . 97939 m) i 1 / 2 = 4 . 61033 m / s . v y = - v b sin θ = - (4 . 61033 m / s) sin 48 . 0128 = - 3 . 42683 m / s . v x = v b cos θ = (4 . 61033 m / s) cos 48 . 0128 = 3 . 08415 m / s . Δ y = y b = 4 . 5 m = v y t + 1 2 g t 2 , so g 2 t 2 + v y t - Δ y = 0 and v 2 y + 2 g Δ y = ( - 3 . 42683 m / s) 2 + 2 (9 . 8 m / s 2 ) (4 . 5 m) = 99 . 9432 m 2 / s 2 , t = 1 g v y ± q v 2 y + 2 g Δ y = - 3 . 42683 m / s + p 99 . 9432 m 2 / s 2 9 . 8 m / s 2 = 0 . 670442 s .
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