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Homework 13

# Homework 13 - homework 13 BAUTISTA ALDO Due 4:00 am Version...

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homework 13 – BAUTISTA, ALDO – Due: Feb 22 2006, 4:00 am 1 Version number encoded for clicker entry: V1:1, V2:4, V3:2, V4:4, V5:3. Question 1 Part 1 of 3. 10 points. A block starts at rest and slides down a frictionless track except for a small rough area on a horizontal section of the 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 . μ =0 . 2 1 . 4 m 441 g h 2 . 1 m 4 . 78 m 9 . 81 m / s 2 v At what height h above the ground is the block released? Correct answer: 5 . 10005 m (tolerance ± 1 %). Explanation: Let : x = 4 . 78 m , g = 9 . 81 m / s 2 , m = 441 g , μ = 0 . 2 , = 1 . 4 m , h 2 = - 2 . 1 m , h = h 1 - h 2 , and v x = v . μ m h h 1 h 2 x 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 ‘ . (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: From energy conservation Eqs. 1, 2, 3, and 4, we have 1 2 m v 2 x = m g ( h - h 2 ) - μ m g ‘ v 2 x = 2 g h 1 - 2 μ g ‘ h 1 = v 2 x 2 g + μ ‘ (7) h 2 = - 1 2 g t 2 (6) x = v x t . (5) Using Eq. 6 and substituting t = x v x from Eq. 5, we have h 2 = - 1 2 g x v x 2 , so v 2 x = - g x 2 2 h 2 . (8)

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homework 13 – BAUTISTA, ALDO – Due: Feb 22 2006, 4:00 am 2 Using Eq. 6 and substituting v 2 x from Eq. 8, we have h 1 = - g x 2 2 h 2 2 g + μ ‘ = - x 2 4 h 2 + μ ‘ (9) = - (4 . 78 m) 2 4 ( - 2 . 1 m) + (0 . 2) (1 . 4 m) = 3 . 00005 m , and h = h 1 - h 2 = (3 . 00005 m) - ( - 2 . 1 m) = 5 . 10005 m . Question 2 Part 2 of 3. 10 points. What is the the speed of the block when it leaves the track? Correct answer: 7 . 30529 m / s (tolerance ± 1 %). Explanation: From Eq. 8, we have v x = s - g x 2 2 h 2 = s - (9 . 81 m / s 2 ) (4 . 78 m) 2 2 ( - 2 . 1 m) = 7 . 30529 m / s . Alternate Solution: From Eq. 7, we have v x = p 2 g ( h 1 - μ ‘ ) = 2 (9 . 81 m / s 2 ) h (3 . 00005 m) - (0 . 2) (1 . 4 m) i 1 / 2 = 7 . 30529 m / s . Question 3 Part 3 of 3. 10 points.
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Homework 13 - homework 13 BAUTISTA ALDO Due 4:00 am Version...

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