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Unformatted text preview: 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 . =0 . 2 1 . 4 m 441 g h 2 . 1m 4 . 78 m 9 . 81m / 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 , = 0 . 2 , = 1 . 4 m , h 2 = 2 . 1 m , h = h 1 h 2 , and v x = v . 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 = m g . (4) Choosing the point where the block leaves the track as the origin of the coordinate system, x = v x t (5) h 2 = 1 2 g 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 ) m g v 2 x = 2 g h 1 2 g h 1 = v 2 x 2 g + (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) 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 + = x 2 4 h 2 + (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 ....
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 Spring '08
 Turner
 Friction, Work

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