This preview shows pages 1–3. Sign up to view the full content.
This preview has intentionally blurred sections. Sign up to view the full version.
View Full Document
Unformatted text preview: practice 06 ALIBHAI, ZAHID Due: Feb 25 2007, 4:00 am 1 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 . 3 1 . 2 m b b b b b b b b b b b b 500 g h 1 . 9m 3 . 42 m 9 . 81m / s 2 v At what height h above the ground is the block released? Correct answer: 3 . 799 m (tolerance 1 %). Explanation: Let : x = 3 . 42 m , g = 9 . 81 m / s 2 , m = 500 g , = 0 . 3 , = 1 . 2 m , h 2 = 1 . 9 m , h = h 1 h 2 , and v x = v . b b b b b b b b b b b b 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 mv 2 (2) U g = mg h (3) W = mg . (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 mv 2 x = mg ( h h 2 ) mg 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 parenleftbigg x v x parenrightbigg 2 , so v 2 x = g x 2 2 h 2 . (8) 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) = (3 . 42 m) 2 4 ( 1 . 9 m) + (0 . 3) (1 . 2 m) = 1 . 899 m , and h = h 1 h 2 = (1 . 899 m) ( 1 . 9 m) = 3 . 799 m . practice 06 ALIBHAI, ZAHID Due: Feb 25 2007, 4:00 am 2 Question 2 part 2 of 3 10 points What is the the speed of the block when it leaves the track? Correct answer: 5 . 49501 m / s (tolerance 1 %). Explanation: From Eq. 8, we have v x = radicalBigg g x 2 2 h 2 = radicalBigg (9 . 81 m / s 2 ) (3 . 42 m) 2 2 ( 1 . 9 m) = 5 . 49501 m / s . Alternate Solution: From Eq. 7, we have v x = radicalbig 2 g ( h 1 ) = braceleftbigg 2 (9 . 81 m / s 2 ) bracketleftBig (1 . 899 m) (0 . 3) (1 . 2 m) bracketrightBig bracerightbigg 1 / 2 = 5 . 49501 m / s . Question 3 part 3 of 3 10 points What is the total speed of the block when it hits the ground? Correct answer: 8 . 21421 m / s (tolerance 1 %). Explanation: Basic Concept: K f = U i , since v i = 0 m/s and h f = 0 m. Solution: 1 2 mv 2 f = mg h v f = radicalbig 2 g ( h + ) (10) = braceleftbigg 2 (9 . 81 m / s 2 ) bracketleftBig 3 . 799 m + (0 . 3) (1 . 2 m) bracketrightBig bracerightbigg 1 / 2 = 8 . 21421 m / s , where h = h 1 + h 2 . Alternate Solution: v y = radicalbig 2 g h 2 (11) = braceleftbigg 2 (9 . 81 m / s 2 ) ( 1 . 9 m) bracerightbigg 1 / 2 = 6 . 10557 m / s , so v f = radicalBig v 2 x + v 2 y (12) = braceleftbigg (5 . 49501 m / s) 2 + (6 . 10557 m / s) 2 bracerightbigg 1 / 2 = 8 . 21421 m / s ....
View
Full
Document
 Spring '08
 Turner
 Friction, Work

Click to edit the document details