Practice Homework 6 Solutions

# Practice Homework 6 Solutions - practice 06 ALIBHAI, ZAHID...

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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 ....
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## Practice Homework 6 Solutions - practice 06 ALIBHAI, ZAHID...

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