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Unformatted text preview: homework 06 GADHIA, TEJAS Due: Feb 28 2007, 4:00 am 1 Question 1 part 1 of 2 10 points While skiing in Jackson, Wyoming, your friend Ben (75 kg) started his descent down the bunny run, 11 m above the bottom of the run. If he started at rest and converted all of his gravitational potential energy into kinetic energy, what is Bens kinetic energy at the bottom of the bunny run? Correct answer: 8085 J (tolerance 1 %). Explanation: Let : m = 75 kg , g = 9 . 8 m / s 2 , and h = 11 m . KE = PE = mg h = (75 kg) (9 . 8 m / s 2 ) (11 m) = 8085 J . Question 2 part 2 of 2 10 points What is his final velocity? Correct answer: 14 . 6833 m / s (tolerance 1 %). Explanation: v = radicalbigg 2 KE m = radicalBigg 2 (8085 J) 75 kg = 14 . 6833 m / s . Question 3 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 . 1 m b b b b b b b b b b b b 525 g h 1 . 9m 4 . 06 m 9 . 81m / s 2 v At what height h above the ground is the block released? Correct answer: 4 . 39889 m (tolerance 1 %). Explanation: Let : x = 4 . 06 m , g = 9 . 81 m / s 2 , m = 525 g , = 0 . 3 , = 1 . 1 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) homework 06 GADHIA, TEJAS Due: Feb 28 2007, 4:00 am 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) = (4 . 06 m) 2 4 ( 1 . 9 m) + (0 . 3) (1 . 1 m) = 2 . 49889 m , and h = h 1 h 2 = (2 . 49889 m) ( 1 . 9 m) = 4 . 39889 m . Question 4 part 2 of 3 10 points What is the the speed of the block when it leaves the track? Correct answer: 6 . 52332 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 ) (4 . 06 m) 2 2 ( 1 . 9 m) = 6 . 52332 m / s ....
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This note was uploaded on 03/04/2010 for the course PHY 303K taught by Professor Turner during the Spring '08 term at University of Texas at Austin.
 Spring '08
 Turner
 Physics, Work

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