hw 2 tsoi - homework 06 GADHIA, TEJAS Due: Feb 28 2007,...

Info iconThis preview shows pages 1–3. Sign up to view the full content.

View Full Document Right Arrow Icon

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon
This is the end of the preview. Sign up to access the rest of the document.

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 ....
View Full Document

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.

Page1 / 8

hw 2 tsoi - homework 06 GADHIA, TEJAS Due: Feb 28 2007,...

This preview shows document pages 1 - 3. Sign up to view the full document.

View Full Document Right Arrow Icon
Ask a homework question - tutors are online