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: hinojosa (jlh3938) homework 16 Turner (58185) 1 This printout should have 11 questions. Multiplechoice questions may continue on the next column or page find all choices before answering. 001 (part 1 of 2) 10.0 points A crate is pulled by a force (parallel to the incline) up a rough incline. The crate has an initial speed shown in the figure below. The crate is pulled a distance of 7 . 94 m on the incline by a 150 N force. The acceleration of gravity is 9 . 8 m / s 2 . 1 3 k g = . 3 8 1 5 N 1 . 2 6 m / s 35 a) What is the change in kinetic energy of the crate? Correct answer: 355 . 581 J. Explanation: Let : F = 150 N , d = 7 . 94 m , = 35 , m = 13 kg , g = 9 . 8 m / s 2 , = 0 . 308 , and v = 1 . 26 m / s . F N N mg v The workenergy theorem with nonconser vative forces reads W fric + W appl + W gravity = K To find the work done by friction we need the normal force on the block from Newtons law summationdisplay F y = N mg cos = 0 N = mg cos . Thus W fric = mg d cos = (0 . 308) (13 kg) (9 . 8 m / s 2 ) (7 . 94 m) cos35 = 255 . 214 J . The work due to the applied force is W appl = F d = (150 N) (7 . 94 m) = 1191 J , and the work due to gravity is W grav = mg d sin = (13 kg) (9 . 8 m / s 2 ) (7 . 94 m) sin35 = 580 . 205 J , so that K = W fric + W appl + W grav = ( 255 . 214 J) + (1191 J) + ( 580 . 205 J) = 355 . 581 J . hinojosa (jlh3938) homework 16 Turner (58185) 2 002 (part 2 of 2) 10.0 points b) What is the speed of the crate after it is pulled the 7 . 94 m? Correct answer: 7 . 50282 m / s. Explanation: Since 1 2 m ( v 2 f v 2 i ) = K v 2 f v 2 i = 2 K m v f = radicalbigg 2 K m + v 2 i = radicalBigg 2(355 . 581 J) 13 kg + (1 . 26 m / s) 2 = 7 . 50282 m / s . 003 (part 1 of 3) 10.0 points A block starts at rest and slides down a fric tionless track. It leaves the track horizontally, flies through the air, and subsequently strikes the ground. b b b b b b b b b b b b 412 g 4 . 1m 2 . 3m x v What is the speed of the ball when it leaves the track? The acceleration of gravity is 9 . 81 m / s 2 . Correct answer: 5 . 94273 m / s. Explanation: Let : g = 9 . 81 m / s 2 , m = 412 g , and h 1 = 1 . 8 m ....
View
Full
Document
 Spring '08
 Turner
 Physics, Force, Work

Click to edit the document details