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Unformatted text preview: brown (twb493) – ohw 14 – turner – (57340) 1 This printout should have 14 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 8 . 01 m on the incline by a 150 N force. The acceleration of gravity is 9 . 8 m / s 2 . 1 2 k g μ = . 3 8 1 5 N 1 . 7 1 m / s 29 ◦ a) What is the change in kinetic energy of the crate? Correct answer: 491 . 069 J. Explanation: Let : F = 150 N , d = 8 . 01 m , θ = 29 ◦ , m = 12 kg , g = 9 . 8 m / s 2 , μ = 0 . 308 , and v = 1 . 71 m / s . F μN N mg v θ The workenergy theorem with nonconser vative forces reads W fric + W appl + W gravity = Δ K To find the work done by friction we need the normal force on the block from Newton’s law summationdisplay F y = N − mg cos θ = 0 ⇒ N = mg cos θ . Thus W fric = − μmg d cos θ = − (0 . 308) (12 kg) (9 . 8 m / s 2 ) × (8 . 01 m) cos29 ◦ = − 253 . 752 J . The work due to the applied force is W appl = F d = (150 N) (8 . 01 m) = 1201 . 5 J , and the work due to gravity is W grav = − mg d sin θ = − (12 kg) (9 . 8 m / s 2 ) × (8 . 01 m) sin29 ◦ = − 456 . 679 J , so that Δ K = W fric + W appl + W grav = ( − 253 . 752 J) + (1201 . 5 J) + ( − 456 . 679 J) = 491 . 069 J . 002 (part 2 of 2) 10.0 points b) What is the speed of the crate after it is pulled the 8 . 01 m? Correct answer: 9 . 207 m / s. Explanation: Since 1 2 m ( v 2 f − v 2 i ) = Δ K v 2 f − v 2 i = 2 Δ K m brown (twb493) – ohw 14 – turner – (57340) 2 v f = radicalbigg 2 Δ K m + v 2 i = radicalBigg 2(491 . 069 J) 12 kg + (1 . 71 m / s) 2 = 9 . 207 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 408 g 3 . 7m 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 . 24099 m / s. Explanation: Let : g = − 9 . 81 m / s 2 , m = 408 g , and h 1 = 1 . 4 m . b b b b b b b b b b b b m h h 1 h 2 x v Choose the point where the block leaves the track as the origin of the coordinate system. While on the ramp, K b = U t 1 2 mv 2 x = − mg h 1 v 2 x = − 2 g h 1 v x = radicalbig − 2 g h 1 = radicalBig − 2 ( − 9 . 81 m / s 2 ) (1 . 4 m) = 5 . 24099 m / s . 004 (part 2 of 3) 10.0 points What horizontal distance does the block travel in the air?...
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This note was uploaded on 04/13/2011 for the course PHY 303K taught by Professor Turner during the Spring '08 term at University of Texas.
 Spring '08
 Turner
 Physics, Force

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