# hw14(1) - lenoir(wml297 – homework 14 – Turner...

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Unformatted text preview: lenoir (wml297) – homework 14 – Turner – (58220) 1 This print-out should have 11 questions. Multiple-choice 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 5 . 59 m on the incline by a 150 N force. The acceleration of gravity is 9 . 8 m / s 2 . 1 k g μ = . 2 7 4 1 5 N 1 . 3 1 m / s 27 ◦ a) What is the change in kinetic energy of the crate? Correct answer: 456 . 052 J. Explanation: Let : F = 150 N , d = 5 . 59 m , θ = 27 ◦ , m = 10 kg , g = 9 . 8 m / s 2 , μ = 0 . 274 , and v = 1 . 31 m / s . F μN N mg v θ The work-energy 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 . 274) (10 kg) (9 . 8 m / s 2 ) × (5 . 59 m) cos27 ◦ = − 133 . 742 J . The work due to the applied force is W appl = F d = (150 N) (5 . 59 m) = 838 . 5 J , and the work due to gravity is W grav = − mg d sin θ = − (10 kg) (9 . 8 m / s 2 ) × (5 . 59 m) sin27 ◦ = − 248 . 705 J , so that Δ K = W fric + W appl + W grav = ( − 133 . 742 J) + (838 . 5 J) + ( − 248 . 705 J) = 456 . 052 J . 002 (part 2 of 2) 10.0 points b) What is the speed of the crate after it is pulled the 5 . 59 m? Correct answer: 9 . 63984 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 lenoir (wml297) – homework 14 – Turner – (58220) 2 v f = radicalbigg 2 Δ K m + v 2 i = radicalBigg 2(456 . 052 J) 10 kg + (1 . 31 m / s) 2 = 9 . 63984 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 569 g 5m 2 . 2m 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: 7 . 41188 m / s. Explanation: Let : g = − 9 . 81 m / s 2 , m = 569 g , and h 1 = 2 . 8 m ....
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## This note was uploaded on 04/15/2010 for the course PHY 12343 taught by Professor Turner during the Spring '10 term at University of Texas.

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hw14(1) - lenoir(wml297 – homework 14 – Turner...

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