# h14 - hernandez(ejh742 homework 14 Turner(58120 1 This...

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hernandez (ejh742) – homework 14 – Turner – (58120) 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 6 . 97 m on the incline by a 150 N force. The acceleration of gravity is 9 . 8 m / s 2 . 13 kg μ = 0 . 338 150 N 1 . 69 m / s 24 a) What is the change in kinetic energy of the crate? Correct answer: 410 . 138 J. Explanation: Let : F = 150 N , d = 6 . 97 m , θ = 24 , m = 13 kg , g = 9 . 8 m / s 2 , μ = 0 . 338 , and v = 1 . 69 m / s . F μ N N m g 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 − m g cos θ = 0 ⇒ N = m g cos θ . Thus W fric = μ m g d cos θ = (0 . 338) (13 kg) (9 . 8 m / s 2 ) × (6 . 97 m) cos 24 = 274 . 188 J . The work due to the applied force is W appl = F d = (150 N) (6 . 97 m) = 1045 . 5 J , and the work due to gravity is W grav = m g d sin θ = (13 kg) (9 . 8 m / s 2 ) × (6 . 97 m) sin 24 = 361 . 173 J , so that Δ K = W fric + W appl + W grav = ( 274 . 188 J) + (1045 . 5 J) + ( 361 . 173 J) = 410 . 138 J . 002 (part 2 of 2) 10.0 points b) What is the speed of the crate after it is pulled the 6 . 97 m? Correct answer: 8 . 12123 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

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hernandez (ejh742) – homework 14 – Turner – (58120) 2 v f = radicalbigg 2 Δ K m + v 2 i = radicalBigg 2(410 . 138 J) 13 kg + (1 . 69 m / s) 2 = 8 . 12123 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. 418 g 3 . 3 m 1 . 8 m 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 . 42494 m / s. Explanation: Let : g = 9 . 81 m / s 2 , m = 418 g , and h 1 = 1 . 5 m . m h h 1 h 2 3 . 3 m 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 m v 2 x = m g h 1 v 2 x = 2 g h 1 v x = radicalbig 2 g h 1 = radicalBig 2 ( 9 . 81 m / s 2 ) (1 . 5 m) = 5 . 42494 m / s . 004 (part 2 of 3) 10.0 points What horizontal distance does the block travel in the air?
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