HW 5 solution_pdf

# HW 5 solution_pdf - ghei (mg42382) HW 5 opyrchal (11130) 1...

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Unformatted text preview: ghei (mg42382) HW 5 opyrchal (11130) 1 This print-out should have 9 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 9 . 27 m on the incline by a 150 N force. The acceleration of gravity is 9 . 8 m / s 2 . 1 4 k g = . 3 7 4 1 5 N 1 . 2 8 m / s 21 a) What is the change in kinetic energy of the crate? Correct answer: 490 . 636 J. Explanation: Let : F = 150 N , d = 9 . 27 m , = 21 , m = 14 kg , g = 9 . 8 m / s 2 , = 0 . 374 , and v = 1 . 28 m / s . F N N m g v The work-energy 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 m g cos = 0 N = m g cos . Thus W fric = m g d cos = (0 . 374) (14 kg) (9 . 8 m / s 2 ) (9 . 27 m) cos21 = 444 . 076 J . The work due to the applied force is W appl = F d = (150 N) (9 . 27 m) = 1390 . 5 J , and the work due to gravity is W grav = m g d sin = (14 kg) (9 . 8 m / s 2 ) (9 . 27 m) sin21 = 455 . 788 J , so that K = W fric + W appl + W grav = ( 444 . 076 J) + (1390 . 5 J) + ( 455 . 788 J) = 490 . 636 J ....
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## This note was uploaded on 05/02/2011 for the course PHYSICS 111 taught by Professor Wang during the Spring '09 term at NJIT.

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HW 5 solution_pdf - ghei (mg42382) HW 5 opyrchal (11130) 1...

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