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Unformatted text preview: Abd Elhai, Mohamed – Homework 6 – Due: Mar 3 2008, 11:00 pm – Inst: Dr Kevin Dean 1 This printout should have 21 questions. Multiplechoice questions may continue on the next column or page – find all choices before answering. The due time is Central time. This is ONLINE HOMEWORK No 6 It is due 09:00 Tuesday morning, 4 March (Abu Dhabi time)  (same as 23:00 Monday night in Texas). 001 (part 1 of 2) 7 points A simple Atwood’s machine uses two masses m 1 and m 2 . Starting from rest, the speed of the two masses is 1 . 8 m / s at the end of 9 . 8 s. At that time, the kinetic energy of the system is 34 J and each mass has moved a distance of 8 . 82 m. The acceleration of gravity is 9 . 81 m / s 2 . 8 . 82 m m 1 m 2 Find the value of heavier mass. Correct answer: 10 . 6903 kg. Explanation: Let : v = 1 . 8 m / s , t = 9 . 8 s , K = 34 J , ‘ = 8 . 82 m , and g = 9 . 81 m / s 2 . The total kinetic energy of the system is K = 1 2 ( m 1 + m 2 ) v 2 m 1 + m 2 = 2 K v 2 = 2(34 J) (1 . 8 m / s) 2 = 20 . 9877 kg a = µ m 1 m 2 m 1 + m 2 ¶ g X F = ( m 1 + m 2 ) a ( m 1 m 2 ) g = ( m 1 + m 2 ) a a = µ m 1 m 2 m 1 + m 2 ¶ g The velocity is v = at = ( m 1 m 2 ) g t m 1 + m 2 m 1 m 2 = ( m 1 + m 2 ) v g t = (20 . 9877 kg)(1 . 8 m / s) (9 . 81 m / s 2 )(9 . 8 s) = 0 . 392954 kg . m 1 + m 2 = 20 . 9877 kg m 1 m 2 = 0 . 392954 kg m 1 = 20 . 9877 kg + 0 . 392954 kg 2 = 10 . 6903 kg . 002 (part 2 of 2) 2 points Find the value of lighter mass. Correct answer: 10 . 2974 kg. Explanation: m 2 = 20 . 9877 kg . 392954 kg 2 = 10 . 2974 kg . keywords: 003 (part 1 of 3) 7 points A 2 . 94 kg block is pushed 1 . 5 m up a vertical wall with constant speed by a constant force of magnitude F applied at an angle of 57 . 6 ◦ with the horizontal. The acceleration of gravity is 9 . 8 m / s 2 . Abd Elhai, Mohamed – Homework 6 – Due: Mar 3 2008, 11:00 pm – Inst: Dr Kevin Dean 2 2 . 94 kg F 5 7 . 6 ◦ If the coefficient of kinetic friction between the block and wall is 0 . 514, find the work done by F . Correct answer: 64 . 1401 J. Explanation: Given : m = 2 . 94 kg , μ = 0 . 514 , θ = 57 . 6 ◦ , and Δ y = 1 . 5 m . F 5 7 . 6 ◦ v mg f k n The block is in equilibrium horizontally, so X F x = F cos θ n = 0 , so that n = F cos θ Since the block moves with constant velocity, X F y = F sin θ mg f k = 0 F sin θ mg μF cos θ = 0 F (sin θ μ cos θ ) = mg F...
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This note was uploaded on 02/16/2010 for the course PHYSICS 191262 taught by Professor Najafzadeh during the Spring '09 term at The Petroleum Institute.
 Spring '09
 NAJAFZADEH
 mechanics, Work

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