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HS6 - Adams Russell Homework 6 Due midnight Inst Opyrchal H...

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Adams, Russell – Homework 6 – Due: Oct 19 2006, midnight – Inst: Opyrchal, H 1 This print-out should have 13 questions. Multiple-choice questions may continue on the next column or page – find all choices before answering. The due time is Central time. 001 (part 1 of 1) 10 points Lee pushes horizontally with a force of 41 N on a 30 kg mass for 21 m across a floor. Calculate the amount of work Lee did. Correct answer: 861 J. Explanation: The mass of the object is ignored since there is no friction present. Thus W = Fd = (41 N) (30 kg) = 861 J keywords: 002 (part 1 of 2) 10 points A particle moving in the xy plane undergoes a displacement ~s = ( s x ˆ ı + s y ˆ ), with s x = 4 . 56 m, s y = 6 . 68 m, while a constant force ~ F = ( F x ˆ ı + F y ˆ ), with F x = 5 . 73 N, F y = 4 . 26 N, acts on the particle. Calculate the magnitude of the displace- ment. Correct answer: 8 . 08802 m. Explanation: The magnitude of ~s is given by s = q s 2 x + s 2 y = q (4 . 56 m) 2 + (6 . 68 m) 2 = 8 . 08802 m . 003 (part 2 of 2) 10 points Calculate the work done by ~ F . Correct answer: 54 . 5856 J. Explanation: The work done by the force ~ F is equal to the scalar product of ~ F and the displacement ~s : W = ~ F · ~s = [(5 . 73 N)ˆ ı + (4 . 26 N)ˆ ] · [(4 . 56 m)ˆ ı + (6 . 68 m)ˆ ] = (5 . 73 N) (4 . 56 m) (ˆ ı · ˆ ı ) + (4 . 26 N) (6 . 68 m) (ˆ · ˆ ) + [(5 . 73 N) (6 . 68 m) + (4 . 26 N) (4 . 56 m)] (ˆ ı · ˆ ) = 54 . 5856 J . keywords: 004 (part 1 of 2) 10 points The force required to stretch a Hooke’s-law spring varies from 0 N to 49 . 4 N as we stretch the spring by moving one end 10 . 9 cm from its unstressed position.
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