OH-06-solutions

# OH-06-solutions - obaidi (aao476) – OH-06 – Bradley –...

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Unformatted text preview: obaidi (aao476) – OH-06 – Bradley – (191) 1 This print-out should have 18 questions. Multiple-choice questions may continue on the next column or page – find all choices before answering. This is Online Homework No. 06. It is due before 06:00 AM on Tuesday 23 September, Abu Dhabi time (09:00 PM on 22 Sept, Texas time) unless notified otherwise. 001 (part 1 of 2) 5.0 points A force vector F = F x ˆ ı + F y ˆ acts on a particle that undergoes a displacement of vectors = s x ˆ ı + s y ˆ . Let: F x = 9 N, F y =- 5 N, s x = 5 m, and s y = 1 m. Find the work done by the force on the particle. Correct answer: 40 J. Explanation: The work is given by W = vector F · vectors = F x s x + F y s y = (9 N) (5 m) + (- 5 N) (1 m) = 40 J . 002 (part 2 of 2) 5.0 points Find the angle between vector F and vectors . Correct answer: 40 . 3646 ◦ . Explanation: Since W = vector F · vectors = vextendsingle vextendsingle vextendsingle vector F vextendsingle vextendsingle vextendsingle | vectors | cos θ . The first and the third sides can be solved for θ to give θ = cos − 1 W vextendsingle vextendsingle vextendsingle vector F vextendsingle vextendsingle vextendsingle | vectors | . First of all vextendsingle vextendsingle vextendsingle vector F vextendsingle vextendsingle vextendsingle = radicalBig F 2 x + F 2 y = radicalBig (9 N) 2 + (- 5 N) 2 = 10 . 2956 N and | vectors | = radicalBig s 2 x + s 2 y = radicalbig 5 m 2 + 1 m 2 = 5 . 09902 m . Thus θ = cos − 1 bracketleftbigg 40 J (10 . 2956 N) (5 . 09902 m) bracketrightbigg = 40 . 3646 ◦ . 003 (part 1 of 3) 4.0 points Note: in this problem the incline angle is measured from horizontal. Starting from rest, a 5 . 12 kg block slides 2 . 46 m down a rough 38 ◦ incline. The coeffi- cient of kinetic friction between the block and the incline is 0 . 427. The acceleration of gravity is 9 . 8 m / s 2 . Find the work done by the force of gravity. Correct answer: 75 . 9929 J. Explanation: Given : m = 5 . 12 kg , d = 2 . 46 m , and θ = 38 ◦ . The force of gravity is W = mg , directed downward. The angle between the force and the displacement is θ ′ = 90 ◦- 38 ◦ = 52 ◦ . The work done by gravity is W gravity = mg cos θ ′ d = (5 . 12 kg) ( 9 . 8 m / s 2 ) · (cos 52 ◦ ) (2 . 46 m) = 75 . 9929 J ....
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## This note was uploaded on 10/04/2008 for the course PYHSICIS 191 taught by Professor Bradly during the Spring '08 term at Abilene Christian University.

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OH-06-solutions - obaidi (aao476) – OH-06 – Bradley –...

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