hw6 - karna (pk4534) HW 6 coker (58245) 1 This print-out...

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Unformatted text preview: karna (pk4534) HW 6 coker (58245) 1 This print-out should have 17 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 force vector F = F x + F y acts on a particle that undergoes a displacement of vectors = s x + s y . Let: F x = 6 N, F y =- 5 N, s x = 5 m, and s y = 2 m. Find the work done by the force on the particle. Correct answer: 20 J. Explanation: The work is given by W = vector F vectors = F x s x + F y s y = (6 N) (5 m) + (- 5 N) (2 m) = 20 J . 002 (part 2 of 2) 10.0 points Find the angle between vector F and vectors . Correct answer: 61 . 607 . Explanation: Since W = vector F vectors = vextendsingle vextendsingle vextendsingle vector F vextendsingle vextendsingle vextendsingle | vectors | cos . The first and the third sides can be solved for to give = 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 (6 N) 2 + (- 5 N) 2 = 7 . 81025 N and | vectors | = radicalBig s 2 x + s 2 y = radicalbig 5 m 2 + 2 m 2 = 5 . 38516 m . Thus = cos 1 bracketleftbigg 20 J (7 . 81025 N) (5 . 38516 m) bracketrightbigg = 61 . 607 . 003 (part 1 of 3) 10.0 points Sally applies a horizontal force of 605 N with a rope to drag a wooden crate across a floor with a constant speed. The rope tied to the crate is pulled at an angle of 46 relative to the floor. The acceleration of gravity is 9 . 8 m / s 2 . m F 605 N 4 6 How much force is exerted by the rope on the crate? Correct answer: 870 . 932 N. Explanation: Given : F h = 605 N , and = 46 . The horizontal component of the force exerted by the rope is defined by cos = F h F so that F = F h cos = 605 N cos 46 = 870 . 932 N . 004 (part 2 of 3) 10.0 points What work is done by Sally if the crate is moved 29 m? Correct answer: 17545 J. karna (pk4534) HW 6 coker (58245) 2 Explanation: The motion is in the direction of the hori- zontal component, so the work Sally does is W = F h d = (605 N) (29 m) = 17545 J ....
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This note was uploaded on 10/29/2009 for the course PHY 303K taught by Professor Turner during the Spring '08 term at University of Texas at Austin.

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hw6 - karna (pk4534) HW 6 coker (58245) 1 This print-out...

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