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Homework 08b-solutions

# Homework 08b-solutions - kaplan(hmk378 Homework 08b...

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kaplan (hmk378) – Homework 08b – Weathers – (17104) 1 This print-out should have 11 questions. Multiple-choice questions may continue on the next column or page – fnd all choices beFore answering. 001 (part 1 oF 2) 10.0 points A Force v F = F x ˆ ı + F y ˆ acts on a particle that undergoes a displacement oF vs = s x ˆ ı + s y ˆ where F x = 2 N, F y = 3 N, s x = 6 m, and s y = 3 m. ±ind the work done by the Force on the particle. Correct answer: 3 J. Explanation: The work is given by W = v F · = F x s x + F y s y = (2 N) (6 m) + ( 3 N) (3 m) = 3 J . 002 (part 2 oF 2) 10.0 points ±ind the angle between v F and . Correct answer: 82 . 875 . Explanation: Since W = v F · = v v v v F v v v | | cos θ θ = cos 1 W v v v v F v v v | | . v v v v F v v v = r F 2 x + F 2 y = r (2 N) 2 + ( 3 N) 2 = 3 . 60555 N and | | = r s 2 x + s 2 y = R 6 m 2 + 3 m 2 = 6 . 7082 m , so θ = cos 1 b 3 J (3 . 60555 N) (6 . 7082 m) B = 82 . 875 . 003 10.0 points A shopper in a supermarket pushes a cart with a Force oF 32 . 3 N directed at an angle oF 23 . 7 downward From the horizontal. ±ind the work done by the shopper as she moves down a 51 . 6 m length oF aisle. Correct answer: 1 . 52612 kJ. Explanation: Given : F shopper = 32 . 3 N , θ = 23 . 7 , and d = 51 . 6 m . The Force is applied at an angle to the displacement, so the net work is W shopper = F shopper d cos θ = (32 . 3 N) (51 . 6 m) (cos 23 . 7 ) p 1 kJ 1000 J P = 1 . 52612 kJ .

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Homework 08b-solutions - kaplan(hmk378 Homework 08b...

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