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Homework_08

# Homework_08 - Nguyen Thanh Homework 8 Due Oct 2 2007 7:00...

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Nguyen, Thanh – Homework 8 – Due: Oct 2 2007, 7:00 pm – Inst: D Weathers 1 This print-out should have 11 questions. Multiple-choice questions may continue on the next column or page – find all choices before answering. The due time is Central time. Solve problems 8 and 9 using work and energy concepts. Assume in problem 9 that the force applied to the car is constant. 001 (part 1 of 1) 10 points A cheerleader lifts his 28 . 3 kg partner straight up off the ground a distance of 0 . 27 m before releasing her. The acceleration of gravity is 9 . 8 m / s 2 . If he does this 31 times, how much work has he done? Correct answer: 2321 . 34 J. Explanation: The work done in lifting the cheerleader once is W 1 = m g h = (28 . 3 kg)(9 . 8 m / s 2 )(0 . 27 m) = 74 . 8818 J . The work required to lift her n = 31 times is W = n W 1 = (31)(74 . 8818 J) = 2321 . 34 J . keywords: 002 (part 1 of 1) 10 points A shopper in a supermarket pushes a cart with a force of 32 . 2 N directed at an angle of 26 . 7 downward from the horizontal. Find the work done by the shopper as she moves down a 48 . 5 m length of aisle. Correct answer: 1 . 39518 kJ. Explanation: Given : F shopper = 32 . 2 N , θ = 26 . 7 , and d = 48 . 5 m . The force is applied at an angle to the displacement, so the net work is W shopper = F shopper d cos θ = (32 . 2 N) (48 . 5 m) (cos 26 . 7 ) 1 kJ 1000 J = 1 . 39518 kJ . keywords: 003 (part 1 of 2) 10 points A force ~ F = F x ˆ ı + F y ˆ acts on a particle that undergoes a displacement of ~s = s x ˆ ı + s y ˆ .

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