solutions 8 - Patel Kinal – Homework 8 – Due Oct 2 2007...

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Unformatted text preview: Patel, Kinal – 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 32 . 7 kg partner straight up off the ground a distance of 0 . 842 m before releasing her. The acceleration of gravity is 9 . 8 m / s 2 . If he does this 21 times, how much work has he done? Correct answer: 5666 . 37 J. Explanation: The work done in lifting the cheerleader once is W 1 = mg h = (32 . 7 kg)(9 . 8 m / s 2 )(0 . 842 m) = 269 . 827 J . The work required to lift her n = 21 times is W = nW 1 = (21)(269 . 827 J) = 5666 . 37 J . keywords: 002 (part 1 of 1) 10 points A shopper in a supermarket pushes a cart with a force of 37 N directed at an angle of 24 . 9 ◦ downward from the horizontal. Find the work done by the shopper as she moves down a 48 . 8 m length of aisle. Correct answer: 1 . 63776 kJ. Explanation: Given : F shopper = 37 N , θ = 24 . 9 ◦ , and d = 48 . 8 m . The force is applied at an angle to the displacement, so the net work is W shopper = F shopper d cos θ = (37 N)(48 . 8 m)(cos24 . 9 ◦ ) µ 1 kJ 1000 J ¶ = 1 . 63776 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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solutions 8 - Patel Kinal – Homework 8 – Due Oct 2 2007...

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