Homework 7 Sol. - Abd Elhai, Mohamed Homework 7 Due: Mar 11...

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Abd Elhai, Mohamed – Homework 7 – Due: Mar 11 2008, 8:00 pm – Inst: Dr Kevin Dean 1 This print-out should have 20 questions. Multiple-choice questions may continue on the next column or page – fnd all choices beFore answering. The due time is Central time. This is ONLINE HOMEWORK No. 7. It is due by 06:00 AM Abu Dhabi Time on Tuesday 11 March(8:00 PM Monday March 10, Texas time) 001 (part 1 oF 1) 5 points A 3 . 37 kg block is released From A at height 4 . 72 m on a Frictionless track shown. The radius oF the track is 2 . 19 m. The acceleration oF gravity is 9 . 8 m / s 2 . R P A h Determine the magnitude oF the accelera- tion For the block at P. Correct answer: 24 . 6727 m / s 2 . Explanation: ±rom conservation oF energy ( K + U ) A = ( K + U ) P we obtain mg h = mv 2 2 + mg R . ThereFore v = p 2 g ( h - R ) = 7 . 04187 m / s . Then the radial acceleration at P is a r = v 2 R = 22 . 6429 m / s 2 and the tangential acceleration is a t = g = 9 . 8 m / s 2 . So, the total acceleration at P is a = q a 2 t + a 2 r = q (9 . 8 m / s 2 ) 2 + (22 . 6429 m / s 2 ) 2 = 24 . 6727 m / s 2 . keywords: 002 (part 1 oF 1) 5 points A 33 g block is released From rest and slides down a Frictionless track that begins 3 m above the horizontal, as shown in the fgure. At the bottom oF the track, where the surFace is horizontal, the block strikes and sticks to a light spring with a spring constant oF 18 N / m. The acceleration oF gravity is 9 . 8 m / s 2 . 3 m 33 g 18 N / m ±ind the maximum distance the spring is compressed. Correct answer: 0 . 328329 m. Explanation: Let : m = 33 g = 0 . 033 kg , h i = 3 m , and k = 18 N / m . ±rom conservation oF mechanical energy, we have mg h i = 1 2 k x 2 x = r 2 mg h i k = s 2(0 . 033 kg)(9 . 8 m / s 2 )(3 m) 18 N / m = 0 . 328329 m . keywords: 003 (part 1 oF 2) 5 points ApendulumconsistsoFasphereoFmass1 . 5kg attached to a light cord oF length 10 . 2 m as in
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Abd Elhai, Mohamed – Homework 7 – Due: Mar 11 2008, 8:00 pm – Inst: Dr Kevin Dean 2 the fgure below. The sphere is released From rest when the cord makes a 34 angle with the vertical, and the pivot at P is Frictionless. The acceleration oF gravity is 9 . 8 m / s 2 . 9 . 8 m / s 2 1 . 5 kg 10 . 2 m 34 ±ind the speed oF the sphere when it is at the lowest point. Correct answer: 5 . 84627 m / s. Explanation: Let : m = 1 . 5 kg , = 10 . 2 m , θ i = 34 , and g = 9 . 8 m / s 2 . The only Force that does work on the sphere is the Force oF gravity, since the Force oF ten- sion is always perpendicular to each element oF the displacement and hence does no work. Since the Force oF gravity is a conservative Force, the total mechanical energy is constant. ThereFore, as the pendulum swings, there is a continuous transFer between potential and ki- netic energy. At the instant the pendulum is released, the energy is entirely potential en- ergy. At the lowest point, the pendulum has kinetic energy but has lost some potential en- ergy. The pendulum swings to the opposite
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This note was uploaded on 02/16/2010 for the course PHYSICS 191262 taught by Professor Najafzadeh during the Spring '09 term at The Petroleum Institute.

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Homework 7 Sol. - Abd Elhai, Mohamed Homework 7 Due: Mar 11...

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