hw9_solns - Answer Key Homework 9 David McIntyre 45123 This...

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Answer, Key – Homework 9 – David McIntyre – 45123 – May 10, 2004 1 This print-out should have 16 questions. Multiple-choice questions may continue on the next column or page – find all choices before making your selection. The due time is Central time. Chapter 8 problems. 001 (part 1 of 1) 0 points A commonly used unit is the kilowatt hour. The physical quantity measured in kilowatt hours is 1. None of these 2. current. 3. power. 4. force. 5. work (energy). correct Explanation: A Watt, which measures power, is a Joule per second. When power is multiplied by time, the result is in Joules, which measures energy or work. 1 kilowatt hr = 10 3 J s (3600 s) = 3600 × 10 3 J 002 (part 1 of 2) 0 points An elevator has a mass of 1040 kg and car- ries a maximum load of 700 kg. A constant frictional force of 4050 N retards its motion upward. The acceleration of gravity is 9 . 8 m / s 2 . What must be the minimum power deliv- ered by the motor to lift the elevator at a constant speed of 2 . 86 m / s? Correct answer: 60351 . 7 W. Explanation: The motor must supply the force ~ T that pulls the elevator upward. From Newton’s second law and from the fact that the acceler- ation is zero (since v is constant), we obtain T - f - M g = 0 , where M = m e + m l = 1740 kg is the total mass (elevator plus load). Therefore, T = f + M g = 4050 N + (1740 kg) (9 . 8 m / s 2 ) = 21102 N . Using the equation P = d W d t = ~ F · ~v and the fact that ~ T is in the same direction as ~v gives P = ~ T · ~v = T v = (21102 N) (2 . 86 m / s) = 60351 . 7 W . 003 (part 2 of 2) 0 points What power must the motor deliver at a in- stantaneous speed of 2 . 86 m / s if the elevator is designed to provide an upward acceleration of 1 . 11 m / s 2 ? Correct answer: 65875 . 5 W. Explanation: Applying Newton’s second law to the ele- vator gives T - f - M g = M a . Thus T = M ( a + g ) + f = (1740 kg) (1 . 11 m / s 2 + 9 . 8 m / s 2 ) + 4050 N = 23033 . 4 N . and the required instantaneous power is P = T v = (23033 . 4 N) (2 . 86 m / s) = 65875 . 5 W . 004 (part 1 of 3) 4 points A 1680 kg car accelerates uniformly from rest to 10 m / s in 3 . 08 s.
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Answer, Key – Homework 9 – David McIntyre – 45123 – May 10, 2004 2 Find the work done on the car in this time. Correct answer: 84000 J. Explanation: The work is given by the change in kinetic energy of the car W = Δ K = 1 2 m v 2 = 1 2 (1680 kg) (10 m / s) 2 = 84000 J .
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