This preview shows pages 1–3. Sign up to view the full content.
This preview has intentionally blurred sections. Sign up to view the full version.
View Full Document
Unformatted text preview: Answer, Key Homework 9 David McIntyre 45123 May 10, 2004 1 This printout should have 16 questions. Multiplechoice 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 Newtons 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 = dW dt = ~ 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 Newtons 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. Answer, Key Homework 9 David McIntyre 45123 May 10, 2004 2 Find the work done on the car in this time....
View Full
Document
 Spring '08
 RIJSSENBEEK
 Work

Click to edit the document details