This preview shows pages 1–2. 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: Morby, Grant – Homework 13 – Due: Feb 22 2006, noon – Inst: Drummond 1 This print-out should have 12 questions. Multiple-choice questions may continue on the next column or page – find all choices before answering. The due time is Central time. 001 (part 1 of 2) 10 points Assume your mass is 92 kg. The acceleration due to gravity is 9 . 8 m / s 2 . How much work against gravity do you do when you climb a flight of stairs 2 m high? Correct answer: 1803 . 2 J. Explanation: Let : m = 92 kg , g = 9 . 8 m / s 2 , and h = 2 m . W = mg h = (92 kg)(9 . 8 m / s 2 )(2 m) = 1803 . 2 J . 002 (part 2 of 2) 10 points Consider the energy consumed by a 200 W light bulb in an hour. How many flights of stairs would you have to climb to equal the work of the lightbulb? Correct answer: 399 . 29 . Explanation: For the light bulb: E = P t = (200 W)(3600 s) = 720000 J , n = E W = 720000 J 1803 . 2 J = 399 . 29 . 003 (part 1 of 1) 10 points A block of mass m is pushed a distance D up an inclined plane by a horizontal force F . The plane is inclined at an angle θ with respect to the horizontal. The block starts from rest and the coefficient of kinetic friction is μ k . m D μ k F θ What is the final speed of the block? 1. v = r 2 m ( F cos θ- mg sin θ + μ k N ) D 2. v = r 2 m ( F sin θ + μ k N ) D 3. v = r 2 m ( F cos θ- mg sin θ ) D 4. v = r 2 m ( F cos θ + mg sin θ- μ k N ) D 5. v = r 2 m ( F cos θ- μ k N ) D 6. v = r 2 m ( F sin θ- μ k N ) D 7. v = r 2 m ( F cos θ- mg sin θ- μ k N ) D correct 8. v = r 2 m ( F cos θ + mg sin θ ) D Explanation: The work done by gravity is W grav = mg D cos(90 ◦ + θ ) =- mg D sin θ ....
View Full Document