hmwk05 - Su(ycs73 HW05 Tsoi(58020 This print-out should...

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Su (ycs73) – HW05 – Tsoi – (58020) 1 This print-out should have 23 questions. Multiple-choice questions may continue on the next column or page – find all choices before answering. 001 (part 1 of 2) 10.0 points A 500-N crate needs to be lifted 1 meter ver- tically in order to get it into the back of a pickup truck. What gives the crate a greater potential energy? 1. lift it straight up into the truck 2. slide it up a frictionless inclined plane 3. Unable to determine 4. Either correct Explanation: The change in height is the same for either method, so the new potential energies are the same. 002 (part 2 of 2) 10.0 points What is the advantage of using the inclined plane? 1. less force correct 2. less distance 3. less total energy 4. more power Explanation: An inclined plane can exchange an in- creased distance for less force. 003 (part 1 of 2) 10.0 points Assume your mass is 74 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 . 6 m high? Correct answer: 1885 . 52 J. Explanation: Let : m = 74 kg , g = 9 . 8 m / s 2 , and h = 2 . 6 m . W = m g h = (74 kg) (9 . 8 m / s 2 ) (2 . 6 m) = 1885 . 52 J . 004 (part 2 of 2) 10.0 points Consider the energy consumed by a 100 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: 190 . 929. Explanation: For the light bulb: E = P t = (100 W) (3600 s) = 3 . 6 × 10 5 J , n = E W = 3 . 6 × 10 5 J 1885 . 52 J = 190 . 929 . 005 10.0 points When an automobile moves with constant ve- locity, the power developed is used to over- come the frictional forces exerted by the air and the road. If the engine develops 20 hp, what total frictional force acts on the car at 140 mph? One horsepower equals 746 W, and one mile is 1609 m. Correct answer: 238 . 444 N. Explanation: Apply P = E t = Fd t = Fv F = P v = 20 hp · 746 W hp 140 mph · 1609 m mi · h 3600 s = 238 . 444 N 006 (part 1 of 3) 10.0 points A block of mass m is pushed a distance D up

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Su (ycs73) – HW05 – Tsoi – (58020) 2 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 If N is the normal force, what is the work done by friction? 1. W = - μ k ( N - m g cos θ ) D 2. W = - μ k N D correct 3. W = - μ k ( N + m g cos θ ) D 4. W = + μ k N D 5. W = + μ k ( N - m g cos θ ) D 6. W = 0 7. W = + μ k ( N + m g cos θ ) D Explanation: The force of friction has a magnitude F friction = μ k N . Since it is in the direc- tion opposite to the motion, we get W friction = - F friction D = - μ k N D. 007 (part 2 of 3) 10.0 points What is the work done by the normal force N ? 1. W = N D sin θ 2. W = N D cos θ 3. W = 0 correct 4. W = N D 5. W = -N D 6. W = ( N + m g cos θ + F sin θ ) D 7. W = ( m g cos θ + F sin θ - N ) D 8. W = ( N - m g cos θ - F sin θ ) D Explanation: The normal force makes an angle of 90 with the displacement, so the work done by it is zero.
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