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Unformatted text preview: Perez (nap563) – HW05 – TSOI – (58160) 1 This print-out should have 25 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- 2 . 5589 kg raindrop falls vertically at con- stant speed under the influence of gravity and air resistance. After the drop has fallen 50 . 7 m, what is the work done by gravity? The acceleration of gravity is 9 . 8 m / s 2 . Correct answer:- 1271 . 42 J. Explanation: The work is W = Force · Distance For the raindrop falling a distance h , the work done by the force of gravity mg is W g = mg h = (- 2 . 5589 kg) (9 . 8 m / s 2 ) (50 . 7 m) =- 1271 . 42 J . 002 (part 2 of 2) 10.0 points What is the work done by air resistance? Correct answer: 1271 . 42 J. Explanation: Since the raindrop remains at constant speed, the acceleration of the raindrop is zero, and Newton’s second law yields summationdisplay F y = F air- mg = 0 . It is evident that the force due to air resistance is equal in magnitude to the force due to gravity, but directed upward. The work done by air resistance can then be found, W air = F air h =- mg h =- (- 2 . 5589 kg) (9 . 8 m / s 2 ) × (50 . 7 m) = 1271 . 42 J . keywords: 003 10.0 points A cheerleader lifts his 60 . 6 kg partner straight up off the ground a distance of 0 . 516 m before releasing her. The acceleration of gravity is 9 . 8 m / s 2 . If he does this 29 times, how much work has he done? Correct answer: 8886 . 82 J. Explanation: The work done in lifting the cheerleader once is W 1 = mg h = (60 . 6 kg)(9 . 8 m / s 2 )(0 . 516 m) = 306 . 442 J . The work required to lift her n = 29 times is W = nW 1 = (29)(306 . 442 J) = 8886 . 82 J . 004 (part 1 of 3) 10.0 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 If N is the normal force, what is the work done by friction? 1. W = 0 2. W =- μ k ( N - mg cos θ ) D 3. W =- μ k ( N + mg cos θ ) D 4. W = + μ k N D Perez (nap563) – HW05 – TSOI – (58160) 2 5. W = + μ k ( N + mg cos θ ) D 6. W =- μ k N D correct 7. W = + μ k ( N - mg 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. 005 (part 2 of 3) 10.0 points What is the work done by the normal force N ? 1. W =-N D 2. W = 0 correct 3. W = N D sin θ 4. W = N D cos θ 5. W = ( N + mg cos θ + F sin θ ) D 6. W = N D 7. W = ( N - mg cos θ- F sin θ ) D 8. W = ( mg cos θ + F sin θ- N ) 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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This note was uploaded on 03/17/2010 for the course PHY 303K taught by Professor Turner during the Spring '08 term at University of Texas.
- Spring '08