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HW #6 Solutions - homework 06 FONTENOT BRIAN Due 4:00 am...

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homework 06 – FONTENOT, BRIAN – Due: Feb 25 2008, 4:00 am 1 Question 1, chap 7, sect 1. part 1 of 3 10 points Assume you are on a planet similar to Earth where the acceleration of gravity g 10 m / s 2 . A plane 50 m in length is inclined at an angle of 16 . 3 as shown. A block of weight 500 N is placed at the top of the plane and allowed to slide down. 500 N μ = 0 . 5 v 50 m 16 . 3 48 m 14 m The mass of the block is most nearly 1. m 25 kg 2. m 26 kg 3. m 39 kg 4. m 10 kg 5. m 52 kg 6. m 15 kg 7. m 20 kg 8. m 13 kg 9. m 50 kg correct Explanation: Let : W = 500 N and g 10 m / s 2 . s θ x y The mass of the block is given by m = W g = 500 N 10 m / s 2 = 50 kg . Question 2, chap 7, sect 1. part 2 of 3 10 points Calculate the magnitude of the normal force exerted on the block by the plane. 1. bardbl vector N bardbl ≈ 240 N 2. bardbl vector N bardbl ≈ 160 N 3. bardbl vector N bardbl ≈ 360 N 4. bardbl vector N bardbl ≈ 480 N correct 5. bardbl vector N bardbl ≈ 200 N 6. bardbl vector N bardbl ≈ 400 N 7. bardbl vector N bardbl ≈ 120 N 8. bardbl vector N bardbl ≈ 80 N Explanation: Let : x = 48 m and y = 14 m . s = radicalbig x 2 + y 2 = radicalBig (48 m) 2 + (14 m) 2 = 50 m . cos θ = x s = 48 50 . The normal force is the force perpendicular to the surface. It is given by N = W cos θ = W x s = (500 N) parenleftbigg 48 m 50 m parenrightbigg 480 N . Question 3, chap 7, sect 1. part 3 of 3 10 points
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homework 06 – FONTENOT, BRIAN – Due: Feb 25 2008, 4:00 am 2 Calculate the work done on the block by the gravitational force during the 50 m slide down the plane in J. 1. W 1750 J 2. W 3690 J 3. W 7000 J correct 4. W 10400 J 5. W 1350 J 6. W 650 J 7. W 3750 J 8. W 5850 J 9. W 2400 J Explanation: Let : d = 50 m . sin θ = y s = 14 50 . Work is given by W = vector F · vector d , where vector d is the displacement vector and vector F is the force act- ing on the system. The component of the force pointing in the direction of the displace- ment is F bardbl = W sin θ . The work done by the gravitational force, then, is W = F bardbl d = W d sin θ = W d y s = (500 N) (50 m) parenleftbigg 14 m 50 m parenrightbigg = 7000 J . Question 4, chap 7, sect 1. part 1 of 3 10 points Sally applies a horizontal force of 706 N with a rope to drag a wooden crate across a floor with a constant speed. The rope tied to the crate is pulled at an angle of 42 relative to the floor. The acceleration of gravity is 9 . 8 m / s 2 . m F 706 N 42 How much force is exerted by the rope on the crate? Correct answer: 950 . 017 N (tolerance ± 1 %). Explanation: Given : F h = 706 N , and θ = 42 . The horizontal component of the force exerted by the rope is defined by cos θ = F h F so that F = F h cos θ = 706 N cos 42 = 950 . 017 N . Question 5, chap 7, sect 1. part 2 of 3 10 points What work is done by Sally if the crate is moved 65 . 4 m? Correct answer: 46172 . 4 J (tolerance ± 1 %). Explanation: The motion is in the direction of the hori- zontal component, so the work Sally does is W = F h d = (706 N) (65 . 4 m) = 46172 . 4 J .
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