exam 4 phy - Version 015 – Exam 4 – chelikowsky...

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Unformatted text preview: Version 015 – Exam 4 – chelikowsky – (59005) 1 This print-out should have 22 questions. Multiple-choice questions may continue on the next column or page – find all choices before answering. 001 10.0 points An automobile engine develops a torque of 520 N · m and is rotating at a speed of 1000 rev / min. What horsepower does the engine generate? 1 hp = 746 W. 1. 72.995 2. 117.915 3. 99.6663 4. 185.295 5. 207.755 6. 315.844 7. 140.375 8. 277.943 9. 154.413 10. 65.9763 Correct answer: 72 . 995 hp. Explanation: Let : τ = 520 N · m and ω = 1000 rev / min . P = τ ω = (520 N · m) (1000 rev / min) × parenleftbigg 2 π 1 rev parenrightbiggparenleftbigg 1 min 60 s parenrightbiggparenleftbigg 1 hp 746 W parenrightbigg = 72 . 995 hp . 002 10.0 points Two ropes support a load of 340 kg. The two ropes are perpendicular to each other, and the tension in the first rope is 1 . 98 times that of the second rope. Find the tension in the second rope. The acceleration of gravity is 9 . 8 m / s 2 . 1. 1378.42 2. 2751.28 3. 1418.43 4. 1338.39 5. 1231.62 6. 3081.45 7. 2424.03 8. 1990.73 9. 1502.12 10. 1308.59 Correct answer: 1502 . 12 N. Explanation: Let : m = 340 kg , f = 1 . 98 , and g = 9 . 8 m / s 2 . T 1 = f T 2 . The load is not moving, so the net force ex- erted on it must be zero. Let θ be the angle between the first rope and the vertical direc- tion; when the net force on the load is zero, its horizontal component and its vertical compo- nent should have a sum of zero. For the horizontal direction, T 1 sin θ − T 2 cos θ = 0 f T 2 sin θ − T 2 cos θ = 0 f sin θ = cos θ tan θ = 1 f θ = tan − 1 parenleftbigg 1 f parenrightbigg = tan − 1 parenleftbigg 1 1 . 98 parenrightbigg = 26 . 7961 ◦ . The vertical components of the net force should also have a sum of zero: T 1 cos θ + T 2 sin θ = M g f T 2 cos θ + T 2 sin θ = M g T 2 ( f cos θ + sin θ ) = M g T 2 = M g f cos θ + sin θ = (340 kg) (9 . 8 m / s 2 ) 1 . 98 cos 26 . 7961 ◦ + sin 26 . 7961 ◦ = 1502 . 12 N . Version 015 – Exam 4 – chelikowsky – (59005) 2 003 (part 1 of 2) 10.0 points A uniform solid disk and a uniform hoop are placed side by side at the top of an incline of height h . If they are released from rest and roll with- out slipping, determine their speeds when they reach the bottom. ( d = disk, h = hoop) 1. v d = radicalbigg 1 2 hg , v h = radicalbigg 1 3 hg 2. v d = radicalbigg 2 3 hg , v h = radicalbig 3 hg 3. v d = radicalbig hg , v h = radicalbig 2 hg 4. v d = radicalbig 2 hg , v h = radicalbig hg 5. v d = radicalbigg 4 3 hg , v h = radicalbig hg correct Explanation: Because they roll without slipping, v = r ω and the total kinetic energy is K = 1 2 mv 2 + 1 2 I ω 2 , so for the disk, since I = 1 2 mr 2 , K = 1 2 mv 2 d + 1 4 mv 2 d = 3 4 mv 2 d ....
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This note was uploaded on 07/24/2010 for the course PHY 59005 taught by Professor Chelikowsky during the Fall '09 term at University of Texas.

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exam 4 phy - Version 015 – Exam 4 – chelikowsky...

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