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Unformatted text preview: Version 026 – TEST01 – tsoi – (92940) 1 This printout should have 15 questions. Multiplechoice questions may continue on the next column or page – find all choices before answering. 001 10.0 points A 888 kg block is pushed on the slope of a 32 ◦ frictionless inclined plane to give it an initial speed of 50 cm / s along the slope when the block is 1 . 5 m from the bottom of the incline. The acceleration of gravity is 9 . 8 m / s 2 . 8 8 8 k g 1 . 5 m 32 ◦ What is the speed of the block at the bot tom of the plane? 1. 3.8459 2. 3.36035 3. 3.97865 4. 5.21834 5. 4.35265 6. 3.78572 7. 4.1071 8. 3.93111 9. 4.56848 10. 4.76804 Correct answer: 3 . 97865 m / s. Explanation: Let : m = 888 kg , θ = 32 ◦ , and v = 50 cm / s = 0 . 5 m / s . Note: It doesn’t make any difference whether the block is initially pushed up or down the slope. Consider the free body diagram for the block M g s i n θ N = M g c o s θ M g We therefore have a x = g sin θ . The final speed is v 2 = v 2 + 2 a x ( x − x ) = v 2 + 2 g sin θ ( x − x ) = (0 . 5 m / s) 2 + 2 (9 . 8 m / s 2 ) sin 32 ◦ (1 . 5 m) = 15 . 8296 m 2 / s 2 , so v = radicalBig 15 . 8296 m 2 / s 2 = 3 . 97865 m / s . 002 10.0 points This question is typical on some driver’s li cense exams: A car moving at 49 km / h skids 17 m with locked brakes. How far will the car skid with locked brakes at 147 km / h? 1. 64.0 2. 117.0 3. 108.0 4. 126.0 5. 135.0 6. 100.0 7. 56.0 8. 81.25 9. 48.0 10. 153.0 Correct answer: 153 m. Explanation: 147 km / h 49 km / h = 3 At 3 times the speed, it has 9 times the KE parenleftbigg 1 2 mv 2 parenrightbigg and will skid 9 times as far: 153 m. Since the frictional force is about the same in both cases, the distance has to be 9 times as great for 9 times as much work done by the pavement on the car. Version 026 – TEST01 – tsoi – (92940) 2 003 10.0 points Take the mass of the Earth to be 5 . 98 × 10 24 kg. If the Earth’s gravitational force causes a falling 68 kg student to accelerate downward at 9 . 8 m / s 2 , determine the upward accelera tion of the Earth during the student’s fall. 1. 1.50769e22 2. 3.93311e23 3. 8.19398e23 4. 1.24548e22 5. 9.01338e23 6. 5.73579e23 7. 6.71906e23 8. 1.27826e22 9. 9.17726e23 10. 1.11438e22 Correct answer: 1 . 11438 × 10 − 22 m / s 2 . Explanation: By Newton’s third law, the force F se ex erted on the student by the earth is equal in magnitude and opposite in direction to the force F es exerted on the earth by the student....
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This note was uploaded on 07/09/2011 for the course PHY 302 taught by Professor Staff during the Summer '08 term at University of Texas.
 Summer '08
 Staff
 Physics, Friction

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