This preview has intentionally blurred sections. Sign up to view the full version.
View Full Document
Unformatted text preview: Version 010/AAACC – Test #2 – Antoniewicz – (57420) 1 This printout should have 17 questions. Multiplechoice questions may continue on the next column or page – find all choices before answering. 001 (part 1 of 3) 4.0 points A particle moving along the xaxis has a po tential energy U ( x ), as shown in the accom panying graph. Scale factors: a = 2 . 6 m, b = 76 J. a 2 a 3 a 4 a 5 a 6 a 1 b 2 b 3 b 4 b 5 b U [J] x [m] C A B What is the force exerted on the particle when x = 3 a (point A , the lowest point on the curve)? 1. correct 2. − 9 . 74359 N 3. 29 . 2308 N 4. − 29 . 2308 N 5. 9 . 74359 N 6. Undetermined, since the magnitude of the force is zero. Explanation: From the definition of Potential Energy, we know F = − d U dx . In other words, the force on a particle at any particular point is given by the negative of the slope of the potential energy function at that point. At x = 3 a , the potential energy has a min imum; the slope of the function is zero at that point. Therefore there are no forces on a par ticle at that point. ( x = 3 a is an equilibrium point.) 002 (part 2 of 3) 3.0 points What direction is the force exerted on the particle when x = 5 a (point B )? 1. To the right. 2. Undetermined, since the magnitude of the force is zero. 3. To the left. correct Explanation: If the potential energy is increasing, the force is negative and vice versa . 003 (part 3 of 3) 3.0 points If the particle has a mass of 6 kg and is released from rest at x = 5 a , what is the particle’s velocity when it reaches x = a ? 1. 2.71679 2. 3.27327 3. 4.34613 4. 4.68623 5. 2.51312 6. 3.13928 7. 3.72104 8. 2.66557 9. 5.03322 10. 2.47339 Correct answer: 5 . 03322 m / s. Explanation: Apply Conservation of Energy: U init + K init = U final + K final U B + 0 = U C + 1 2 mv 2 Solving for v gives: v = radicalbigg 2 m ( U B − U C ) = radicalbigg 2 m (3 b − 2 b ) = radicalbigg 2 m b Version 010/AAACC – Test #2 – Antoniewicz – (57420) 2 = radicalBigg 2 (6 kg) (76 J) = 5 . 03322 m / s . 004 10.0 points A bucket full of water is rotated in a verti cal circle of radius 1 . 574 m (the approximate length of a person’s arm). What must be the minimum speed of the pail at the top of the circle if no water is to spill out? 1. 2.32585 2. 3.5265 3. 2.9031 4. 3.13675 5. 3.44497 6. 3.92749 7. 3.34831 8. 3.00919 9. 2.82959 10. 2.0647 Correct answer: 3 . 92749 m / s. Explanation: If we analyze the forces on the water at the top of a vertical circle (using the notation F ji for a two body force on ˆ i from ˆ j ), we see that N pail,water + W earth,water = ma centripetal = m v 2 r . To minimize the velocity, we minimize the lhs of the equation. Since we can’t change the weight of the water, we use the lowest Normal force that we can, 0. Then, mg = m v 2 r and so, v = √ g r = radicalBig (9 . 8 m / s 2 ) (1 . 574 m) = 3 . 92749 m / s ....
View
Full Document
 Spring '08
 Turner
 Physics, Force, Friction, Potential Energy, Correct Answer

Click to edit the document details