{[ promptMessage ]}

Bookmark it

{[ promptMessage ]}

Physics Test #2 solutions

Physics Test #2 solutions - Version 010/AAACC – Test#2...

Info iconThis preview shows pages 1–3. Sign up to view the full content.

View Full Document Right Arrow Icon

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full Document Right Arrow Icon
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: Version 010/AAACC – Test #2 – Antoniewicz – (57420) 1 This print-out should have 17 questions. Multiple-choice 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 x-axis 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

{[ snackBarMessage ]}

Page1 / 10

Physics Test #2 solutions - Version 010/AAACC – Test#2...

This preview shows document pages 1 - 3. Sign up to view the full document.

View Full Document Right Arrow Icon bookmark
Ask a homework question - tutors are online