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Unformatted text preview: baum (lmb2768) – HW05 – distler – (56295) 1 This printout should have 21 questions. Multiplechoice questions may continue on the next column or page – find all choices before answering. There are various ways to do part 2 of the ”Wind and Pendulum” problem. The easiest is to find the angle at which the three forces on the ball (tension, gravity and F) add up to zero. In part 1 of ”Pulling Two Masses”, you are asked about the ”work done against fric tion”. This is very funny phraseology. The friction does negative work, and what you are being askedfor is the magnitude (a positive number) of the work done by friction. 001 10.0 points A block sliding on a horizontal surface has an initial speed of 0.5 m/s. The block travels a distance of 1 m as it slows to a stop. What distance would the block have trav eled if its initial speed had been 1 m/s? 1. 1 m 2. 2 m 3. 3 m 4. 4 m correct 5. more information is needed to answer the question 6. 0.5 m Explanation: W nc = f k L = 0 1 2 mv 2 μmg L = 1 2 mv 2 L = v 2 2 μg So the distance is proportional to the square of the initial speed. 002 (part 1 of 2) 10.0 points A ball having mass 1 . 1 kg is connected by a string of length 0 . 8 m to a pivot point and held in place in a vertical position. A constant wind force of magnitude 17 . 85 N blows from left to right. Pivot Pivot F F (a) (b) H m m L L If the mass is released from the vertical po sition, what maximum height above its initial position will it attain? Assume that the string does not break in the process. The accelera tion of gravity is 9 . 8 m / s 2 . Correct answer: 1 . 1724 m. Explanation: Let : m = 1 . 1 kg , L = 0 . 8 m , F = 17 . 85 N , and g = 9 . 8 m / s 2 . Take the original point to be the one where the ball is released and the final point where its upward swing stops at height H . Then the horizontal displacement is x = L 2 ( L H ) 2 Since the force is purely horizontal, it does work W = F · d s = F 2 LH H 2 . Applying the workenergy theorem, ( K + U ) i + W wind = ( K + U ) f 0 + 0 + F 2 LH H 2 = 0 + mg H . H = 2 L 1 + mg F 2 · F 2 F 2 = 2 LF 2 F 2 + ( mg ) 2 baum (lmb2768) – HW05 – distler – (56295) 2 = 2 (0 . 8 m) (17 . 85 N) 2 (17 . 85 N) 2 + (1 . 1 kg) 2 (9 . 8 m / s 2 ) 2 = 1 . 1724 m . 003 (part 2 of 2) 10.0 points What will be the equilibrium height of the mass? Correct answer: 0 . 38643 m. Explanation: Considering components of the force equi librium F = 0, T sin θ + F = 0 and T cos θ mg = 0 . Dividing these, tan θ = F mg θ = arctan F mg = arctan 17 . 85 N (1 . 1 kg) (9 . 8 m / s 2 ) = 58 . 8713 ◦ , so the equilibrium height is h = L (1 cos θ ) = (0 . 8 m)(1 cos 58 . 8713 ◦ ) = . 38643 m ....
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This note was uploaded on 11/30/2011 for the course PHYSICS 302K taught by Professor Irenepolycarpou during the Spring '09 term at University of Texas.
 Spring '09
 IRENEPOLYCARPOU
 Physics

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